What are the vector components of 3 i - 2 j + 5 k along the axes?

3/(√(38)), -2/(√(38)), 5/(√(38))

Vectors for CUET Maths

Q: How do you find the magnitude and direction cosines of a vector?

For a = a₁ i + a₂ j + a₃ k, a = √(a₁^2 + a₂^2 + a₃^2). The unit vector along it is a = a a, and its components are the direction cosines l, m, n, with l^2 + m^2 + n^2 = 1. The components themselves are direction ratios.

Q: How do you find the vector joining two points, and the point that divides it?

PQ = q - p: end minus start. The point dividing PQ internally in the ratio m: n is m q + n pm + n; externally it is m q - n pm - n. The midpoint is p + q2.

Q: What is the scalar product, and what does it give?

a · b = a b cos θ = a₁b₁ + a₂b₂ + a₃b₃. It gives the angle between two vectors, tests perpendicularity ( a · b = 0 ), and the projection of a on b, a · b b.

Q: What is the vector product, and how does it give an area?

a × b is the vector of magnitude a b sin θ, perpendicular to both. Work it out as the determinant with rows i, j, k, then a, then b, remembering the minus sign on the j term. a × b is the area of the parallelogram on a and b; half of it is the triangle.

About 5 min to learn · 125 questions

Vectors are one of the largest chapters of CUET Maths Section B1, and one of the most mechanical: magnitudes, components, the dot product and the cross product, each a short calculation. Learn each rule here, from magnitude, direction cosines and the types of vectors to adding them, the section formula, the scalar product with angles and projections and the vector product with areas, with worked examples, time-saving shortcuts and the traps CUET sets, then practise on free questions in CUET's own formats, each fully solved.

Syllabus area: Section B1: Mathematics

On this page · 10 sections

01A quantity with a direction
02Vectors and their types
03Addition, components and the section formula
04The scalar product
05The vector product
06How CUET asks this chapter
07Exam shortcuts
08Common mistakes
09Quick revision
10Check your understanding

A quantity with a direction

A drone flies 6 km east, 2 km north and climbs 3 km. Its displacement is the vector $6\hat{i} + 2\hat{j} + 3\hat{k}$ , and how far it is from where it started is that vector's magnitude. Vectors carry both a size and a direction, and this chapter is about computing with them.

Vectors and their types

How do you find the magnitude and direction cosines of a vector?

For $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ , $\lvert \vec{a} \rvert = \sqrt{a_1^{2} + a_2^{2} + a_3^{2}}$ . The unit vector along it is $\hat{a} = \frac{\vec{a}}{\lvert \vec{a} \rvert}$ , and its components are the direction cosines $l, m, n$ , with $l^{2} + m^{2} + n^{2} = 1$ . The components themselves are direction ratios.

Example 1 · The drone

How far is the drone from its start, and what are the direction cosines of its displacement $6\hat{i} + 2\hat{j} + 3\hat{k}$ ?

$\sqrt{36 + 4 + 9} = 7$ km.
Direction cosines: $\frac{6}{7}, \frac{2}{7}, \frac{3}{7}$ .

Answer

7 km, in the direction $\frac{6}{7}, \frac{2}{7}, \frac{3}{7}$

Types: the zero vector has magnitude 0 and no fixed direction; a unit vector has magnitude 1; equal vectors have the same magnitude and direction; collinear vectors are multiples of each other, of either sign.

Addition, components and the section formula

How do you find the vector joining two points, and the point that divides it?

$\overrightarrow{PQ} = \vec{q} - \vec{p}$ : end minus start. The point dividing $PQ$ internally in the ratio $m : n$ is $\frac{m\vec{q} + n\vec{p}}{m + n}$ ; externally it is $\frac{m\vec{q} - n\vec{p}}{m - n}$ . The midpoint is $\frac{\vec{p} + \vec{q}}{2}$ .

Example 2 · A midpoint

Find the midpoint of $P(2, 4, -1)$ and $Q(6, -2, 5)$ .

$\frac{(2 + 6, 4 - 2, -1 + 5)}{2}$ .
$= (4, 1, 2)$ .

Answer

(4, 1, 2)

The scalar product

What is the scalar product, and what does it give?

$\vec{a} \cdot \vec{b} = \lvert \vec{a} \rvert \lvert \vec{b} \rvert \cos \theta = a_1b_1 + a_2b_2 + a_3b_3$ . It gives the angle between two vectors, tests perpendicularity ( $\vec{a} \cdot \vec{b} = 0$ ), and the projection of $\vec{a}$ on $\vec{b}$ , $\frac{\vec{a} \cdot \vec{b}}{\lvert \vec{b} \rvert}$ .

Example 3 · An angle

Find the angle between $\hat{i} + 2\hat{j} + 2\hat{k}$ and $2\hat{i} + 2\hat{j} + \hat{k}$ .

$\vec{a} \cdot \vec{b} = 2 + 4 + 2 = 8$ ; both magnitudes are 3.
$\cos \theta = \frac{8}{9}$ .

Answer

\theta = \cos^{-1}\frac{8}{9}

Watch out

$\vec{a} \cdot \vec{b} = 0$ does not mean one of the vectors is zero; it means they are perpendicular, as $\hat{i}$ and $\hat{j}$ are.

The vector product

What is the vector product, and how does it give an area?

$\vec{a} \times \vec{b}$ is the vector of magnitude $\lvert \vec{a} \rvert \lvert \vec{b} \rvert \sin \theta$ , perpendicular to both. Work it out as the determinant with rows $\hat{i}, \hat{j}, \hat{k}$ , then $\vec{a}$ , then $\vec{b}$ , remembering the minus sign on the $\hat{j}$ term. $\lvert \vec{a} \times \vec{b} \rvert$ is the area of the parallelogram on $\vec{a}$ and $\vec{b}$ ; half of it is the triangle.

Example 4 · A parallelogram

Find the area of the parallelogram with adjacent sides $3\hat{i} + \hat{k}$ and $2\hat{i} + \hat{j}$ .

$\vec{a} \times \vec{b} = \hat{i}(0 - 1) - \hat{j}(0 - 2) + \hat{k}(3 - 0) = -\hat{i} + 2\hat{j} + 3\hat{k}$ .
$\lvert \vec{a} \times \vec{b} \rvert = \sqrt{1 + 4 + 9} = \sqrt{14}$ .

Answer

$\sqrt{14}$ square units

Key idea

$\lvert \vec{a} \times \vec{b} \rvert^{2} + (\vec{a} \cdot \vec{b})^{2} = \lvert \vec{a} \rvert^{2}\lvert \vec{b} \rvert^{2}$ , because $\sin^{2} \theta + \cos^{2} \theta = 1$ . Given the magnitudes and one product, it gives the other.

How CUET asks this chapter

Basics: magnitude, a unit vector, a vector of given magnitude along another, direction cosines, the angle a vector makes with an axis.
Points: the vector joining two points, the section formula, collinear points.
Dot product: a value, an angle, a $\lambda$ for perpendicularity, a projection, $\lvert \vec{a} + \vec{b} \rvert$ from magnitudes.
Cross product: a value, an area, a unit vector perpendicular to two vectors, a $\lambda$ for parallel vectors.

Exam shortcuts

$\hat{i} \cdot \hat{i} = 1$ , $\hat{i} \cdot \hat{j} = 0$ ; $\hat{i} \times \hat{j} = \hat{k}$ , $\hat{j} \times \hat{k} = \hat{i}$ , $\hat{k} \times \hat{i} = \hat{j}$ .
$\lvert \vec{a} \pm \vec{b} \rvert^{2} = \lvert \vec{a} \rvert^{2} \pm 2\vec{a} \cdot \vec{b} + \lvert \vec{b} \rvert^{2}$ .
Parallel vectors: $\vec{a} \times \vec{b} = \vec{0}$ , components in proportion.
Triangle $ABC$ : area $= \frac{1}{2}\lvert \overrightarrow{AB} \times \overrightarrow{AC} \rvert$ .

Common mistakes

Adding components instead of squaring them for a magnitude.
Taking $\vec{p} - \vec{q}$ for $\overrightarrow{PQ}$ .
Dividing by the wrong magnitude in a projection.
Dropping the minus sign of the $\hat{j}$ term in a cross product, or forgetting the $\frac{1}{2}$ for a triangle.

Quick revision

Basics

$\lvert \vec{a} \rvert = \sqrt{a_1^{2} + a_2^{2} + a_3^{2}}$ ; $\hat{a} = \frac{\vec{a}}{\lvert \vec{a} \rvert}$
$l^{2} + m^{2} + n^{2} = 1$

Dot product

$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{\lvert \vec{a} \rvert \lvert \vec{b} \rvert}$
Projection on $\vec{b}$ : $\frac{\vec{a} \cdot \vec{b}}{\lvert \vec{b} \rvert}$

Cross product

Perpendicular to both; $\vec{a} \times \vec{b} = -\vec{b} \times \vec{a}$
Parallelogram $\lvert \vec{a} \times \vec{b} \rvert$ , triangle half

Next in Section B1, lines in space: three-dimensional geometry.

Next · 09Quick revision

Check your understanding

125 questions, grouped by type, easy to hard. Pick an answer and check it. The worked solution opens when you ask for it.

Vectors and their typesQ1 of 125Q1

Vectors and their types

28 questions

Question 1

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What is the magnitude of $2\hat{i} - 3\hat{j} + 6\hat{k}$ ?

5

Correct answerYour answer

7

Correct answerYour answer

11

Correct answerYour answer

49

Correct answerYour answer

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AnswerB · $7$

$\lvert a\hat{i} + b\hat{j} + c\hat{k} \rvert = \sqrt{a^{2} + b^{2} + c^{2}}$ .
$\sqrt{4 + 9 + 36} = 7$ .

Worth noting

5 adds the components, $2 - 3 + 6$ ; the magnitude adds their squares under a root.

Question 2

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What is the unit vector in the direction of $\hat{i} + \hat{j} - \hat{k}$ ?

\frac{1}{3}(\hat{i} + \hat{j} - \hat{k})

Correct answerYour answer

\sqrt{3}(\hat{i} + \hat{j} - \hat{k})

Correct answerYour answer

\hat{i} + \hat{j} - \hat{k}

Correct answerYour answer

\frac{1}{\sqrt{3}}(\hat{i} + \hat{j} - \hat{k})

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AnswerD · $\frac{1}{\sqrt{3}}(\hat{i} + \hat{j} - \hat{k})$

Divide the vector by its magnitude, $\sqrt{1 + 1 + 1} = \sqrt{3}$ .
$\hat{a} = \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} - \hat{k})$ .

Worth noting

$\frac{1}{3}$ divides by the sum of the squares, not by its square root.

Question 3

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What are the direction cosines of $2\hat{i} - \hat{j} + 2\hat{k}$ ?

2, -1, 2

Correct answerYour answer

\frac{2}{9}, -\frac{1}{9}, \frac{2}{9}

Correct answerYour answer

\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}

Correct answerYour answer

\frac{2}{3}, \frac{1}{3}, \frac{2}{3}

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AnswerC · $\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}$

The magnitude is $\sqrt{4 + 1 + 4} = 3$ .
Divide each component by 3: $\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}$ .

Worth noting

$2, -1, 2$ are direction ratios; the cosines are those divided by the magnitude.

Question 4

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What is the magnitude of $\hat{i} - 4\hat{j} + 8\hat{k}$ ?

5

Correct answerYour answer

9

Correct answerYour answer

13

Correct answerYour answer

81

Correct answerYour answer

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AnswerB · $9$

$\lvert a\hat{i} + b\hat{j} + c\hat{k} \rvert = \sqrt{a^{2} + b^{2} + c^{2}}$ .
$\sqrt{1 + 16 + 64} = 9$ .

Worth noting

81 is the sum of the squares before the square root is taken.

Question 5

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What is the unit vector in the direction of $3\hat{i} - 4\hat{k}$ ?

\frac{1}{5}(3\hat{i} - 4\hat{k})

Correct answerYour answer

\frac{1}{25}(3\hat{i} - 4\hat{k})

Correct answerYour answer

\frac{1}{7}(3\hat{i} - 4\hat{k})

Correct answerYour answer

5(3\hat{i} - 4\hat{k})

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AnswerA · $\frac{1}{5}(3\hat{i} - 4\hat{k})$

$\lvert 3\hat{i} - 4\hat{k} \rvert = \sqrt{9 + 16} = 5$ .
Divide by 5: $\frac{1}{5}(3\hat{i} - 4\hat{k})$ .

Worth noting

$\frac{1}{7}$ divides by $3 + 4$ , the sizes of the components added, not by the magnitude.

Question 6

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What are the direction cosines of $6\hat{i} - 2\hat{j} + 3\hat{k}$ ?

6, -2, 3

Correct answerYour answer

\frac{6}{49}, -\frac{2}{49}, \frac{3}{49}

Correct answerYour answer

\frac{6}{7}, \frac{2}{7}, \frac{3}{7}

Correct answerYour answer

\frac{6}{7}, -\frac{2}{7}, \frac{3}{7}

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AnswerD · $\frac{6}{7}, -\frac{2}{7}, \frac{3}{7}$

The magnitude is $\sqrt{36 + 4 + 9} = 7$ .
Divide each component by 7, keeping its sign.

Worth noting

$6, -2, 3$ are direction ratios; the cosines are those divided by the magnitude.

Question 7

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What is the magnitude of $3\hat{i} - 4\hat{j} + 12\hat{k}$ ?

11

Correct answerYour answer

13

Correct answerYour answer

19

Correct answerYour answer

169

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AnswerB · $13$

Square each component, add, and take the square root.
$\sqrt{9 + 16 + 144} = \sqrt{169} = 13$ .

Worth noting

19 adds the sizes of the components, $3 + 4 + 12$ , instead of their squares.

Question 8

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What is the magnitude of $4\hat{i} + 4\hat{j} - 7\hat{k}$ ?

1

Correct answerYour answer

9

Correct answerYour answer

15

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81

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AnswerB · $9$

Square each component, add, and take the square root.
$\sqrt{16 + 16 + 49} = \sqrt{81} = 9$ .

Worth noting

1 adds the components with their signs, $4 + 4 - 7$ .

Question 9

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Which of these is a unit vector?

\frac{1}{3}(\hat{i} + \hat{j} + \hat{k})

Correct answerYour answer

\frac{\sqrt{2}}{2}(\hat{i} - \hat{j} + \hat{k})

Correct answerYour answer

\frac{1}{5}(3\hat{i} + 4\hat{j} - 5\hat{k})

Correct answerYour answer

\frac{1}{7}(2\hat{i} + 3\hat{j} - 6\hat{k})

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AnswerD · $\frac{1}{7}(2\hat{i} + 3\hat{j} - 6\hat{k})$

A unit vector has magnitude 1.
$\lvert 2\hat{i} + 3\hat{j} - 6\hat{k} \rvert = \sqrt{4 + 9 + 36} = 7$ , so $\frac{1}{7}(2\hat{i} + 3\hat{j} - 6\hat{k})$ has magnitude 1; the others have magnitudes $\frac{1}{\sqrt{3}}$ , $\sqrt{\frac{3}{2}}$ and $\sqrt{2}$ .

