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Three-Dimensional Geometry for CUET Maths

About 5 min to learn · 120 questions

Three-dimensional geometry in CUET Maths Section B1 is all about lines in space: their direction, their equations, the angle between two of them and how close two lines come. It builds directly on vectors, so most questions take a few lines of working. Learn each rule here, from the direction cosines of a line to its vector and Cartesian equations, the angle between two lines, skew lines and the shortest distance between them, 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
  1. 01Lines in space
  2. 02Direction cosines of a line
  3. 03The equation of a line
  4. 04The angle between two lines
  5. 05Skew lines and the shortest distance
  6. 06How CUET asks this chapter
  7. 07Exam shortcuts
  8. 08Common mistakes
  9. 09Quick revision
  10. 10Check your understanding

Lines in space

Two straight wires are strung across a hall at different heights. They are not parallel, and they never meet. How close do they come? In three dimensions two lines can be skew: neither parallel nor intersecting. This chapter writes down a line, measures the angle between two lines, and finds that closest distance.

Direction cosines of a line

How do you find the direction cosines of the line through two points?

The direction ratios of the line through (x1,y1,z1)(x_1, y_1, z_1) and (x2,y2,z2)(x_2, y_2, z_2) are x2−x1,y2−y1,z2−z1x_2 - x_1, y_2 - y_1, z_2 - z_1. Divide them by (x2−x1)2+(y2−y1)2+(z2−z1)2\sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2} + (z_2 - z_1)^{2}} to get the direction cosines.

Example 1 · Two points

Find the direction cosines of the line from (2,3,1)(2, 3, 1) to (4,4,3)(4, 4, 3).

  1. Direction ratios 2,1,22, 1, 2, with 4+1+4=3\sqrt{4 + 1 + 4} = 3.

  2. Direction cosines 23,13,23\frac{2}{3}, \frac{1}{3}, \frac{2}{3}.

Answer
23,13,23\frac{2}{3}, \frac{1}{3}, \frac{2}{3}

The equation of a line

What are the vector and Cartesian equations of a line?

Through the point a⃗\vec{a} parallel to b⃗\vec{b}: r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda\vec{b}. In Cartesian form, through (x1,y1,z1)(x_1, y_1, z_1) with direction ratios a,b,ca, b, c:

x−x1a=y−y1b=z−z1c.\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}.

Through two points, use their difference as the direction.

Example 2 · From a point and a direction

Write the line through (1,−1,2)(1, -1, 2) parallel to 3i^+2j^−k^3\hat{i} + 2\hat{j} - \hat{k} in both forms.

  1. Vector: r⃗=(i^−j^+2k^)+λ(3i^+2j^−k^)\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - \hat{k}).

  2. Cartesian: x−13=y+12=z−2−1\frac{x - 1}{3} = \frac{y + 1}{2} = \frac{z - 2}{-1}.

Answer

Both describe the same line.

Watch out

y+1y + 1 means y−(−1)y - (-1): the point has y1=−1y_1 = -1. Read the signs off the numerators carefully.

The angle between two lines

How do you find the angle between two lines?

Only the directions matter. With direction ratios a1,b1,c1a_1, b_1, c_1 and a2,b2,c2a_2, b_2, c_2,

cos⁡θ=∣a1a2+b1b2+c1c2∣a12+b12+c12a22+b22+c22.\cos \theta = \frac{\lvert a_1a_2 + b_1b_2 + c_1c_2 \rvert}{\sqrt{a_1^{2} + b_1^{2} + c_1^{2}}\sqrt{a_2^{2} + b_2^{2} + c_2^{2}}}.

The lines are perpendicular when a1a2+b1b2+c1c2=0a_1a_2 + b_1b_2 + c_1c_2 = 0, and parallel when the ratios are proportional.

Example 3 · An angle

Find the angle between lines with direction ratios 1,1,01, 1, 0 and 1,0,11, 0, 1.

  1. cos⁡θ=122=12\cos \theta = \frac{1}{\sqrt{2}\sqrt{2}} = \frac{1}{2}.

  2. θ=60∘\theta = 60^\circ.

Answer
60∘60^\circ

Skew lines and the shortest distance

How do you find the shortest distance between two lines?

For r⃗=a⃗1+λb⃗1\vec{r} = \vec{a}_1 + \lambda\vec{b}_1 and r⃗=a⃗2+μb⃗2\vec{r} = \vec{a}_2 + \mu\vec{b}_2, the shortest distance runs along b⃗1×b⃗2\vec{b}_1 \times \vec{b}_2:

d=∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣∣b⃗1×b⃗2∣.d = \frac{\lvert (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) \rvert}{\lvert \vec{b}_1 \times \vec{b}_2 \rvert}.