Worth noting

$\frac{1}{3}(\hat{i} + \hat{j} + \hat{k})$ has components adding to 1, but its magnitude is $\frac{1}{\sqrt{3}}$ .

Question 10

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The vectors $(x + 1)\hat{i} + 3\hat{j} - 2\hat{k}$ and $4\hat{i} + (y - 2)\hat{j} + z\hat{k}$ are equal. What is $x + y + z$ ?

2

Correct answerYour answer

6

Correct answerYour answer

8

Correct answerYour answer

10

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AnswerB · $6$

Equal vectors have equal components: $x + 1 = 4$ , $3 = y - 2$ and $-2 = z$ .
$x = 3$ , $y = 5$ , $z = -2$ , so $x + y + z = 6$ .

Worth noting

10 takes $z = 2$ , dropping the minus sign of the $\hat{k}$ component.

Question 11

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Which of these vectors has the same direction as $3\hat{i} - 6\hat{j} + 2\hat{k}$ ?

-3\hat{i} + 6\hat{j} - 2\hat{k}

Correct answerYour answer

3\hat{i} + 6\hat{j} + 2\hat{k}

Correct answerYour answer

9\hat{i} - 18\hat{j} + 2\hat{k}

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6\hat{i} - 12\hat{j} + 4\hat{k}

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AnswerD · $6\hat{i} - 12\hat{j} + 4\hat{k}$

Same direction means a positive multiple: $\vec{b} = \lambda\vec{a}$ with $\lambda > 0$ .
$6\hat{i} - 12\hat{j} + 4\hat{k} = 2(3\hat{i} - 6\hat{j} + 2\hat{k})$ ; $-3\hat{i} + 6\hat{j} - 2\hat{k}$ is a negative multiple, and the other two are not multiples at all.

Worth noting

$-3\hat{i} + 6\hat{j} - 2\hat{k}$ is collinear, but its multiple is $-1$ : it points the opposite way.

Question 12

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Read the two statements and choose the correct option.

I. $\hat{i} + \hat{j}$ and $-\hat{i} - \hat{j}$ are collinear vectors.

II. Two equal vectors must have the same initial point.

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

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AnswerA · Only I

I: $-\hat{i} - \hat{j} = -1(\hat{i} + \hat{j})$ , a scalar multiple, so they are collinear. True.
II: equal vectors need only the same magnitude and the same direction; their initial points can differ. False.

Worth noting

Opposite directions do not stop two vectors from being collinear.

Question 13

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Which of these triples can be the direction cosines of a line?

\frac{1}{2}, \frac{1}{2}, \frac{1}{2}

Correct answerYour answer

1, 1, 1

Correct answerYour answer

\frac{1}{3}, \frac{1}{3}, \frac{1}{3}

Correct answerYour answer

\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}

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AnswerD · $\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}$

Direction cosines satisfy $l^{2} + m^{2} + n^{2} = 1$ .
$\frac{1}{2} + \frac{1}{4} + \frac{1}{4} = 1$ ; the others give $\frac{3}{4}$ , 3 and $\frac{1}{3}$ .

Worth noting

$\frac{1}{3}, \frac{1}{3}, \frac{1}{3}$ adds to 1, but it is the squares that must add to 1.

Question 14

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A line makes an angle of $45^\circ$ with the $x$ -axis and $60^\circ$ with the $y$ -axis, and an acute angle with the $z$ -axis. What is that angle?

60^\circ

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75^\circ

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45^\circ

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90^\circ

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AnswerA · $60^\circ$

$\cos^{2} 45^\circ + \cos^{2} 60^\circ + \cos^{2} \gamma = 1$ : $\frac{1}{2} + \frac{1}{4} + \cos^{2} \gamma = 1$ .
$\cos^{2} \gamma = \frac{1}{4}$ , and for an acute angle $\cos \gamma = \frac{1}{2}$ : $\gamma = 60^\circ$ .

Worth noting

$75^\circ$ makes the angles add to $180^\circ$ ; it is the squares of the cosines that add to 1.

Question 15

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Which of the following statements are true?

I. The zero vector has no definite direction.

II. $2\hat{i} - 3\hat{j} + 4\hat{k}$ and $-4\hat{i} + 6\hat{j} - 8\hat{k}$ are collinear.

III. Two vectors with the same magnitude are equal.

IV. A unit vector has magnitude 1.

All four: I, II, III and IVCorrect answerYour answerI and IV onlyCorrect answerYour answerI, II and IV onlyCorrect answerYour answerII, III and IV onlyCorrect answerYour answer

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AnswerC · I, II and IV only

I and IV are definitions. II: the second is $-2$ times the first, so they are collinear.
III is false: equal vectors need the same magnitude and the same direction.

Worth noting

Collinear vectors can point opposite ways; a negative multiple still lies on the same line.

Question 16

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What is the vector of magnitude 9 in the direction of $2\hat{i} + \hat{j} - 2\hat{k}$ ?

6\hat{i} + 3\hat{j} - 6\hat{k}

Correct answerYour answer

18\hat{i} + 9\hat{j} - 18\hat{k}

Correct answerYour answer

2\hat{i} + \hat{j} - 2\hat{k}

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-6\hat{i} - 3\hat{j} + 6\hat{k}

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AnswerA · $6\hat{i} + 3\hat{j} - 6\hat{k}$

$\lvert 2\hat{i} + \hat{j} - 2\hat{k} \rvert = 3$ , so the unit vector is $\frac{1}{3}(2\hat{i} + \hat{j} - 2\hat{k})$ .
$9 \times \frac{1}{3}(2\hat{i} + \hat{j} - 2\hat{k}) = 6\hat{i} + 3\hat{j} - 6\hat{k}$ .

Worth noting

$18\hat{i} + 9\hat{j} - 18\hat{k}$ multiplies by 9 without first making the vector a unit vector; its magnitude is 27.

Question 17

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Which pair of vectors is collinear?

2\hat{i} - 3\hat{j} + \hat{k}

and

-4\hat{i} + 6\hat{j} - 2\hat{k}

Correct answerYour answer

\hat{i} + \hat{j}

and

\hat{j} + \hat{k}

Correct answerYour answer

2\hat{i} + \hat{j}

and

\hat{i} + 2\hat{j}

Correct answerYour answer

\hat{i} - \hat{j} + \hat{k}

and

\hat{i} + \hat{j} - \hat{k}

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AnswerA · $2\hat{i} - 3\hat{j} + \hat{k}$ and $-4\hat{i} + 6\hat{j} - 2\hat{k}$

Collinear vectors are scalar multiples of each other: their components are in one ratio.
$-4\hat{i} + 6\hat{j} - 2\hat{k} = -2(2\hat{i} - 3\hat{j} + \hat{k})$ ; no other pair has one ratio.

Worth noting

$2\hat{i} + \hat{j}$ and $\hat{i} + 2\hat{j}$ use the same numbers, but in a different order, so the ratios are 2 and $\frac{1}{2}$ .

Question 18

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A line makes equal acute angles with the three coordinate axes. What is the cosine of each angle?

\frac{1}{3}

Correct answerYour answer

\frac{\sqrt{3}}{3}

Correct answerYour answer

\frac{\sqrt{2}}{2}

Correct answerYour answer

\frac{\sqrt{3}}{2}

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AnswerB · $\frac{\sqrt{3}}{3}$

With all three equal to $l$ : $l^{2} + l^{2} + l^{2} = 1$ .
$l^{2} = \frac{1}{3}$ , and acute angles give $l = \frac{1}{\sqrt{3}}$ .

Worth noting

$\frac{1}{3}$ solves $3l = 1$ ; it is the squares of the cosines that add to 1.

Question 19

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$\vec{a} = 4\hat{i} - 4\hat{j} + 2\hat{k}$ . For which values of $\lambda$ is $\lambda\vec{a}$ a unit vector?

\lambda = \pm\frac{1}{6}

Correct answerYour answer

\lambda = \frac{1}{6}

Correct answerYour answer

\lambda = \pm\frac{1}{36}

Correct answerYour answer

\lambda = \pm 6

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AnswerA · $\lambda = \pm\frac{1}{6}$

$\lvert \vec{a} \rvert = \sqrt{16 + 16 + 4} = 6$ , and $\lvert \lambda\vec{a} \rvert = \lvert \lambda \rvert \lvert \vec{a} \rvert = 6\lvert \lambda \rvert$ .
$6\lvert \lambda \rvert = 1$ gives $\lambda = \pm\frac{1}{6}$ .

Worth noting

$\frac{1}{6}$ alone misses the negative value: $-\frac{1}{6}\vec{a}$ also has magnitude 1.

Question 20

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The direction cosines of $\vec{r}$ are $\frac{2}{7}, -\frac{3}{7}, \frac{6}{7}$ , and its $\hat{k}$ component is 12. What is $\lvert \vec{r} \rvert$ ?

7

Correct answerYour answer

\frac{72}{7}

Correct answerYour answer

12

Correct answerYour answer

14

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AnswerD · $14$

Each component is the magnitude times its direction cosine: $r_3 = \frac{6}{7}\lvert \vec{r} \rvert$ .
$12 = \frac{6}{7}\lvert \vec{r} \rvert$ , so $\lvert \vec{r} \rvert = 14$ .

Worth noting

$\frac{72}{7}$ multiplies 12 by the direction cosine instead of dividing by it.

Question 21

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Match each vector in List I with the angle it makes with the $x$ -axis in List II.

List I	List II
A. $\hat{i} + \hat{j} + \sqrt{2}\hat{k}$	I. $\frac{\pi}{4}$
B. $\sqrt{2}\hat{i} - \hat{j} + \hat{k}$	II. $\frac{\pi}{2}$
C. $-\hat{i} + \hat{j} - \sqrt{2}\hat{k}$	III. $\frac{\pi}{3}$
D. $2\hat{j} - \hat{k}$	IV. $\frac{2\pi}{3}$

A-III, B-I, C-IV, D-IICorrect answerYour answerA-IV, B-I, C-III, D-IICorrect answerYour answerA-I, B-III, C-IV, D-IICorrect answerYour answerA-III, B-IV, C-I, D-IICorrect answerYour answer

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AnswerA · A-III, B-I, C-IV, D-II

$\cos \alpha = \frac{a_1}{\lvert \vec{a} \rvert}$ , and the first three vectors each have magnitude $\sqrt{1 + 1 + 2} = 2$ .
A: $\cos \alpha = \frac{1}{2}$ , $\frac{\pi}{3}$ ; B: $\frac{\sqrt{2}}{2}$ , $\frac{\pi}{4}$ ; C: $-\frac{1}{2}$ , $\frac{2\pi}{3}$ ; D has no $\hat{i}$ part, so $\frac{\pi}{2}$ .

Worth noting

A negative $\hat{i}$ component gives an obtuse angle: $\cos \alpha = -\frac{1}{2}$ is $\frac{2\pi}{3}$ , not $\frac{\pi}{3}$ .

Question 22

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$\lvert \vec{a} \rvert = 3$ and $-4 \le \lambda \le 2$ . What is the range of $\lvert \lambda\vec{a} \rvert$ ?

[-12, 6]

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[6, 12]

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[0, 12]

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[0, 6]

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AnswerC · $[0, 12]$

$\lvert \lambda\vec{a} \rvert = 3\lvert \lambda \rvert$ , and on $[-4, 2]$ , $\lvert \lambda \rvert$ takes every value from $0$ (at $\lambda = 0$ ) to $4$ (at $\lambda = -4$ ).
So $\lvert \lambda\vec{a} \rvert$ runs over $[0, 12]$ .

Worth noting

$[-12, 6]$ multiplies the ends by 3 without the modulus; a magnitude is never negative.

Question 23

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Assertion (A): The vector $3\hat{i} - 2\hat{j} + 6\hat{k}$ makes an obtuse angle with the $y$ -axis.

Reason (R): The cosine of the angle a vector makes with the $y$ -axis is its $\hat{j}$ component divided by its magnitude.

Both A and R are true, and R is the correct explanation of A.Correct answerYour answerBoth A and R are true, but R is not the correct explanation of A.Correct answerYour answerA is true, but R is false.Correct answerYour answerA is false, but R is true.Correct answerYour answer

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AnswerA · Both A and R are true, and R is the correct explanation of A.

R is true: it is the direction cosine $m = \frac{a_2}{\lvert \vec{a} \rvert}$ .
Here $m = -\frac{2}{7} < 0$ , so the angle is obtuse: A is true, and R is the reason.

Question 24

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A vector has direction ratios $1, -2, 2$ and makes an obtuse angle with the $x$ -axis. What are its direction cosines?

\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}

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-\frac{1}{3}, \frac{2}{3}, -\frac{2}{3}

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-\frac{1}{3}, -\frac{2}{3}, -\frac{2}{3}

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-1, 2, -2

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AnswerB · $-\frac{1}{3}, \frac{2}{3}, -\frac{2}{3}$

$\sqrt{1 + 4 + 4} = 3$ , so the direction cosines are $\pm\left(\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}\right)$ .
An obtuse angle with the $x$ -axis needs $l < 0$ : take the minus sign, $-\frac{1}{3}, \frac{2}{3}, -\frac{2}{3}$ .

Worth noting

$\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}$ has $l > 0$ , which is an acute angle with the $x$ -axis.

Question 25

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For $\vec{a} = -2\hat{i} + \hat{j} + 2\hat{k}$ , which of the following statements are true?

I. $\lvert \vec{a} \rvert = 3$ .

II. The direction cosines of $\vec{a}$ are $-2, 1, 2$ .

III. $\vec{a}$ makes an obtuse angle with the $x$ -axis.

IV. $\frac{1}{9}(-2\hat{i} + \hat{j} + 2\hat{k})$ is the unit vector along $\vec{a}$ .

I, II and III onlyCorrect answerYour answerI, III and IV onlyCorrect answerYour answerI and III onlyCorrect answerYour answerI and II onlyCorrect answerYour answer

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AnswerC · I and III only

$\lvert \vec{a} \rvert = \sqrt{4 + 1 + 4} = 3$ , so I is true; the direction cosines are $-\frac{2}{3}, \frac{1}{3}, \frac{2}{3}$ , and II lists the direction ratios.
III: $l = -\frac{2}{3} < 0$ , an obtuse angle, so it is true. IV divides by $9$ , the square of the magnitude, so it is false.

Worth noting

Direction ratios become direction cosines only after dividing by the magnitude.

Question 26

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A vector $\vec{r}$ of magnitude 4 makes an angle $\frac{\pi}{3}$ with the $x$ -axis, an angle $\frac{2\pi}{3}$ with the $y$ -axis and an obtuse angle with the $z$ -axis. What is $\vec{r}$ ?

2\hat{i} - 2\hat{j} + 2\sqrt{2}\hat{k}

Correct answerYour answer

2\hat{i} - 2\hat{j} - 2\sqrt{2}\hat{k}

Correct answerYour answer

2\hat{i} + 2\hat{j} - 2\sqrt{2}\hat{k}

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\frac{1}{2}\hat{i} - \frac{1}{2}\hat{j} - \frac{1}{\sqrt{2}}\hat{k}

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AnswerB · $2\hat{i} - 2\hat{j} - 2\sqrt{2}\hat{k}$

$l = \cos\frac{\pi}{3} = \frac{1}{2}$ and $m = \cos\frac{2\pi}{3} = -\frac{1}{2}$ , so $n^{2} = 1 - \frac{1}{4} - \frac{1}{4} = \frac{1}{2}$ .
The angle with the $z$ -axis is obtuse, so $n = -\frac{1}{\sqrt{2}}$ .
$\vec{r} = 4\left(\frac{1}{2}\hat{i} - \frac{1}{2}\hat{j} - \frac{1}{\sqrt{2}}\hat{k}\right) = 2\hat{i} - 2\hat{j} - 2\sqrt{2}\hat{k}$ .

Worth noting

$2\hat{i} - 2\hat{j} + 2\sqrt{2}\hat{k}$ takes the positive root for $n$ , which is an acute angle with the $z$ -axis.

Question 27

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A vector makes angles $\alpha$ , $\beta$ and $\gamma$ with the $x$ -, $y$ - and $z$ -axes. What is $\cos 2\alpha + \cos 2\beta + \cos 2\gamma$ ?

-3

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-1

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0

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1

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AnswerB · $-1$

$\cos 2\alpha = 2\cos^{2}\alpha - 1$ , so the sum is $2(\cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma) - 3$ .
The squares of the direction cosines add to 1, so the sum is $2(1) - 3 = -1$ .

Worth noting

1 is the value of $\cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma$ ; the double angles bring in the $-3$ .

Question 28

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$\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}$ . A vector $\vec{b}$ is collinear with $\vec{a}$ , has magnitude $3\sqrt{14}$ and makes an acute angle with the $z$ -axis. What is $\vec{b}$ ?

6\hat{i} + 9\hat{j} - 3\hat{k}

Correct answerYour answer

-2\hat{i} - 3\hat{j} + \hat{k}

Correct answerYour answer

6\hat{i} + 9\hat{j} + 3\hat{k}

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-6\hat{i} - 9\hat{j} + 3\hat{k}

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AnswerD · $-6\hat{i} - 9\hat{j} + 3\hat{k}$

$\lvert \vec{a} \rvert = \sqrt{4 + 9 + 1} = \sqrt{14}$ , so a collinear vector of magnitude $3\sqrt{14}$ is $\pm 3\vec{a}$ .
An acute angle with the $z$ -axis needs a positive $\hat{k}$ component; $3\vec{a}$ has $-3$ , so $\vec{b} = -3\vec{a} = -6\hat{i} - 9\hat{j} + 3\hat{k}$ .

Worth noting

$6\hat{i} + 9\hat{j} - 3\hat{k}$ has the right magnitude, but its $\hat{k}$ component is negative: an obtuse angle with the $z$ -axis.

Addition, components and the section formula

30 questions

Question 29

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What is the vector from $P(1, -2, 3)$ to $Q(4, 2, -1)$ ?

-3\hat{i} - 4\hat{j} + 4\hat{k}

Correct answerYour answer

3\hat{i} + 4\hat{j} - 4\hat{k}

Correct answerYour answer

5\hat{i} + 2\hat{k}

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3\hat{i} + 4\hat{j} + 4\hat{k}

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AnswerB · $3\hat{i} + 4\hat{j} - 4\hat{k}$

$\overrightarrow{PQ} =$ (position vector of $Q$ ) $-$ (position vector of $P$ ).
$(4 - 1)\hat{i} + (2 + 2)\hat{j} + (-1 - 3)\hat{k} = 3\hat{i} + 4\hat{j} - 4\hat{k}$ .

Worth noting

$-3\hat{i} - 4\hat{j} + 4\hat{k}$ is $\overrightarrow{QP}$ , the end minus the start taken the wrong way round.

Question 30

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If $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 3\hat{i} - \hat{j} + 2\hat{k}$ , what is $2\vec{a} - \vec{b}$ ?

5\hat{i} + 3\hat{j}

Correct answerYour answer

-2\hat{i} + 3\hat{j} - 3\hat{k}

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-\hat{i} + 5\hat{j}

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-\hat{i} + 5\hat{j} - 4\hat{k}

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AnswerD · $-\hat{i} + 5\hat{j} - 4\hat{k}$

$2\vec{a} = 2\hat{i} + 4\hat{j} - 2\hat{k}$ .
Subtract component by component: $-\hat{i} + 5\hat{j} - 4\hat{k}$ .

Worth noting

$-\hat{i} + 5\hat{j}$ doubles the $\hat{k}$ part of $\vec{a}$ wrongly, $-2 - 2$ taken as 0.

Question 31

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For any triangle $ABC$ , what is $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA}$ ?

\vec{0}

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2\overrightarrow{AC}

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\overrightarrow{AC}

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2\overrightarrow{CA}

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AnswerA · $\vec{0}$

Triangle law: $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$ .
$\overrightarrow{AC} + \overrightarrow{CA} = \vec{0}$ : the path ends where it began.

Worth noting

$2\overrightarrow{AC}$ treats $\overrightarrow{CA}$ as $\overrightarrow{AC}$ ; they point opposite ways.

Question 32

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What is the distance between $P(3, -1, 2)$ and $Q(1, 2, -4)$ ?

\sqrt{21}

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5

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7

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49

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AnswerC · $7$

$\overrightarrow{PQ} = -2\hat{i} + 3\hat{j} - 6\hat{k}$ .
$\lvert \overrightarrow{PQ} \rvert = \sqrt{4 + 9 + 36} = 7$ .

Worth noting

$\sqrt{21}$ adds the position vectors instead of subtracting them.

Question 33

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What are the vector components of $3\hat{i} - 2\hat{j} + 5\hat{k}$ along the axes?

3, -2, 5

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3\hat{i}, 2\hat{j}, 5\hat{k}

Correct answerYour answer

3\hat{i}, -2\hat{j}, 5\hat{k}

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\frac{3}{\sqrt{38}}, -\frac{2}{\sqrt{38}}, \frac{5}{\sqrt{38}}

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AnswerC · $3\hat{i}, -2\hat{j}, 5\hat{k}$

For $a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ , the scalar components are $a_1, a_2, a_3$ and the vector components are $a_1\hat{i}, a_2\hat{j}, a_3\hat{k}$ .
Here the vector components are $3\hat{i}$ , $-2\hat{j}$ and $5\hat{k}$ , each keeping its sign.

Worth noting

$3, -2, 5$ are the scalar components; the vector components carry the unit vectors.

Question 34

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In triangle $ABC$ , $D$ is the midpoint of $BC$ . What is $\overrightarrow{AB} + \overrightarrow{AC}$ ?

\overrightarrow{AD}

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\overrightarrow{BC}

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2\overrightarrow{DA}

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2\overrightarrow{AD}

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AnswerD · $2\overrightarrow{AD}$

Taking $A$ as the origin, $D$ is the midpoint of $BC$ : $\overrightarrow{AD} = \frac{\overrightarrow{AB} + \overrightarrow{AC}}{2}$ .
So $\overrightarrow{AB} + \overrightarrow{AC} = 2\overrightarrow{AD}$ , twice the median.

Worth noting

$\overrightarrow{AD}$ forgets the half in the midpoint formula; the sum is twice the median.

Question 35

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$\vec{a} = \hat{i} - 2\hat{j} + 4\hat{k}$ , $\vec{b} = 3\hat{i} + \hat{j} - \hat{k}$ and $\vec{c} = -2\hat{i} - 2\hat{j} + 3\hat{k}$ . What is $\lvert \vec{a} + \vec{b} + \vec{c} \rvert$ ?

5

Correct answerYour answer

7

Correct answerYour answer

11

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49

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AnswerB · $7$

$\vec{a} + \vec{b} + \vec{c} = (1 + 3 - 2)\hat{i} + (-2 + 1 - 2)\hat{j} + (4 - 1 + 3)\hat{k} = 2\hat{i} - 3\hat{j} + 6\hat{k}$ .
$\lvert \vec{a} + \vec{b} + \vec{c} \rvert = \sqrt{4 + 9 + 36} = 7$ .

Worth noting

11 adds the sizes of the components of the sum, $2 + 3 + 6$ , instead of their squares under a root.

Question 36

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Three vertices of a parallelogram $ABCD$ , taken in order, are $A(1, 2, -1)$ , $B(2, -1, 0)$ and $C(5, 4, 3)$ .

At which point do the diagonals of $ABCD$ meet?

(3, 3, 1)

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(2, 1, 2)

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(\frac{3}{2}, \frac{1}{2}, - \frac{1}{2})

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(\frac{8}{3}, \frac{5}{3}, \frac{2}{3})

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AnswerA · $(3, 3, 1)$

The diagonals of a parallelogram bisect each other, so they meet at the midpoint of $AC$ .
$\frac{(1, 2, -1) + (5, 4, 3)}{2} = (3, 3, 1)$ .

Worth noting

$(2, 1, 2)$ halves $\vec{c} - \vec{a}$ ; the midpoint halves $\vec{a} + \vec{c}$ .

Question 37

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What point divides the join of $P(2, -1, 3)$ and $Q(6, 3, -5)$ internally in the ratio 3 : 1?

(5, 2, -3)

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(3, 0, 1)

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(4, 1, -1)

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(16, 10, -18)

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AnswerA · $(5, 2, -3)$

$\frac{m\vec{q} + n\vec{p}}{m + n}$ with $m = 3$ towards $Q$ and $n = 1$ towards $P$ .
$\frac{3(6, 3, -5) + (2, -1, 3)}{4} = (5, 2, -3)$ .

Worth noting

$(3, 0, 1)$ puts the weight 3 on $P$ ; the point is nearer $Q$ .

Question 38

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If $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 3\hat{i} - \hat{j} + 2\hat{k}$ , what is the unit vector in the direction of $\vec{a} + \vec{b}$ ?

\frac{1}{18}(4\hat{i} + \hat{j} + \hat{k})

Correct answerYour answer

\frac{1}{\sqrt{18}}(-2\hat{i} + 3\hat{j} - 3\hat{k})

Correct answerYour answer

\frac{1}{6}(4\hat{i} + \hat{j} + \hat{k})

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\frac{1}{\sqrt{18}}(4\hat{i} + \hat{j} + \hat{k})

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AnswerD · $\frac{1}{\sqrt{18}}(4\hat{i} + \hat{j} + \hat{k})$

$\vec{a} + \vec{b} = 4\hat{i} + \hat{j} + \hat{k}$ , with magnitude $\sqrt{16 + 1 + 1} = \sqrt{18}$ .
The unit vector is $\frac{1}{\sqrt{18}}(4\hat{i} + \hat{j} + \hat{k})$ .

Worth noting

$\frac{1}{6}$ adds the components, $4 + 1 + 1$ , instead of their squares.

Question 39

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What is the distance between $P(1, -2, 3)$ and $Q(4, 2, -1)$ ?

\sqrt{33}

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\sqrt{41}

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11

Correct answerYour answer

41

Correct answerYour answer

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AnswerB · $\sqrt{41}$

$\overrightarrow{PQ} = 3\hat{i} + 4\hat{j} - 4\hat{k}$ .
$\lvert \overrightarrow{PQ} \rvert = \sqrt{9 + 16 + 16} = \sqrt{41}$ .

Worth noting

41 is the square of the distance.

Question 40

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For which $\lambda$ are $2\hat{i} + 3\hat{j} - \hat{k}$ and $\lambda\hat{i} - 6\hat{j} + 2\hat{k}$ parallel?

-4

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-3

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-1

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4

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AnswerA · $-4$

Parallel vectors have proportional components: $\frac{\lambda}{2} = \frac{-6}{3} = \frac{2}{-1}$ .
The common ratio is $-2$ , so $\lambda = -4$ .

Worth noting

4 takes the ratio as $+2$ , missing that $-6$ and 3 have opposite signs.

Question 41

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The points $A(1, 2, 7)$ , $B(2, 6, 3)$ and $C(3, 10, -1)$ are collinear. In what ratio does $B$ divide $AC$ ?

1 : 2Correct answerYour answer2 : 1Correct answerYour answer1 : 3Correct answerYour answer1 : 1Correct answerYour answer

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AnswerD · 1 : 1

$\overrightarrow{AB} = \hat{i} + 4\hat{j} - 4\hat{k}$ and $\overrightarrow{BC} = \hat{i} + 4\hat{j} - 4\hat{k}$ .
They are equal, so $AB = BC$ : $B$ is the midpoint, ratio 1 : 1.

Worth noting

1 : 2 compares $AB$ with the whole $AC$ , not with $BC$ .

Question 42

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What point divides the join of $A(2, -3, 4)$ and $B(5, 3, -2)$ internally in the ratio 1 : 2?

(4, 1, 0)

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(\frac{7}{2}, 0, 1)

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(-1, -9, 10)

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(3, -1, 2)

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AnswerD · $(3, -1, 2)$

For $m : n$ internally: $\frac{m\vec{b} + n\vec{a}}{m + n}$ , with $m = 1$ on $B$ 's side.
$\frac{(5, 3, -2) + 2(2, -3, 4)}{3} = (3, -1, 2)$ .

Worth noting

$(4, 1, 0)$ puts the 2 on $B$ ; that point is nearer $B$ , in the ratio 2 : 1.

Question 43

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In a parallelogram $ABCD$ , what is $\overrightarrow{AC} + \overrightarrow{BD}$ ?

2\overrightarrow{AB}

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\vec{0}

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2\overrightarrow{AD}

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2\overrightarrow{DA}

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AnswerC · $2\overrightarrow{AD}$

Parallelogram law: $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{AD}$ ; and by the triangle law, $\overrightarrow{BD} = \overrightarrow{AD} - \overrightarrow{AB}$ .
Adding, $\overrightarrow{AB}$ cancels: $\overrightarrow{AC} + \overrightarrow{BD} = 2\overrightarrow{AD}$ .

Worth noting

$2\overrightarrow{AB}$ takes $\overrightarrow{BD} = \overrightarrow{AB} - \overrightarrow{AD}$ , which is $\overrightarrow{DB}$ , the diagonal the other way.

Question 44

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Triangle $ABC$ has vertices $A(3, -1, 2)$ and $B(-2, 4, 1)$ , and its centroid is $G(1, 2, -1)$ . What is $C$ ?

(0, -1, -4)

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(-2, -3, 6)

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(-2, 11, -4)

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(2, 3, -6)

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AnswerD · $(2, 3, -6)$

The centroid's position vector is $\frac{\vec{a} + \vec{b} + \vec{c}}{3}$ , so $\vec{c} = 3\vec{g} - \vec{a} - \vec{b}$ .
$3(1, 2, -1) - (3, -1, 2) - (-2, 4, 1) = (2, 3, -6)$ .