If d=0d = 0 the lines intersect. For parallel lines r⃗=a⃗1+λb⃗\vec{r} = \vec{a}_1 + \lambda\vec{b} and r⃗=a⃗2+μb⃗\vec{r} = \vec{a}_2 + \mu\vec{b}, use d=∣b⃗×(a⃗2−a⃗1)∣∣b⃗∣d = \frac{\lvert \vec{b} \times (\vec{a}_2 - \vec{a}_1) \rvert}{\lvert \vec{b} \rvert}.

Example 4 · The two wires

Find the shortest distance between r⃗=(i^+2k^)+λ(i^+j^)\vec{r} = (\hat{i} + 2\hat{k}) + \lambda(\hat{i} + \hat{j}) and r⃗=(j^+k^)+μ(j^−k^)\vec{r} = (\hat{j} + \hat{k}) + \mu(\hat{j} - \hat{k}).

  1. b⃗1×b⃗2=(i^+j^)×(j^−k^)=−i^+j^+k^\vec{b}_1 \times \vec{b}_2 = (\hat{i} + \hat{j}) \times (\hat{j} - \hat{k}) = -\hat{i} + \hat{j} + \hat{k}, of magnitude 3\sqrt{3}.

  2. a⃗2−a⃗1=−i^+j^−k^\vec{a}_2 - \vec{a}_1 = -\hat{i} + \hat{j} - \hat{k}, and its dot product with −i^+j^+k^-\hat{i} + \hat{j} + \hat{k} is 1+1−1=11 + 1 - 1 = 1.

Answer
d=13d = \frac{1}{\sqrt{3}}

How CUET asks this chapter

  • Direction cosines: of a line through two points, of an axis, from direction ratios, or which triple can be direction cosines.
  • Equations: the line through a point with a direction, or through two points, in either form; whether a point lies on a line.
  • Angles: the angle between two lines, or the kk that makes them perpendicular or parallel.
  • Distance: between skew lines or parallel lines, and whether two lines meet.

Exam shortcuts

  • l2+m2+n2=1l^{2} + m^{2} + n^{2} = 1; the axes have direction cosines 1,0,01, 0, 0, 0,1,00, 1, 0, 0,0,10, 0, 1.
  • Angle: use only the direction ratios, never the points.
  • Lines meet exactly when (a⃗2−a⃗1)⋅(b⃗1×b⃗2)=0(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 0, for non-parallel lines.

Common mistakes

  • Taking the point's coordinates with the wrong sign from the Cartesian form.
  • Using the points in an angle calculation.
  • Stopping at the numerator of the distance formula.
  • Calling parallel lines skew: parallel lines lie in a plane; skew lines never do.

Quick revision

A line

  • r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda\vec{b}
  • x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}

Two lines

  • cos⁡θ\cos \theta from the direction ratios
  • Skew: d=∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣∣b⃗1×b⃗2∣d = \frac{\lvert (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) \rvert}{\lvert \vec{b}_1 \times \vec{b}_2 \rvert}

Last in Section B1, probability when something is already known: conditional probability and Bayes' theorem.

Next · 09Quick revision

Check your understanding

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

Direction cosines of a lineQ1 of 120Q1

Direction cosines of a line

28 questions

Question 1

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What are the direction cosines of the line joining P(1,2,−3)P(1, 2, -3) and Q(3,1,−1)Q(3, 1, -1), directed from PP to QQ?

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AnswerB · 23,−13,23\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}

  1. Direction ratios: 3−1,1−2,−1+3=2,−1,23 - 1, 1 - 2, -1 + 3 = 2, -1, 2, with 4+1+4=3\sqrt{4 + 1 + 4} = 3.

  2. Direction cosines: 23,−13,23\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}.

Worth noting

−23,13,−23-\frac{2}{3}, \frac{1}{3}, -\frac{2}{3} points from QQ to PP, the other way.

Question 2

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What are the direction cosines of the yy-axis?

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AnswerD · 0,1,00, 1, 0

  1. The yy-axis makes angles 90∘90^\circ, 0∘0^\circ, 90∘90^\circ with the three axes.

  2. Their cosines are 0,1,00, 1, 0.

Worth noting

1,0,01, 0, 0 is the xx-axis.

Question 3

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A line makes angles of 90∘90^\circ, 135∘135^\circ and 45∘45^\circ with the xx-, yy- and zz-axes. What are its direction cosines?

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AnswerC · 0,−12,120, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}

  1. Direction cosines are the cosines of the three angles.

  2. cos⁡90∘=0\cos 90^\circ = 0, cos⁡135∘=−12\cos 135^\circ = -\frac{1}{\sqrt{2}}, cos⁡45∘=12\cos 45^\circ = \frac{1}{\sqrt{2}}.

Worth noting

+12+\frac{1}{\sqrt{2}} for 135∘135^\circ misses that cosine is negative in the second quadrant.