Worth noting

$(0, -1, -4)$ uses $\vec{g} - \vec{a} - \vec{b}$ , forgetting that the centroid is the average of three points.

Question 45

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What are the direction cosines of the vector from $A(2, -1, 3)$ to $B(4, 1, 2)$ ?

-\frac{2}{3}, -\frac{2}{3}, \frac{1}{3}

Correct answerYour answer

\frac{2}{3}, \frac{2}{3}, -\frac{1}{3}

Correct answerYour answer

2, 2, -1

Correct answerYour answer

\frac{2}{9}, \frac{2}{9}, -\frac{1}{9}

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AnswerB · $\frac{2}{3}, \frac{2}{3}, -\frac{1}{3}$

$\overrightarrow{AB} = (4 - 2)\hat{i} + (1 + 1)\hat{j} + (2 - 3)\hat{k} = 2\hat{i} + 2\hat{j} - \hat{k}$ , of magnitude $\sqrt{4 + 4 + 1} = 3$ .
Divide each component by 3: $\frac{2}{3}, \frac{2}{3}, -\frac{1}{3}$ .

Worth noting

$-\frac{2}{3}, -\frac{2}{3}, \frac{1}{3}$ belong to $\overrightarrow{BA}$ : the vector from $A$ to $B$ is $B$ minus $A$ .

Question 46

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For any four points $A$ , $B$ , $C$ and $D$ , which of the following statements are true?

I. $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$ .

II. $\overrightarrow{AB} - \overrightarrow{AC} = \overrightarrow{CB}$ .

III. $\overrightarrow{AB} + \overrightarrow{BA} = 2\overrightarrow{AB}$ .

IV. $\lvert \overrightarrow{AB} + \overrightarrow{BD} \rvert = \lvert \overrightarrow{AB} \rvert + \lvert \overrightarrow{BD} \rvert$ .

I, II and IV onlyCorrect answerYour answerI onlyCorrect answerYour answerI, II and III onlyCorrect answerYour answerI and II onlyCorrect answerYour answer

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AnswerD · I and II only

I is the triangle law. II: $\overrightarrow{AB} - \overrightarrow{AC} = (\vec{b} - \vec{a}) - (\vec{c} - \vec{a}) = \vec{b} - \vec{c} = \overrightarrow{CB}$ .
III is false: $\overrightarrow{BA} = -\overrightarrow{AB}$ , so the sum is $\vec{0}$ . IV is false unless $B$ lies on the segment $AD$ : the left side is $\lvert \overrightarrow{AD} \rvert$ , a side of triangle $ABD$ , which is shorter than the other two sides together.

Worth noting

Magnitudes add only for vectors in the same direction; in general $\lvert \vec{a} + \vec{b} \rvert \le \lvert \vec{a} \rvert + \lvert \vec{b} \rvert$ .

Question 47

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Three vertices of a parallelogram $ABCD$ , taken in order, are $A(1, 2, -1)$ , $B(2, -1, 0)$ and $C(5, 4, 3)$ .

What are the coordinates of $D$ ?

(6, 1, 4)

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(-2, -3, -4)

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(4, 7, 2)

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(8, 5, 2)

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AnswerC · $(4, 7, 2)$

In parallelogram $ABCD$ , $\overrightarrow{AD} = \overrightarrow{BC}$ , so $\vec{d} = \vec{a} + \vec{c} - \vec{b}$ .
$(1, 2, -1) + (5, 4, 3) - (2, -1, 0) = (4, 7, 2)$ .

Worth noting

$(6, 1, 4)$ is $\vec{b} + \vec{c} - \vec{a}$ , which makes $ABDC$ the parallelogram, with $BC$ as a diagonal.

Question 48

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Which of these sets of three points is collinear?

(1, 2, 3), (3, 0, 5), (4, -1, 6)

Correct answerYour answer

(1, 1, 1), (2, 3, 2), (4, 7, 5)

Correct answerYour answer

(0, 0, 0), (1, 2, 3), (2, 3, 4)

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(2, 1, 0), (0, 3, 2), (-2, 5, 3)

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AnswerA · $(1, 2, 3), (3, 0, 5), (4, -1, 6)$

Three points $A$ , $B$ , $C$ are collinear when $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel, with components in one ratio.
For $(1, 2, 3), (3, 0, 5), (4, -1, 6)$ : $\overrightarrow{AB} = 2\hat{i} - 2\hat{j} + 2\hat{k}$ and $\overrightarrow{AC} = 3\hat{i} - 3\hat{j} + 3\hat{k} = \frac{3}{2}\overrightarrow{AB}$ .

Worth noting

In $(1, 1, 1), (2, 3, 2), (4, 7, 5)$ the $\hat{i}$ and $\hat{j}$ parts of $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are in the ratio 3, but the $\hat{k}$ parts are not.

Question 49

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Three vertices of a parallelogram $ABCD$ , taken in order, are $A(1, 2, -1)$ , $B(2, -1, 0)$ and $C(5, 4, 3)$ .

What is the length of the diagonal $BD$ ?

6

Correct answerYour answer

6 \sqrt{2}

Correct answerYour answer

2 \sqrt{19}

Correct answerYour answer

72

Correct answerYour answer

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AnswerB · $6 \sqrt{2}$

$\overrightarrow{BD} = \vec{d} - \vec{b} = 2\hat{i} + 8\hat{j} + 2\hat{k}$ .
$\lvert \overrightarrow{BD} \rvert = \sqrt{4 + 64 + 4} = \sqrt{72} = 6\sqrt{2}$ .

Worth noting

$2\sqrt{19}$ adds the position vectors of $B$ and $D$ instead of subtracting them.

Question 50

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The point $R(3, 1, 0)$ divides $PQ$ internally in the ratio 2 : 3, and $P$ is $(1, -1, 2)$ . What is $Q$ ?

(\frac{13}{3}, \frac{7}{3}, - \frac{4}{3})

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(5, 3, -2)

Correct answerYour answer

(6, 4, -3)

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(9, 1, 3)

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AnswerC · $(6, 4, -3)$

$\vec{r} = \frac{2\vec{q} + 3\vec{p}}{5}$ , so $\vec{q} = \frac{5\vec{r} - 3\vec{p}}{2}$ .
$\frac{5(3, 1, 0) - 3(1, -1, 2)}{2} = \frac{(12, 8, -6)}{2} = (6, 4, -3)$ .

Worth noting

$(\frac{13}{3}, \frac{7}{3}, - \frac{4}{3})$ puts the weight 2 on $P$ instead of on $Q$ .

Question 51

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For which $\lambda$ are $A(1, 2, 3)$ , $B(2, 3, 5)$ and $C(\lambda, 5, 9)$ collinear?

3

Correct answerYour answer

4

Correct answerYour answer

5

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6

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AnswerB · $4$

$\overrightarrow{AB} = \hat{i} + \hat{j} + 2\hat{k}$ and $\overrightarrow{AC} = (\lambda - 1)\hat{i} + 3\hat{j} + 6\hat{k}$ must be parallel.
The $\hat{j}$ and $\hat{k}$ parts show $\overrightarrow{AC} = 3\overrightarrow{AB}$ , so $\lambda - 1 = 3$ : $\lambda = 4$ .

Worth noting

3 is the multiplier, $\overrightarrow{AC} = 3\overrightarrow{AB}$ , not $\lambda$ ; $\lambda$ is one more.

Question 52

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What point divides the join of $A(1, 2, 3)$ and $B(3, 4, 5)$ externally in the ratio 2 : 1?

(\frac{7}{3}, \frac{10}{3}, \frac{13}{3})

Correct answerYour answer

(-1, 0, 1)

Correct answerYour answer

(5, 6, 7)

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(2, 2, 2)

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AnswerC · $(5, 6, 7)$

External division: $\frac{m\vec{b} - n\vec{a}}{m - n}$ with $m = 2$ , $n = 1$ .
$2(3, 4, 5) - (1, 2, 3) = (5, 6, 7)$ .

Worth noting

$(\frac{7}{3}, \frac{10}{3}, \frac{13}{3})$ divides internally; an external point lies beyond $B$ .

Question 53

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The points $A(1, -2, 3)$ , $B(3, 0, 1)$ and $C(4, 1, 0)$ are collinear. How does $C$ divide $AB$ ?

Externally, in the ratio 3 : 1Correct answerYour answerInternally, in the ratio 3 : 1Correct answerYour answerExternally, in the ratio 1 : 3Correct answerYour answerInternally, in the ratio 2 : 1Correct answerYour answer

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AnswerA · Externally, in the ratio 3 : 1

$\overrightarrow{AC} = 3\hat{i} + 3\hat{j} - 3\hat{k}$ and $\overrightarrow{CB} = -\hat{i} - \hat{j} + \hat{k}$ , so $\overrightarrow{AC} = -3\overrightarrow{CB}$ .
The lengths are in the ratio $AC : CB = 3 : 1$ , and the minus sign puts $C$ outside $AB$ , beyond $B$ : external division.
Check: $\frac{3(3, 0, 1) - (1, -2, 3)}{3 - 1} = (4, 1, 0)$ .

Worth noting

Internally in the ratio 3 : 1 would put $C$ between $A$ and $B$ , at $\left(\frac{5}{2}, -\frac{1}{2}, \frac{3}{2}\right)$ .

Question 54

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$ABCDEF$ is a regular hexagon. What is $\overrightarrow{AB} + \overrightarrow{AC} + \overrightarrow{AD} + \overrightarrow{AE} + \overrightarrow{AF}$ ?

2\overrightarrow{AD}

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3\overrightarrow{AD}

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\vec{0}

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6\overrightarrow{AD}

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AnswerB · $3\overrightarrow{AD}$

Opposite sides of a regular hexagon are equal and parallel: $\overrightarrow{AB} = \overrightarrow{ED}$ and $\overrightarrow{AF} = \overrightarrow{CD}$ .
By the triangle law, $\overrightarrow{AE} + \overrightarrow{ED} = \overrightarrow{AD}$ and $\overrightarrow{AC} + \overrightarrow{CD} = \overrightarrow{AD}$ .
So the sum is $\overrightarrow{AD} + \overrightarrow{AD} + \overrightarrow{AD} = 3\overrightarrow{AD}$ .

Worth noting

$2\overrightarrow{AD}$ pairs off the four outer vectors but leaves out $\overrightarrow{AD}$ itself.

Question 55

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Three vertices of a parallelogram $ABCD$ , taken in order, are $A(1, 2, -1)$ , $B(2, -1, 0)$ and $C(5, 4, 3)$ .

$N$ is the midpoint of $AD$ , and the segment $BN$ meets the diagonal $AC$ at $G$ . What are the coordinates of $G$ ?

(3, 3, 1)

Correct answerYour answer

(\frac{11}{3}, \frac{10}{3}, \frac{5}{3})

Correct answerYour answer

(\frac{5}{2}, \frac{9}{2}, \frac{1}{2})

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(\frac{7}{3}, \frac{8}{3}, \frac{1}{3})

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AnswerD · $(\frac{7}{3}, \frac{8}{3}, \frac{1}{3})$

The diagonals meet at the midpoint $M$ of $BD$ , so $AM$ and $BN$ are both medians of triangle $ABD$ ; they meet at its centroid.
$G = \frac{\vec{a} + \vec{b} + \vec{d}}{3} = \frac{(7, 8, 1)}{3} = (\frac{7}{3}, \frac{8}{3}, \frac{1}{3})$ , a third of the way from $A$ to $C$ .

Worth noting

$(3, 3, 1)$ is where the diagonals meet; $BN$ crosses $AC$ before that, at the centroid of triangle $ABD$ .

Question 56

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Points $A$ , $B$ and $C$ have position vectors $\vec{a}$ , $\vec{b}$ and $\vec{c}$ , with $2\vec{a} + 3\vec{b} - 5\vec{c} = \vec{0}$ . How does $C$ divide $AB$ ?

Internally, in the ratio 2 : 3Correct answerYour answerExternally, in the ratio 3 : 2Correct answerYour answerInternally, in the ratio 3 : 2Correct answerYour answerExternally, in the ratio 2 : 3Correct answerYour answer

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AnswerC · Internally, in the ratio 3 : 2

$\vec{c} = \frac{2\vec{a} + 3\vec{b}}{5} = \frac{3\vec{b} + 2\vec{a}}{3 + 2}$ .
This is $\frac{m\vec{b} + n\vec{a}}{m + n}$ with $m = 3$ , $n = 2$ : $C$ divides $AB$ internally with $AC : CB = 3 : 2$ .

Worth noting

2 : 3 pairs each weight with its own point; in $\frac{m\vec{b} + n\vec{a}}{m + n}$ the weight on $\vec{b}$ is the $AC$ part of the ratio.

Question 57

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$P$ is $(1, 2, 0)$ and $Q$ is $(7, -4, 6)$ . Match each point in List I with its coordinates in List II.

List I	List II
A. The midpoint of $PQ$	I. $(-5, 8, -6)$
B. The point dividing $PQ$ internally in the ratio 2 : 1	II. $(13, -10, 12)$
C. The point dividing $PQ$ externally in the ratio 2 : 1	III. $(5, -2, 4)$
D. The point dividing $PQ$ externally in the ratio 1 : 2	IV. $(4, -1, 3)$

A-IV, B-III, C-I, D-IICorrect answerYour answerA-III, B-IV, C-II, D-ICorrect answerYour answerA-IV, B-II, C-III, D-ICorrect answerYour answerA-IV, B-III, C-II, D-ICorrect answerYour answer

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AnswerD · A-IV, B-III, C-II, D-I

Internally in $m : n$ : $\frac{m\vec{q} + n\vec{p}}{m + n}$ ; externally: $\frac{m\vec{q} - n\vec{p}}{m - n}$ .
A is $(4, -1, 3)$ . B: $\frac{2(7, -4, 6) + (1, 2, 0)}{3} = (5, -2, 4)$ . C: $2(7, -4, 6) - (1, 2, 0) = (13, -10, 12)$ . D: $\frac{(7, -4, 6) - 2(1, 2, 0)}{-1} = (-5, 8, -6)$ .

Worth noting

The two external points differ: 2 : 1 lies beyond $Q$ , and 1 : 2 lies beyond $P$ .

Question 58

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$\vec{a}$ and $\vec{b}$ are non-collinear vectors. For which $\lambda$ are the points with position vectors $\vec{a} + 2\vec{b}$ , $3\vec{a} - \vec{b}$ and $\lambda\vec{a} + 8\vec{b}$ collinear?

-4

Correct answerYour answer

-3

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3

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5

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AnswerB · $-3$

Call the points $A$ , $B$ , $C$ : $\overrightarrow{AB} = 2\vec{a} - 3\vec{b}$ and $\overrightarrow{AC} = (\lambda - 1)\vec{a} + 6\vec{b}$ .
Collinear means $\overrightarrow{AC} = \mu\overrightarrow{AB}$ ; the $\vec{b}$ parts give $\mu = -2$ , so $\lambda - 1 = -4$ and $\lambda = -3$ .

Worth noting

5 takes $\mu = 2$ , missing that $6$ and $-3$ have opposite signs.

The scalar product

33 questions

Question 59

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What is $(2\hat{i} - \hat{j} + 3\hat{k}) \cdot (\hat{i} + 4\hat{j} - 2\hat{k})$ ?

-8

Correct answerYour answer

-6

Correct answerYour answer

4

Correct answerYour answer

12

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AnswerA · $-8$

Multiply matching components and add.
$2(1) + (-1)(4) + 3(-2) = 2 - 4 - 6 = -8$ .

Worth noting

12 drops the signs: $2 + 4 + 6$ .

Question 60

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What is the angle between $\hat{i} + \hat{j}$ and $\hat{j} + \hat{k}$ ?

\frac{\pi}{6}

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\frac{\pi}{4}

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\frac{\pi}{3}

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\frac{\pi}{2}

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AnswerC · $\frac{\pi}{3}$

$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{\lvert \vec{a} \rvert \lvert \vec{b} \rvert} = \frac{1}{\sqrt{2}\sqrt{2}} = \frac{1}{2}$ .
$\theta = \frac{\pi}{3}$ .

Worth noting

$\frac{\pi}{4}$ takes $\cos \theta = \frac{1}{\sqrt{2}}$ , dividing by one magnitude only.

Question 61

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$\lvert \vec{a} \rvert = 3$ and $\lvert \vec{b} \rvert = 4$ . What is $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b})$ ?

-7

Correct answerYour answer

-1

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7

Correct answerYour answer

25

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AnswerA · $-7$

The cross terms $\vec{a} \cdot \vec{b}$ and $\vec{b} \cdot \vec{a}$ cancel: it is $\lvert \vec{a} \rvert^{2} - \lvert \vec{b} \rvert^{2}$ .
$9 - 16 = -7$ .

Worth noting

$-1$ subtracts the magnitudes, $3 - 4$ , without squaring.

Question 62

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Read the two statements and choose the correct option.

I. If $\vec{a} \cdot \vec{b} = 0$ and neither vector is $\vec{0}$ , the vectors are perpendicular.

II. $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$ , and likewise $\vec{a} \times \vec{b} = \vec{b} \times \vec{a}$ .

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

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AnswerA · Only I

I: $\lvert \vec{a} \rvert \lvert \vec{b} \rvert \cos \theta = 0$ with both magnitudes non-zero forces $\cos \theta = 0$ . True.
II: the scalar product commutes, but $\vec{b} \times \vec{a} = -\vec{a} \times \vec{b}$ . False.

Question 63

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A constant force $\vec{F} = 2\hat{i} + 2\hat{j} + \hat{k}$ moves a particle from $P(1, -1, 2)$ to $Q(3, 2, 1)$ . The work done is $\vec{F} \cdot \overrightarrow{PQ}$ .

What is the displacement $\overrightarrow{PQ}$ ?

-2\hat{i} - 3\hat{j} + \hat{k}

Correct answerYour answer

4\hat{i} + \hat{j} + 3\hat{k}

Correct answerYour answer

2\hat{i} + \hat{j} - \hat{k}

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2\hat{i} + 3\hat{j} - \hat{k}

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AnswerD · $2\hat{i} + 3\hat{j} - \hat{k}$

$\overrightarrow{PQ} = \vec{q} - \vec{p}$ .
$(3 - 1)\hat{i} + (2 + 1)\hat{j} + (1 - 2)\hat{k} = 2\hat{i} + 3\hat{j} - \hat{k}$ .

Worth noting

$2\hat{i} + \hat{j} - \hat{k}$ takes $2 - (-1)$ as 1.

Question 64

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What is $(3\hat{i} - 2\hat{j} + \hat{k}) \cdot (\hat{i} + 4\hat{j} - 5\hat{k})$ ?

-10

Correct answerYour answer

-5

Correct answerYour answer

0

Correct answerYour answer

16

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AnswerA · $-10$

Multiply matching components and add: $3(1) + (-2)(4) + 1(-5)$ .
$3 - 8 - 5 = -10$ .

Worth noting

0 takes $1 \times (-5)$ as $+5$ .

Question 65

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$\lvert \vec{a} \rvert = 4$ , $\lvert \vec{b} \rvert = 3$ , and the angle between $\vec{a}$ and $\vec{b}$ is $\frac{2\pi}{3}$ . What is $\vec{a} \cdot \vec{b}$ ?

- 6 \sqrt{3}

Correct answerYour answer

-6

Correct answerYour answer

6

Correct answerYour answer

6 \sqrt{3}

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AnswerB · $-6$

$\vec{a} \cdot \vec{b} = \lvert \vec{a} \rvert \lvert \vec{b} \rvert \cos \theta = 4 \times 3 \times \cos\frac{2\pi}{3}$ .
$\cos\frac{2\pi}{3} = -\frac{1}{2}$ , so $\vec{a} \cdot \vec{b} = -6$ .

Worth noting

6 takes $\cos\frac{2\pi}{3} = \frac{1}{2}$ ; the angle is obtuse, so its cosine is negative.

Question 66

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Which pair of vectors is perpendicular?

2\hat{i} - \hat{j} + 3\hat{k}

and

\hat{i} + 5\hat{j} + \hat{k}

Correct answerYour answer

\hat{i} + 2\hat{j} - \hat{k}

and

2\hat{i} - \hat{j} + \hat{k}

Correct answerYour answer

3\hat{i} + \hat{j}

and

\hat{i} + 3\hat{j}

Correct answerYour answer

\hat{i} - \hat{j} + \hat{k}

and

-\hat{i} + \hat{j} - \hat{k}

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AnswerA · $2\hat{i} - \hat{j} + 3\hat{k}$ and $\hat{i} + 5\hat{j} + \hat{k}$

Two non-zero vectors are perpendicular exactly when their scalar product is 0.
$2(1) + (-1)(5) + 3(1) = 2 - 5 + 3 = 0$ ; the other pairs give $-1$ , $6$ and $-3$ .

Worth noting

For $\hat{i} + 2\hat{j} - \hat{k}$ and $2\hat{i} - \hat{j} + \hat{k}$ the $\hat{i}$ and $\hat{j}$ products cancel, but the $\hat{k}$ product $-1$ remains.

Question 67

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Read the two statements and choose the correct option.

I. If $\lvert \vec{a} \rvert = \lvert \vec{b} \rvert$ , the projection of $\vec{a}$ on $\vec{b}$ equals the projection of $\vec{b}$ on $\vec{a}$ .

II. If $\vec{a} \cdot \vec{b} < 0$ , the angle between $\vec{a}$ and $\vec{b}$ is acute.

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

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AnswerA · Only I

I: the two projections are $\frac{\vec{a} \cdot \vec{b}}{\lvert \vec{b} \rvert}$ and $\frac{\vec{a} \cdot \vec{b}}{\lvert \vec{a} \rvert}$ , equal when the magnitudes are equal. True.
II: $\vec{a} \cdot \vec{b} < 0$ means $\cos \theta < 0$ , an obtuse angle. False.

Worth noting

The sign of the scalar product is the sign of $\cos \theta$ : negative means obtuse.

Question 68

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For which $\lambda$ are $2\hat{i} + \lambda\hat{j} + \hat{k}$ and $\hat{i} - 2\hat{j} + 3\hat{k}$ perpendicular?

- \frac{5}{2}

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\frac{1}{2}

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\frac{5}{2}

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5

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AnswerC · $\frac{5}{2}$

Perpendicular means the dot product is 0: $2 - 2\lambda + 3 = 0$ .
$\lambda = \frac{5}{2}$ .

Worth noting

$-\frac{5}{2}$ loses the sign of the $-2\lambda$ term.

Question 69

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What is the projection of $2\hat{i} + 3\hat{j} + 2\hat{k}$ on $\hat{i} + 2\hat{j} + \hat{k}$ ?

\frac{5}{3}

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\frac{10 \sqrt{17}}{17}

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\frac{5 \sqrt{6}}{3}

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10

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AnswerC · $\frac{5 \sqrt{6}}{3}$

Projection of $\vec{a}$ on $\vec{b}$ $= \frac{\vec{a} \cdot \vec{b}}{\lvert \vec{b} \rvert}$ .
$\frac{2 + 6 + 2}{\sqrt{6}} = \frac{10}{\sqrt{6}} = \frac{5\sqrt{6}}{3}$ .

Worth noting

$\frac{10}{\sqrt{17}}$ divides by $\lvert \vec{a} \rvert$ , which gives the projection of $\vec{b}$ on $\vec{a}$ .

Question 70

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Which of the following statements are true?

I. $\hat{i} \cdot \hat{i} = 1$ and $\hat{i} \cdot \hat{j} = 0$ .

II. $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$ .

III. If $\vec{a} \cdot \vec{b} = 0$ , then $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$ .

IV. $\vec{a} \cdot \vec{a} = \lvert \vec{a} \rvert^{2}$ .

All four: I, II, III and IVCorrect answerYour answerI and IV onlyCorrect answerYour answerII, III and IV onlyCorrect answerYour answerI, II and IV onlyCorrect answerYour answer

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AnswerD · I, II and IV only

I, II and IV are properties of the dot product.
III is false: $\hat{i} \cdot \hat{j} = 0$ with neither vector zero; they are perpendicular.

Question 71

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$\lvert \vec{a} \rvert = 3$ , $\lvert \vec{b} \rvert = 4$ and $\vec{a} \cdot \vec{b} = 6$ . What is $\lvert \vec{a} + \vec{b} \rvert$ ?

5

Correct answerYour answer

\sqrt{31}

Correct answerYour answer

\sqrt{37}

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7

Correct answerYour answer

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AnswerC · $\sqrt{37}$

$\lvert \vec{a} + \vec{b} \rvert^{2} = \lvert \vec{a} \rvert^{2} + 2\vec{a} \cdot \vec{b} + \lvert \vec{b} \rvert^{2} = 9 + 12 + 16$ .
$\lvert \vec{a} + \vec{b} \rvert = \sqrt{37}$ .

Worth noting

7 adds the magnitudes, which only works when the vectors point the same way.

Question 72

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What is the angle between $\hat{i} - \hat{j}$ and $\hat{j} - \hat{k}$ ?

\frac{\pi}{6}

Correct answerYour answer

\frac{\pi}{3}

Correct answerYour answer

\frac{2 \pi}{3}

Correct answerYour answer

\frac{5 \pi}{6}

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AnswerC · $\frac{2 \pi}{3}$

$\vec{a} \cdot \vec{b} = 0 - 1 + 0 = -1$ , and each magnitude is $\sqrt{2}$ .
$\cos \theta = -\frac{1}{2}$ , so $\theta = \frac{2\pi}{3}$ .

Worth noting

$\frac{\pi}{3}$ drops the minus sign of the scalar product.

Question 73

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A constant force $\vec{F} = 2\hat{i} + 2\hat{j} + \hat{k}$ moves a particle from $P(1, -1, 2)$ to $Q(3, 2, 1)$ . The work done is $\vec{F} \cdot \overrightarrow{PQ}$ .

What is the work done by the force?

-9

Correct answerYour answer

9

Correct answerYour answer

10

Correct answerYour answer

11

Correct answerYour answer

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AnswerB · $9$

$\vec{F} \cdot \overrightarrow{PQ} = 2(2) + 2(3) + 1(-1)$ .
$4 + 6 - 1 = 9$ .

Worth noting

$-9$ uses $\overrightarrow{QP}$ , the displacement in the wrong direction.

Question 74

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For which $\lambda$ are $\lambda\hat{i} + 2\hat{j} + \hat{k}$ and $4\hat{i} - 9\hat{j} + 2\hat{k}$ perpendicular?

-4

Correct answerYour answer

4

Correct answerYour answer

\frac{9}{2}

Correct answerYour answer

5

Correct answerYour answer

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AnswerB · $4$

$4\lambda - 18 + 2 = 0$ .
$\lambda = 4$ .

Worth noting

$\frac{9}{2}$ leaves out the $\hat{k}$ term, $1 \times 2$ .

Question 75

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What is the projection of $3\hat{i} + 4\hat{j}$ on $2\hat{i} - \hat{j} + 2\hat{k}$ ?

\frac{2}{9}

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\frac{2}{5}

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\frac{2}{3}

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2

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AnswerC · $\frac{2}{3}$

$\vec{a} \cdot \vec{b} = 6 - 4 + 0 = 2$ , and $\lvert \vec{b} \rvert = 3$ .
Projection $= \frac{2}{3}$ .

Worth noting

$\frac{2}{5}$ divides by $\lvert \vec{a} \rvert = 5$ ; that is the projection of $\vec{b}$ on $\vec{a}$ .

Question 76

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$\lvert \vec{a} \rvert = 2$ , $\lvert \vec{b} \rvert = 3$ , and the angle between them is $60^\circ$ . What is $\lvert \vec{a} - \vec{b} \rvert$ ?

\sqrt{7}

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\sqrt{13}

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\sqrt{19}

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7

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AnswerA · $\sqrt{7}$

$\vec{a} \cdot \vec{b} = 2 \times 3 \times \cos 60^\circ = 3$ .
$\lvert \vec{a} - \vec{b} \rvert^{2} = 4 + 9 - 2(3) = 7$ , so $\sqrt{7}$ .

Worth noting

$\sqrt{19}$ adds $2\vec{a} \cdot \vec{b}$ ; for $\vec{a} - \vec{b}$ it is taken away.

Question 77

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A constant force $\vec{F} = 2\hat{i} + 2\hat{j} + \hat{k}$ moves a particle from $P(1, -1, 2)$ to $Q(3, 2, 1)$ . The work done is $\vec{F} \cdot \overrightarrow{PQ}$ .

What is the cosine of the angle between $\vec{F}$ and $\overrightarrow{PQ}$ ?

\frac{3}{14}

Correct answerYour answer

\frac{\sqrt{14}}{14}

Correct answerYour answer

\frac{9}{14}

Correct answerYour answer

\frac{3 \sqrt{14}}{14}

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AnswerD · $\frac{3 \sqrt{14}}{14}$

$\lvert \vec{F} \rvert = 3$ and $\lvert \overrightarrow{PQ} \rvert = \sqrt{14}$ .
$\cos \theta = \frac{9}{3\sqrt{14}} = \frac{3}{\sqrt{14}}$ .

Worth noting

$\frac{9}{14}$ divides by $\lvert \overrightarrow{PQ} \rvert^{2}$ instead of $\lvert \vec{F} \rvert \lvert \overrightarrow{PQ} \rvert$ .

Question 78

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What is the projection vector of $\vec{a} = 3\hat{i} + \hat{j} - \hat{k}$ on $\vec{b} = \hat{i} + 2\hat{j} + 2\hat{k}$ ?

\hat{i} + 2\hat{j} + 2\hat{k}

Correct answerYour answer

\frac{3}{11}(3\hat{i} + \hat{j} - \hat{k})

Correct answerYour answer

\frac{1}{9}(\hat{i} + 2\hat{j} + 2\hat{k})

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\frac{1}{3}(\hat{i} + 2\hat{j} + 2\hat{k})

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AnswerD · $\frac{1}{3}(\hat{i} + 2\hat{j} + 2\hat{k})$

The projection vector of $\vec{a}$ on $\vec{b}$ is $\left(\frac{\vec{a} \cdot \vec{b}}{\lvert \vec{b} \rvert}\right)\hat{b} = \frac{\vec{a} \cdot \vec{b}}{\lvert \vec{b} \rvert^{2}}\vec{b}$ .
$\vec{a} \cdot \vec{b} = 3 + 2 - 2 = 3$ and $\lvert \vec{b} \rvert^{2} = 9$ , so it is $\frac{3}{9}\vec{b} = \frac{1}{3}(\hat{i} + 2\hat{j} + 2\hat{k})$ .

Worth noting

$\hat{i} + 2\hat{j} + 2\hat{k}$ multiplies the scalar projection by $\vec{b}$ instead of by the unit vector $\hat{b}$ .

Question 79

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$\vec{a} = 2\hat{i} + \hat{j} + 2\hat{k}$ , $\vec{b} = \hat{i} + 2\hat{j}$ and $\vec{c} = \hat{i} - 2\hat{j} + 3\hat{k}$ . For which $\lambda$ is $\vec{a} + \lambda\vec{b}$ perpendicular to $\vec{c}$ ?

-2

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- \frac{1}{2}

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\frac{1}{2}

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2

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AnswerD · $2$

$(\vec{a} + \lambda\vec{b}) \cdot \vec{c} = \vec{a} \cdot \vec{c} + \lambda\,\vec{b} \cdot \vec{c} = 0$ .
$\vec{a} \cdot \vec{c} = 2 - 2 + 6 = 6$ and $\vec{b} \cdot \vec{c} = 1 - 4 + 0 = -3$ , so $6 - 3\lambda = 0$ and $\lambda = 2$ .

Worth noting

$-2$ loses the minus sign of $\vec{b} \cdot \vec{c} = -3$ .

Question 80

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Assertion (A): For non-zero vectors $\vec{a}$ and $\vec{b}$ , if $\lvert \vec{a} + \vec{b} \rvert = \lvert \vec{a} - \vec{b} \rvert$ , then $\vec{a}$ is perpendicular to $\vec{b}$ .

Reason (R): $\lvert \vec{a} + \vec{b} \rvert^{2} - \lvert \vec{a} - \vec{b} \rvert^{2} = 4\vec{a} \cdot \vec{b}$ .

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AnswerA · Both A and R are true, and R is the correct explanation of A.

R: the squares are $\lvert \vec{a} \rvert^{2} \pm 2\vec{a} \cdot \vec{b} + \lvert \vec{b} \rvert^{2}$ , and their difference is $4\vec{a} \cdot \vec{b}$ . True.
A: equal magnitudes make the left side of R zero, so $\vec{a} \cdot \vec{b} = 0$ and the non-zero vectors are perpendicular. R is the reason.

Question 81

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$\vec{a}$ , $\vec{b}$ and $\vec{c}$ are mutually perpendicular, with $\lvert \vec{a} \rvert = 2$ , $\lvert \vec{b} \rvert = 3$ and $\lvert \vec{c} \rvert = 6$ . What is $\lvert \vec{a} + \vec{b} + \vec{c} \rvert$ ?

\sqrt{11}

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7

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11

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49

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AnswerB · $7$

$\lvert \vec{a} + \vec{b} + \vec{c} \rvert^{2} = \lvert \vec{a} \rvert^{2} + \lvert \vec{b} \rvert^{2} + \lvert \vec{c} \rvert^{2} + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$ .
Every scalar product is 0, so it is $4 + 9 + 36 = 49$ , and $\lvert \vec{a} + \vec{b} + \vec{c} \rvert = 7$ .

Worth noting

11 adds the magnitudes, as if all three vectors pointed the same way.

Question 82

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$\vec{a} = \hat{i} + 2\hat{j} - 2\hat{k}$ and $\vec{b} = 2\hat{i} - 2\hat{j} - \hat{k}$ . Which of the following statements are true?

I. $\vec{a} \cdot \vec{b} = 0$ .

II. The projection of $\vec{a}$ on $\vec{b}$ is 3.

III. $\lvert \vec{a} - \vec{b} \rvert = 3\sqrt{2}$ .

IV. $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ are perpendicular.

I and III onlyCorrect answerYour answerAll four: I, II, III and IVCorrect answerYour answerI, III and IV onlyCorrect answerYour answerIII and IV onlyCorrect answerYour answer

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AnswerC · I, III and IV only

I: $2 - 4 + 2 = 0$ , true; so the projection of $\vec{a}$ on $\vec{b}$ is $\frac{0}{3} = 0$ , and II is false.
III: $\lvert \vec{a} - \vec{b} \rvert^{2} = 9 - 0 + 9 = 18$ , true. IV: $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = \lvert \vec{a} \rvert^{2} - \lvert \vec{b} \rvert^{2} = 9 - 9 = 0$ , true.

Worth noting

IV holds because the magnitudes are equal, not because $\vec{a}$ and $\vec{b}$ are perpendicular.

Question 83

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$\vec{a}$ is a unit vector and $(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 15$ . What is $\lvert \vec{x} \rvert$ ?

\sqrt{14}

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\sqrt{15}

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4

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16

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AnswerC · $4$

The cross terms $\vec{x} \cdot \vec{a}$ and $\vec{a} \cdot \vec{x}$ cancel: $(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = \lvert \vec{x} \rvert^{2} - \lvert \vec{a} \rvert^{2}$ .
$\lvert \vec{x} \rvert^{2} - 1 = 15$ , so $\lvert \vec{x} \rvert^{2} = 16$ and $\lvert \vec{x} \rvert = 4$ .

Worth noting

$\sqrt{14}$ adds the squares, $\lvert \vec{x} \rvert^{2} + 1 = 15$ ; the product is a difference of squares.

Question 84

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What is the angle between $3\hat{i} - 4\hat{k}$ and $2\hat{i} + 2\hat{j} - \hat{k}$ ?

\cos^{-1}\frac{2}{15}

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\cos^{-1}\frac{2}{5}

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\cos^{-1}\frac{2}{3}

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\sin^{-1}\frac{2}{3}

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AnswerC · $\cos^{-1}\frac{2}{3}$

$\vec{a} \cdot \vec{b} = 3(2) + 0(2) + (-4)(-1) = 10$ , $\lvert \vec{a} \rvert = 5$ and $\lvert \vec{b} \rvert = 3$ .
$\cos \theta = \frac{10}{15} = \frac{2}{3}$ , so $\theta = \cos^{-1}\frac{2}{3}$ .

Worth noting

$\cos^{-1}\frac{2}{15}$ takes $(-4)(-1)$ as $-4$ , so the scalar product comes out as 2.

Question 85

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$\vec{a}$ and $\vec{b}$ are unit vectors and $\lvert \vec{a} + \vec{b} \rvert = 1$ . What is the angle between them?

\frac{\pi}{6}

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\frac{\pi}{4}

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\frac{\pi}{3}

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\frac{2 \pi}{3}

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AnswerD · $\frac{2 \pi}{3}$

$\lvert \vec{a} + \vec{b} \rvert^{2} = 1 + 1 + 2\cos \theta = 1$ .
$\cos \theta = -\frac{1}{2}$ , so $\theta = \frac{2\pi}{3}$ .

Worth noting

$\frac{\pi}{3}$ takes $\cos \theta = +\frac{1}{2}$ ; that would make $\lvert \vec{a} + \vec{b} \rvert = \sqrt{3}$ .

Question 86

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A constant force $\vec{F} = 2\hat{i} + 2\hat{j} + \hat{k}$ moves a particle from $P(1, -1, 2)$ to $Q(3, 2, 1)$ . The work done is $\vec{F} \cdot \overrightarrow{PQ}$ .

What is the projection of $\overrightarrow{PQ}$ on $\vec{F}$ ?

1

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\frac{9 \sqrt{14}}{14}

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3

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9

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AnswerC · $3$

Projection of $\overrightarrow{PQ}$ on $\vec{F}$ $= \frac{\vec{F} \cdot \overrightarrow{PQ}}{\lvert \vec{F} \rvert}$ .
$\frac{9}{3} = 3$ .

Worth noting

$\frac{9}{\sqrt{14}}$ divides by $\lvert \overrightarrow{PQ} \rvert$ ; that is the projection of $\vec{F}$ on $\overrightarrow{PQ}$ .

Question 87

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$\vec{a} + \vec{b} + \vec{c} = \vec{0}$ , with $\lvert \vec{a} \rvert = 3$ , $\lvert \vec{b} \rvert = 8$ and $\lvert \vec{c} \rvert = 7$ . What is the angle between $\vec{a}$ and $\vec{b}$ ?

\frac{\pi}{6}

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\frac{\pi}{3}

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\frac{2 \pi}{3}

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\frac{5 \pi}{6}

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AnswerC · $\frac{2 \pi}{3}$

$\vec{c} = -(\vec{a} + \vec{b})$ , so $\lvert \vec{c} \rvert^{2} = \lvert \vec{a} \rvert^{2} + 2\vec{a} \cdot \vec{b} + \lvert \vec{b} \rvert^{2}$ : $49 = 9 + 2\vec{a} \cdot \vec{b} + 64$ .
$\vec{a} \cdot \vec{b} = -12$ , so $\cos \theta = \frac{-12}{3 \times 8} = -\frac{1}{2}$ and $\theta = \frac{2\pi}{3}$ .

Worth noting

$\frac{\pi}{3}$ is the angle inside the triangle the vectors form head to tail; the angle between $\vec{a}$ and $\vec{b}$ is its supplement.

Question 88

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$\vec{a}$ and $\vec{b}$ are unit vectors at an angle $\theta$ . What is $\lvert \vec{a} - \vec{b} \rvert$ ?

2\cos\frac{\theta}{2}

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2 - 2\cos \theta

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\sqrt{2}\sin\frac{\theta}{2}

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2\sin\frac{\theta}{2}

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AnswerD · $2\sin\frac{\theta}{2}$

$\lvert \vec{a} - \vec{b} \rvert^{2} = 1 + 1 - 2\cos \theta = 2(1 - \cos \theta)$ .
$1 - \cos \theta = 2\sin^{2}\frac{\theta}{2}$ , so $\lvert \vec{a} - \vec{b} \rvert = 2\sin\frac{\theta}{2}$ , which is not negative since $\frac{\theta}{2}$ lies in $\left[0, \frac{\pi}{2}\right]$ .

Worth noting

$2\cos\frac{\theta}{2}$ is $\lvert \vec{a} + \vec{b} \rvert$ , where $2\cos \theta$ is added, not taken away.

Question 89

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For which values of $\lambda$ is the angle between $\lambda\hat{i} + 2\hat{j} + \hat{k}$ and $\lambda\hat{i} - 3\hat{j} + 2\hat{k}$ obtuse?

\lambda < -2

\lambda > 2

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0 < \lambda < 2

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-4 < \lambda < 4

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-2 < \lambda < 2

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AnswerD · $-2 < \lambda < 2$

The angle is obtuse when $\vec{a} \cdot \vec{b} < 0$ ; these vectors are never opposite, so $\theta = \pi$ does not arise.
$\vec{a} \cdot \vec{b} = \lambda^{2} - 6 + 2 = \lambda^{2} - 4 < 0$ gives $-2 < \lambda < 2$ .

Worth noting

$\lambda < -2$ or $\lambda > 2$ makes the scalar product positive: that is when the angle is acute.

Question 90

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$\vec{a} = \lambda\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}$ , with $\lambda > 0$ . For which $\lambda$ is $\vec{a} + \vec{b}$ perpendicular to $\vec{a} - \vec{b}$ ?

3

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\sqrt{14}

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\sqrt{19}

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9

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AnswerA · $3$

$(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = \lvert \vec{a} \rvert^{2} - \lvert \vec{b} \rvert^{2}$ , so the two are perpendicular exactly when $\lvert \vec{a} \rvert = \lvert \vec{b} \rvert$ .
$\lambda^{2} + 4 + 1 = 4 + 1 + 9$ , so $\lambda^{2} = 9$ , and $\lambda > 0$ gives $\lambda = 3$ .

Worth noting

9 is $\lambda^{2}$ ; $\lambda$ is its square root.

Question 91

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$\vec{a} + \vec{b} + \vec{c} = \vec{0}$ , with $\lvert \vec{a} \rvert = 2$ , $\lvert \vec{b} \rvert = 3$ and $\lvert \vec{c} \rvert = 4$ . What is $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$ ?

-29

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- \frac{29}{2}

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0

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\frac{29}{2}

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AnswerB · $- \frac{29}{2}$

Square $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ : $\lvert \vec{a} \rvert^{2} + \lvert \vec{b} \rvert^{2} + \lvert \vec{c} \rvert^{2} + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$ .
$4 + 9 + 16 = 29$ , so the sum is $-\frac{29}{2}$ .

Worth noting

$\frac{29}{2}$ loses the sign when the squares move to the other side.

The vector product

34 questions

Question 92

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Assertion (A): $\hat{i} \times \hat{j} = \hat{k}$ and $\hat{j} \times \hat{i} = -\hat{k}$ .

Reason (R): Swapping the order of a vector product changes its sign.

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AnswerA · Both A and R are true, and R is the correct explanation of A.

A is true: $\hat{i}, \hat{j}, \hat{k}$ form a right-handed system.
R is true and gives $\hat{j} \times \hat{i} = -(\hat{i} \times \hat{j})$ , the second half of A.

Question 93

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A triangle has vertices $A(1, 1, 1)$ , $B(2, 3, 3)$ and $C(3, 1, 2)$ .

What is $\overrightarrow{AB}$ ?

-\hat{i} - 2\hat{j} - 2\hat{k}

Correct answerYour answer

\hat{i} + 2\hat{j} + 2\hat{k}

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3\hat{i} + 4\hat{j} + 4\hat{k}

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\hat{i} + 2\hat{j} + 4\hat{k}

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AnswerB · $\hat{i} + 2\hat{j} + 2\hat{k}$

$\overrightarrow{AB}$ = position vector of $B$ − position vector of $A$ .
$(2 - 1)\hat{i} + (3 - 1)\hat{j} + (3 - 1)\hat{k} = \hat{i} + 2\hat{j} + 2\hat{k}$ .

Worth noting

$-\hat{i} - 2\hat{j} - 2\hat{k}$ is $\overrightarrow{BA}$ , $A$ minus $B$ .

Question 94

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$\lvert \vec{a} \rvert = 3$ , $\lvert \vec{b} \rvert = 4$ , and the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$ . What is $\lvert \vec{a} \times \vec{b} \rvert$ ?

3

Correct answerYour answer

6

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6 \sqrt{3}

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12

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AnswerB · $6$

$\lvert \vec{a} \times \vec{b} \rvert = \lvert \vec{a} \rvert \lvert \vec{b} \rvert \sin \theta = 3 \times 4 \times \sin\frac{\pi}{6}$ .
$\sin\frac{\pi}{6} = \frac{1}{2}$ , so $\lvert \vec{a} \times \vec{b} \rvert = 6$ .

Worth noting

$6\sqrt{3}$ uses $\cos\frac{\pi}{6}$ ; the vector product carries the sine.

Question 95

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What is $(2\hat{i} - 3\hat{k}) \times \hat{j}$ ?

3\hat{i} + 2\hat{k}

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-3\hat{i} - 2\hat{k}

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-3\hat{i} + 2\hat{k}

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3\hat{i} - 2\hat{k}

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AnswerA · $3\hat{i} + 2\hat{k}$

$(2\hat{i} - 3\hat{k}) \times \hat{j} = 2(\hat{i} \times \hat{j}) - 3(\hat{k} \times \hat{j})$ .
$\hat{i} \times \hat{j} = \hat{k}$ and $\hat{k} \times \hat{j} = -\hat{i}$ , so it is $2\hat{k} + 3\hat{i} = 3\hat{i} + 2\hat{k}$ .

Worth noting

$-3\hat{i} - 2\hat{k}$ is $\hat{j} \times (2\hat{i} - 3\hat{k})$ , the product in the other order.

Question 96

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A parallelogram has adjacent sides $\vec{a} = 2\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = 2\hat{i} + 3\hat{j} - 2\hat{k}$ .

What is the area of the parallelogram?

3 square unitsCorrect answerYour answer6 square unitsCorrect answerYour answer12 square unitsCorrect answerYour answer144 square unitsCorrect answerYour answer

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AnswerC · 12 square units

The area is $\lvert \vec{a} \times \vec{b} \rvert$ .
$\sqrt{64 + 64 + 16} = \sqrt{144} = 12$ .

Worth noting

6 halves the vector product, which gives the triangle on $\vec{a}$ and $\vec{b}$ , not the parallelogram.

Question 97

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What is $(2\hat{i} - \hat{j} + \hat{k}) \times (\hat{i} + 3\hat{j} - 2\hat{k})$ ?

\hat{i} - 5\hat{j} - 7\hat{k}

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-\hat{i} + 5\hat{j} + 7\hat{k}

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-\hat{i} - 5\hat{j} + 7\hat{k}

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2\hat{i} - 3\hat{j} - 2\hat{k}

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AnswerB · $-\hat{i} + 5\hat{j} + 7\hat{k}$

Expand $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 3 & -2 \end{vmatrix}$ .
$\hat{i}(2 - 3) - \hat{j}(-4 - 1) + \hat{k}(6 + 1) = -\hat{i} + 5\hat{j} + 7\hat{k}$ .

Worth noting

$-\hat{i} - 5\hat{j} + 7\hat{k}$ forgets the minus sign in front of the $\hat{j}$ term.

Question 98

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What is the area of the parallelogram with adjacent sides $\hat{i} + 2\hat{j}$ and $\hat{j} + 3\hat{k}$ ?

2 square unitsCorrect answerYour answer

\frac{\sqrt{46}}{2}

square unitsCorrect answerYour answer

\sqrt{46}

square unitsCorrect answerYour answer46 square unitsCorrect answerYour answer

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AnswerC · $\sqrt{46}$ square units

$\vec{a} \times \vec{b} = 6\hat{i} - 3\hat{j} + \hat{k}$ .
Area $= \lvert \vec{a} \times \vec{b} \rvert = \sqrt{36 + 9 + 1} = \sqrt{46}$ .

Worth noting

2 is $\vec{a} \cdot \vec{b}$ , the scalar product; area needs the vector product.

Question 99

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For which $\lambda$ is $(3\hat{i} - 6\hat{j} + 9\hat{k}) \times (\hat{i} + \lambda\hat{j} + 3\hat{k}) = \vec{0}$ ?

-6

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-3

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-2

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2

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AnswerC · $-2$

The vector product is zero exactly when the vectors are parallel: $\frac{1}{3} = \frac{\lambda}{-6} = \frac{3}{9}$ .
$\lambda = -2$ .

Worth noting

2 loses the sign: the $\hat{j}$ part of the first vector is $-6$ .

Question 100

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Which of the following statements are true?

I. $\hat{i} \times \hat{j} = \hat{k}$ and $\hat{j} \times \hat{i} = -\hat{k}$ .

II. $\vec{a} \times \vec{a} = \vec{0}$ for every $\vec{a}$ .

III. $\vec{a} \times \vec{b} = \vec{b} \times \vec{a}$ .

IV. $\vec{a} \times \vec{b}$ is perpendicular to both $\vec{a}$ and $\vec{b}$ .

I, II and IV onlyCorrect answerYour answerAll four: I, II, III and IVCorrect answerYour answerI and IV onlyCorrect answerYour answerII, III and IV onlyCorrect answerYour answer

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AnswerA · I, II and IV only

I, II and IV are properties of the vector product.
III is false: $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$ ; the order matters.

Worth noting

The scalar product commutes; the vector product changes sign.

Question 101

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$\lvert \vec{a} \rvert = 2$ , $\lvert \vec{b} \rvert = 6$ and $\lvert \vec{a} \times \vec{b} \rvert = 6\sqrt{3}$ . What is $\lvert \vec{a} \cdot \vec{b} \rvert$ ?

6

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6 \sqrt{3}

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12

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36

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AnswerA · $6$

$\lvert \vec{a} \times \vec{b} \rvert^{2} + (\vec{a} \cdot \vec{b})^{2} = \lvert \vec{a} \rvert^{2}\lvert \vec{b} \rvert^{2} = 144$ .
$(\vec{a} \cdot \vec{b})^{2} = 144 - 108 = 36$ , so $\lvert \vec{a} \cdot \vec{b} \rvert = 6$ .

Worth noting

36 is $(\vec{a} \cdot \vec{b})^{2}$ , before the square root.

Question 102

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$\lvert \vec{a} \rvert = \sqrt{3}$ , $\lvert \vec{b} \rvert = 2$ and $\lvert \vec{a} \times \vec{b} \rvert = 3$ . What is the acute angle between $\vec{a}$ and $\vec{b}$ ?

\frac{\pi}{6}

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\frac{\pi}{4}

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\frac{\pi}{3}

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\frac{\pi}{2}

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AnswerC · $\frac{\pi}{3}$

$\lvert \vec{a} \times \vec{b} \rvert = \lvert \vec{a} \rvert \lvert \vec{b} \rvert \sin \theta$ : $3 = 2\sqrt{3}\sin \theta$ .
$\sin \theta = \frac{\sqrt{3}}{2}$ , so $\theta = \frac{\pi}{3}$ .

Worth noting

$\frac{\pi}{6}$ reads $\frac{\sqrt{3}}{2}$ as a cosine; the vector product carries the sine.

Question 103

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Which vector is perpendicular to both $\hat{i} + \hat{j}$ and $\hat{j} + \hat{k}$ ?

\hat{i} + \hat{j} + \hat{k}

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\hat{i} - \hat{j} - \hat{k}

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2\hat{i} - \hat{j} + \hat{k}

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\hat{i} - \hat{j} + \hat{k}

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AnswerD · $\hat{i} - \hat{j} + \hat{k}$

$\vec{a} \times \vec{b}$ is perpendicular to both.
$\hat{i}(1 - 0) - \hat{j}(1 - 0) + \hat{k}(1 - 0) = \hat{i} - \hat{j} + \hat{k}$ ; its dot product with each is 0.

Worth noting

$\hat{i} + \hat{j} + \hat{k}$ forgets the minus sign on the $\hat{j}$ term.

Question 104

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A triangle has vertices $A(1, 1, 1)$ , $B(2, 3, 3)$ and $C(3, 1, 2)$ .

What is $\overrightarrow{AB} \cdot \overrightarrow{AC}$ ?

0

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2

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4

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6

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AnswerC · $4$

$\overrightarrow{AC} = 2\hat{i} + \hat{k}$ .
$1(2) + 2(0) + 2(1) = 4$ .

Worth noting

2 stops after the first product, $1 \times 2$ .

Question 105

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A triangle has vertices $A(1, 1, 1)$ , $B(2, 3, 3)$ and $C(3, 1, 2)$ .

What is $\cos A$ , the cosine of the angle at $A$ ?

\frac{2 \sqrt{5}}{15}

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\frac{4}{9}

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\frac{4 \sqrt{5}}{15}

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\frac{4}{5}

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AnswerC · $\frac{4 \sqrt{5}}{15}$

$\lvert \overrightarrow{AB} \rvert = 3$ and $\lvert \overrightarrow{AC} \rvert = \sqrt{5}$ .
$\cos A = \frac{4}{3\sqrt{5}} = \frac{4 \sqrt{5}}{15}$ .

Worth noting

$\frac{4}{9}$ divides by $\lvert \overrightarrow{AB} \rvert^{2}$ instead of the product of the two magnitudes.

Question 106

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What is $(\hat{i} + 2\hat{j} - \hat{k}) \times (2\hat{i} - \hat{j} + 3\hat{k})$ ?

-5\hat{i} + 5\hat{j} + 5\hat{k}

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5\hat{i} + 5\hat{j} - 5\hat{k}

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7\hat{i} - 5\hat{j} - 5\hat{k}

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5\hat{i} - 5\hat{j} - 5\hat{k}

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AnswerD · $5\hat{i} - 5\hat{j} - 5\hat{k}$

$\hat{i}(2 \cdot 3 - (-1)(-1)) - \hat{j}(1 \cdot 3 - (-1) \cdot 2) + \hat{k}(1 \cdot (-1) - 2 \cdot 2)$ .
$= 5\hat{i} - 5\hat{j} - 5\hat{k}$ .

Worth noting

$-5\hat{i} + 5\hat{j} + 5\hat{k}$ is $\vec{b} \times \vec{a}$ , the product in the other order.

Question 107

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Assertion (A): If $\vec{a} \times \vec{b} = \vec{0}$ and $\vec{a} \cdot \vec{b} = 0$ , then $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$ .

Reason (R): For non-zero vectors, $\vec{a} \times \vec{b} = \vec{0}$ means they are parallel, and $\vec{a} \cdot \vec{b} = 0$ means they are perpendicular.

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AnswerA · Both A and R are true, and R is the correct explanation of A.

R is true: $\lvert \vec{a} \rvert \lvert \vec{b} \rvert \sin \theta = 0$ forces $\sin \theta = 0$ , and $\lvert \vec{a} \rvert \lvert \vec{b} \rvert \cos \theta = 0$ forces $\cos \theta = 0$ .
No angle has both sine and cosine 0, so two non-zero vectors cannot do both: one of them is $\vec{0}$ . A is true, and R explains it.

Question 108

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Read the two statements and choose the correct option.

I. The vector product of two unit vectors is always a unit vector.

II. If $\vec{a} \times \vec{b} = \vec{a} \times \vec{c}$ and $\vec{a} \ne \vec{0}$ , then $\vec{b} = \vec{c}$ .

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

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AnswerD · Neither I nor II

I: $\lvert \hat{a} \times \hat{b} \rvert = \sin \theta$ , which is 1 only for perpendicular vectors; $\hat{i} \times \hat{i} = \vec{0}$ . False.
II: $\vec{a} \times (\vec{b} - \vec{c}) = \vec{0}$ only says $\vec{b} - \vec{c}$ is parallel to $\vec{a}$ ; for example $\hat{i} \times \hat{j} = \hat{i} \times (\hat{i} + \hat{j})$ . False.

Worth noting

A vector product cannot be cancelled: $\vec{b} - \vec{c}$ can be any multiple of $\vec{a}$ .

Question 109

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$\lvert \vec{a} \rvert = \lvert \vec{b} \rvert$ , $\lvert \vec{a} \times \vec{b} \rvert = 12$ and $\vec{a} \cdot \vec{b} = 5$ . What is $\lvert \vec{a} \rvert$ ?

\sqrt{13}

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\sqrt{17}

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13

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169

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AnswerA · $\sqrt{13}$

$\lvert \vec{a} \times \vec{b} \rvert^{2} + (\vec{a} \cdot \vec{b})^{2} = \lvert \vec{a} \rvert^{2}\lvert \vec{b} \rvert^{2} = \lvert \vec{a} \rvert^{4}$ .
$144 + 25 = 169$ , so $\lvert \vec{a} \rvert^{2} = 13$ and $\lvert \vec{a} \rvert = \sqrt{13}$ .

Worth noting

13 is $\lvert \vec{a} \rvert^{2}$ : with equal magnitudes, 169 is the fourth power of $\lvert \vec{a} \rvert$ .

Question 110

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For non-zero vectors $\vec{a}$ and $\vec{b}$ , $\lvert \vec{a} \times \vec{b} \rvert = \sqrt{3}\,(\vec{a} \cdot \vec{b})$ . What is the angle between them?

\frac{\pi}{6}

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\frac{\pi}{4}

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\frac{\pi}{3}

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\frac{2 \pi}{3}

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AnswerC · $\frac{\pi}{3}$

$\lvert \vec{a} \rvert \lvert \vec{b} \rvert \sin \theta = \sqrt{3}\lvert \vec{a} \rvert \lvert \vec{b} \rvert \cos \theta$ , so $\tan \theta = \sqrt{3}$ .
With $\theta$ in $[0, \pi]$ and $\cos \theta > 0$ , $\theta = \frac{\pi}{3}$ .

Worth noting

$\frac{\pi}{6}$ solves $\tan \theta = \frac{1}{\sqrt{3}}$ , with the sine and cosine swapped.

Question 111

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$\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$ . Which of the following statements are true?

I. $\vec{a} \times \vec{b} = -\hat{i} + 5\hat{j} + 3\hat{k}$ .

II. $\vec{b} \times \vec{a} = \hat{i} - 5\hat{j} - 3\hat{k}$ .

III. The triangle with sides $\vec{a}$ and $\vec{b}$ has area $\frac{\sqrt{35}}{2}$ .

IV. $\lvert \vec{a} \times \vec{b} \rvert = \lvert \vec{a} \rvert \lvert \vec{b} \rvert$ .

I and II onlyCorrect answerYour answerAll four: I, II, III and IVCorrect answerYour answerI, III and IV onlyCorrect answerYour answerI, II and III onlyCorrect answerYour answer

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AnswerD · I, II and III only

$\vec{a} \times \vec{b} = \hat{i}((-1)(-1) - 2(1)) - \hat{j}(1(-1) - 2(2)) + \hat{k}(1(1) - (-1)(2)) = -\hat{i} + 5\hat{j} + 3\hat{k}$ : I is true, and II, its negative, is true too.
III: $\frac{1}{2}\sqrt{1 + 25 + 9} = \frac{\sqrt{35}}{2}$ , true. IV: $\lvert \vec{a} \rvert \lvert \vec{b} \rvert = \sqrt{6}\sqrt{6} = 6 \ne \sqrt{35}$ , because $\vec{a} \cdot \vec{b} = -1 \ne 0$ ; false.

Worth noting

$\lvert \vec{a} \times \vec{b} \rvert = \lvert \vec{a} \rvert \lvert \vec{b} \rvert$ only when the vectors are perpendicular.

Question 112

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A parallelogram has adjacent sides $\vec{a} = 2\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = 2\hat{i} + 3\hat{j} - 2\hat{k}$ .

If $\theta$ is the angle between $\vec{a}$ and $\vec{b}$ , what is $\tan \theta$ ?

\frac{1}{4}

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\frac{4 \sqrt{17}}{17}

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4

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12

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AnswerC · $4$

$\tan \theta = \frac{\lvert \vec{a} \times \vec{b} \rvert}{\vec{a} \cdot \vec{b}}$ , since $\lvert \vec{a} \rvert \lvert \vec{b} \rvert$ cancels from $\sin \theta$ and $\cos \theta$ .
$\vec{a} \cdot \vec{b} = 4 + 3 - 4 = 3$ , so $\tan \theta = \frac{12}{3} = 4$ .

Worth noting

$\frac{1}{4}$ puts the scalar product on top, which gives $\cot \theta$ .

Question 113

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$(\lambda\hat{i} + 2\hat{j} - \hat{k}) \times (\hat{i} - \hat{j} + \hat{k}) = \hat{i} - 3\hat{j} + \mu\hat{k}$ . What is $\lambda + \mu$ ?

-6

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-2

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2

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6

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AnswerB · $-2$

$\hat{i}(2 \cdot 1 - (-1)(-1)) - \hat{j}(\lambda \cdot 1 - (-1) \cdot 1) + \hat{k}(\lambda(-1) - 2 \cdot 1) = \hat{i} - (\lambda + 1)\hat{j} - (\lambda + 2)\hat{k}$ .
Matching: $\lambda + 1 = 3$ , so $\lambda = 2$ , and $\mu = -(\lambda + 2) = -4$ : $\lambda + \mu = -2$ .

Worth noting

6 works out the $\hat{k}$ part the wrong way round, $2 \cdot 1 - \lambda(-1)$ , which gives $\mu = 4$ .

Question 114

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A parallelogram has adjacent sides $\vec{a} = 2\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = 2\hat{i} + 3\hat{j} - 2\hat{k}$ .

What is $\vec{a} \times \vec{b}$ ?

-8\hat{i} - 8\hat{j} + 4\hat{k}

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8\hat{i} - 8\hat{j} - 4\hat{k}

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-8\hat{i} + 8\hat{j} + 4\hat{k}

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-8\hat{i} + 8\hat{j} - 4\hat{k}

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AnswerC · $-8\hat{i} + 8\hat{j} + 4\hat{k}$

$\hat{i}(1(-2) - 2(3)) - \hat{j}(2(-2) - 2(2)) + \hat{k}(2(3) - 1(2))$ .
$= -8\hat{i} + 8\hat{j} + 4\hat{k}$ .

Worth noting

$-8\hat{i} - 8\hat{j} + 4\hat{k}$ forgets the minus sign in front of the $\hat{j}$ term.

Question 115

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$\lvert \vec{a} \rvert = 2$ , $\lvert \vec{b} \rvert = 5$ and $\vec{a} \cdot \vec{b} = 6$ . What is $\lvert \vec{a} \times \vec{b} \rvert$ ?

4

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8

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\sqrt{94}

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64

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AnswerB · $8$

$\lvert \vec{a} \times \vec{b} \rvert^{2} + (\vec{a} \cdot \vec{b})^{2} = \lvert \vec{a} \rvert^{2}\lvert \vec{b} \rvert^{2}$ , since $\sin^{2} \theta + \cos^{2} \theta = 1$ .
$\lvert \vec{a} \times \vec{b} \rvert^{2} = 100 - 36 = 64$ , so $\lvert \vec{a} \times \vec{b} \rvert = 8$ .

Worth noting

64 is $\lvert \vec{a} \times \vec{b} \rvert^{2}$ , before the square root.

Question 116

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Which is a unit vector perpendicular to both $\hat{i} + \hat{j} + \hat{k}$ and $\hat{i} - \hat{j}$ ?

\frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})

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\frac{1}{6}(\hat{i} + \hat{j} - 2\hat{k})

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\frac{1}{\sqrt{6}}(\hat{i} - \hat{j} + 2\hat{k})

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\frac{1}{\sqrt{6}}(\hat{i} + \hat{j} - 2\hat{k})

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AnswerD · $\frac{1}{\sqrt{6}}(\hat{i} + \hat{j} - 2\hat{k})$

$\vec{a} \times \vec{b} = \hat{i} + \hat{j} - 2\hat{k}$ is perpendicular to both; its magnitude is $\sqrt{1 + 1 + 4} = \sqrt{6}$ .
Dividing gives $\frac{1}{\sqrt{6}}(\hat{i} + \hat{j} - 2\hat{k})$ ; its negative works too.

Worth noting

$\frac{1}{6}(\hat{i} + \hat{j} - 2\hat{k})$ divides by the sum of the squares, not its square root.

Question 117

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Match each expression in List I with its value in List II.

List I	List II
A. $\hat{j} \times \hat{k}$	I. $\vec{0}$
B. $\hat{k} \cdot \hat{k}$	II. $\hat{i}$
C. $\hat{i} \times \hat{i}$	III. $-\hat{i}$
D. $\hat{k} \times \hat{j}$	IV. $1$

A-IV, B-II, C-I, D-IIICorrect answerYour answerA-II, B-IV, C-III, D-ICorrect answerYour answerA-IV, B-II, C-III, D-ICorrect answerYour answerA-II, B-IV, C-I, D-IIICorrect answerYour answer

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AnswerD · A-II, B-IV, C-I, D-III

Going round $\hat{i} \to \hat{j} \to \hat{k} \to \hat{i}$ gives $+$ ; against it, $-$ : $\hat{j} \times \hat{k} = \hat{i}$ , $\hat{k} \times \hat{j} = -\hat{i}$ .
$\hat{k} \cdot \hat{k} = 1$ and $\hat{i} \times \hat{i} = \vec{0}$ .

Question 118

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What is the area of the triangle with vertices $A(1, 0, 1)$ , $B(2, 2, 3)$ and $C(0, 3, 1)$ ?

\sqrt{13}

square unitsCorrect answerYour answer

\frac{\sqrt{65}}{2}

square unitsCorrect answerYour answer

\sqrt{65}

square unitsCorrect answerYour answer32.5 square unitsCorrect answerYour answer

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AnswerB · $\frac{\sqrt{65}}{2}$ square units

$\overrightarrow{AB} = \hat{i} + 2\hat{j} + 2\hat{k}$ , $\overrightarrow{AC} = -\hat{i} + 3\hat{j}$ , and $\overrightarrow{AB} \times \overrightarrow{AC} = -6\hat{i} - 2\hat{j} + 5\hat{k}$ .
Area $= \frac{1}{2}\sqrt{36 + 4 + 25} = \frac{\sqrt{65}}{2}$ .

Worth noting

$\sqrt{65}$ is the parallelogram on $\overrightarrow{AB}$ and $\overrightarrow{AC}$ ; the triangle is half of it.

Question 119

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What is the area of the triangle with vertices $A(1, 0, 2)$ , $B(3, 1, 0)$ and $C(2, 4, 1)$ ?

\frac{7 \sqrt{2}}{2}

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7

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7 \sqrt{2}

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\frac{49}{2}

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AnswerA · $\frac{7 \sqrt{2}}{2}$

$\overrightarrow{AB} = 2\hat{i} + \hat{j} - 2\hat{k}$ , $\overrightarrow{AC} = \hat{i} + 4\hat{j} - \hat{k}$ , and $\overrightarrow{AB} \times \overrightarrow{AC} = 7\hat{i} + 7\hat{k}$ .
Area $= \frac{1}{2}\sqrt{49 + 49} = \frac{7\sqrt{2}}{2}$ .

Worth noting

$7\sqrt{2}$ is the parallelogram on $AB$ and $AC$ ; the triangle is half of it.

Question 120

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The diagonals of a parallelogram are $3\hat{i} + \hat{j} - 2\hat{k}$ and $\hat{i} - 3\hat{j} + 4\hat{k}$ . What is its area?

\frac{5 \sqrt{3}}{2}

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5 \sqrt{3}

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10 \sqrt{3}

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20 \sqrt{3}

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AnswerB · $5 \sqrt{3}$

$\vec{d_1} \times \vec{d_2} = -2\hat{i} - 14\hat{j} - 10\hat{k}$ , of magnitude $\sqrt{4 + 196 + 100} = 10\sqrt{3}$ .
From the diagonals, area $= \frac{1}{2}\lvert \vec{d_1} \times \vec{d_2} \rvert = 5\sqrt{3}$ .

Worth noting

$10\sqrt{3}$ uses the side formula on the diagonals; with diagonals the half is needed.

Question 121

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A triangle has vertices $A(1, 1, 1)$ , $B(2, 3, 3)$ and $C(3, 1, 2)$ .

What is the area of triangle $ABC$ ?

\frac{\sqrt{21}}{2}

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\frac{\sqrt{29}}{2}

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\sqrt{29}

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\frac{29}{2}

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AnswerB · $\frac{\sqrt{29}}{2}$

$\overrightarrow{AB} \times \overrightarrow{AC} = 2\hat{i} + 3\hat{j} - 4\hat{k}$ , of magnitude $\sqrt{4 + 9 + 16} = \sqrt{29}$ .
Area $= \frac{\sqrt{29}}{2}$ .

Worth noting

$\sqrt{29}$ is the parallelogram, twice the triangle.

Question 122

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$\vec{a} = \hat{i} + \hat{j} + \lambda\hat{k}$ and $\vec{b} = \hat{i} - \hat{j}$ , with $\lambda > 0$ . The parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ has area $2\sqrt{3}$ square units. What is $\lambda$ ?

2

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\sqrt{6}

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4

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\sqrt{22}

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AnswerA · $2$

$\vec{a} \times \vec{b} = \hat{i}(0 + \lambda) - \hat{j}(0 - \lambda) + \hat{k}(-1 - 1) = \lambda\hat{i} + \lambda\hat{j} - 2\hat{k}$ .
Area $= \sqrt{2\lambda^{2} + 4} = 2\sqrt{3}$ , so $2\lambda^{2} + 4 = 12$ , $\lambda^{2} = 4$ and $\lambda = 2$ .

Worth noting

$\sqrt{22}$ halves the vector product as for a triangle; a parallelogram's area is the whole $\lvert \vec{a} \times \vec{b} \rvert$ .

Question 123

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$\vec{a} = \hat{i} + \hat{j} + \hat{k}$ . A vector $\vec{b}$ has $\vec{a} \times \vec{b} = -\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{a} \cdot \vec{b} = 3$ . What is $\vec{b}$ ?

\hat{j} + 2\hat{k}

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\hat{i} + \hat{j} + \hat{k}

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2\hat{i} + \hat{j}

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2\hat{i} + \hat{k}

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AnswerC · $2\hat{i} + \hat{j}$

Let $\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}$ . Then $\vec{a} \cdot \vec{b} = x + y + z = 3$ .
$\vec{a} \times \vec{b} = (z - y)\hat{i} + (x - z)\hat{j} + (y - x)\hat{k}$ , so $z - y = -1$ , $x - z = 2$ and $y - x = -1$ .
So $y = x - 1$ and $z = x - 2$ ; then $3x - 3 = 3$ gives $x = 2$ , and $\vec{b} = 2\hat{i} + \hat{j}$ .

Worth noting

$\hat{j} + 2\hat{k}$ solves $\vec{b} \times \vec{a} = -\hat{i} + 2\hat{j} - \hat{k}$ , the product in the other order.

Question 124

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$\lvert \vec{a} \rvert = 2$ , $\lvert \vec{b} \rvert = 3$ , and the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$ . What is the area of the parallelogram with adjacent sides $2\vec{a} + \vec{b}$ and $\vec{a} + 3\vec{b}$ ?

3 square unitsCorrect answerYour answer15 square unitsCorrect answerYour answer18 square unitsCorrect answerYour answer21 square unitsCorrect answerYour answer

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AnswerB · 15 square units

$(2\vec{a} + \vec{b}) \times (\vec{a} + 3\vec{b}) = 2\vec{a} \times \vec{a} + 6\vec{a} \times \vec{b} + \vec{b} \times \vec{a} + 3\vec{b} \times \vec{b}$ .
$\vec{a} \times \vec{a} = \vec{b} \times \vec{b} = \vec{0}$ and $\vec{b} \times \vec{a} = -\vec{a} \times \vec{b}$ , so it is $5\vec{a} \times \vec{b}$ .
$\lvert \vec{a} \times \vec{b} \rvert = 2 \times 3 \times \sin\frac{\pi}{6} = 3$ , so the area is $5 \times 3 = 15$ .

Worth noting

21 adds $\vec{b} \times \vec{a}$ as if it were $+\vec{a} \times \vec{b}$ ; swapping the order changes the sign.

Question 125

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A parallelogram has adjacent sides $\vec{a} = 2\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = 2\hat{i} + 3\hat{j} - 2\hat{k}$ .

A vector $\vec{c}$ of magnitude 6 is perpendicular to both $\vec{a}$ and $\vec{b}$ and makes an acute angle with the $z$ -axis. What is $\vec{c}$ ?

4\hat{i} - 4\hat{j} - 2\hat{k}

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- \frac{2}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{1}{3}\hat{k}

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-48\hat{i} + 48\hat{j} + 24\hat{k}

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-4\hat{i} + 4\hat{j} + 2\hat{k}

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AnswerD · $-4\hat{i} + 4\hat{j} + 2\hat{k}$

$\vec{a} \times \vec{b} = -8\hat{i} + 8\hat{j} + 4\hat{k}$ is perpendicular to both, with magnitude 12; the unit vectors along it are $\pm\frac{1}{3}(-2\hat{i} + 2\hat{j} + \hat{k})$ .
Magnitude 6 gives $\pm(-4\hat{i} + 4\hat{j} + 2\hat{k})$ ; an acute angle with the $z$ -axis needs a positive $\hat{k}$ component, so $\vec{c} = -4\hat{i} + 4\hat{j} + 2\hat{k}$ .

Worth noting

$4\hat{i} - 4\hat{j} - 2\hat{k}$ is perpendicular to both and has magnitude 6, but its $\hat{k}$ component is negative: an obtuse angle with the $z$ -axis.

NTA's Mathematics / Applied Mathematics syllabus has three sections: Section A1, which every candidate answers, then Section B1 (Mathematics) or Section B2 (Applied Mathematics), of which a candidate answers one. Each section lists units and topics but no chapters, so the chapters here are ours. Where both B sections list a topic that Section A also lists, its questions sit once, in Section A.