Question 4

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What are the direction cosines of the line x4=y−1−4=z+27\frac{x}{4} = \frac{y - 1}{-4} = \frac{z + 2}{7}?

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AnswerD · 49,−49,79\frac{4}{9}, -\frac{4}{9}, \frac{7}{9}

  1. The direction ratios are the denominators, 4,−4,74, -4, 7, with length 16+16+49=9\sqrt{16 + 16 + 49} = 9.

  2. Direction cosines: 49,−49,79\frac{4}{9}, -\frac{4}{9}, \frac{7}{9}.

Worth noting

0,1,−20, 1, -2 comes from the point on the line, not its direction.

Question 5

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A line makes equal acute angles with the three coordinate axes. What are its direction cosines?

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AnswerA · 13,13,13\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}

  1. Equal angles give l=m=nl = m = n, and l2+m2+n2=1l^{2} + m^{2} + n^{2} = 1 then gives 3l2=13l^{2} = 1.

  2. The angles are acute, so l=m=n=13l = m = n = \frac{1}{\sqrt{3}}.

Worth noting

13\frac{1}{3} each makes l+m+n=1l + m + n = 1; it is the squares that add to 1.

Question 6

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A line has direction ratios 3,−1,43, -1, 4. Which of these is also a set of its direction ratios?

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AnswerB · −6,2,−8-6, 2, -8

  1. Direction ratios are fixed only up to a non-zero multiple.

  2. −6,2,−8-6, 2, -8 is −2-2 times 3,−1,43, -1, 4; none of the others is one multiple of it.

Worth noting

6,2,86, 2, 8 doubles the sizes but drops the signs.

Question 7

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A line has direction cosines 32,0,−12\frac{\sqrt{3}}{2}, 0, -\frac{1}{2}. What angles does it make with the xx-, yy- and zz-axes?

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AnswerD · π6,π2,2π3\frac{\pi}{6}, \frac{\pi}{2}, \frac{2\pi}{3}

  1. Each angle is the inverse cosine of its direction cosine: cos⁡−132=π6\cos^{-1}\frac{\sqrt{3}}{2} = \frac{\pi}{6} and cos⁡−10=π2\cos^{-1} 0 = \frac{\pi}{2}.

  2. cos⁡−1(−12)=2π3\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}, an obtuse angle, because the cosine is negative.

Worth noting

π3\frac{\pi}{3} for the zz-axis drops the minus sign of −12-\frac{1}{2}.

Question 8

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A hall's corner is the origin OO, with the floor edges along the xx- and yy-axes and the vertical edge along the zz-axis; lengths are in metres. A straight rod runs from OO to the point P(6,2,9)P(6, 2, 9).

What are the direction cosines of the rod, taken from OO to PP?

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AnswerB · 611,211,911\frac{6}{11}, \frac{2}{11}, \frac{9}{11}

  1. The rod is 36+4+81=121=11\sqrt{36 + 4 + 81} = \sqrt{121} = 11 long.

  2. Its direction cosines are 611,211,911\frac{6}{11}, \frac{2}{11}, \frac{9}{11}.

Worth noting

617,217,917\frac{6}{17}, \frac{2}{17}, \frac{9}{17} divides by 6+2+9=176 + 2 + 9 = 17, not by the length 11.

Question 9

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A line makes an angle of π4\frac{\pi}{4} with each of the xx- and yy-axes. What angle does it make with the zz-axis?

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AnswerD · π2\frac{\pi}{2}

  1. cos⁡2γ=1−cos⁡2π4−cos⁡2π4=1−12−12=0\cos^{2}\gamma = 1 - \cos^{2}\frac{\pi}{4} - \cos^{2}\frac{\pi}{4} = 1 - \frac{1}{2} - \frac{1}{2} = 0.

  2. cos⁡γ=0\cos \gamma = 0, so γ=π2\gamma = \frac{\pi}{2}.

Worth noting

π4\frac{\pi}{4} assumes the line makes the same angle with every axis.

Question 10

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A line has direction ratios −18,12,−4-18, 12, -4. What are its direction cosines?

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AnswerA · −911,611,−211-\frac{9}{11}, \frac{6}{11}, -\frac{2}{11}

  1. 324+144+16=484=22\sqrt{324 + 144 + 16} = \sqrt{484} = 22.

  2. −1822,1222,−422=−911,611,−211-\frac{18}{22}, \frac{12}{22}, -\frac{4}{22} = -\frac{9}{11}, \frac{6}{11}, -\frac{2}{11}.

Worth noting

−9,6,−2-9, 6, -2 only halves the ratios; they are still ratios, not cosines.

Question 11

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Read the two statements and choose the correct option.

I. A line has infinitely many sets of direction ratios but only two sets of direction cosines.

II. If l,m,nl, m, n are direction cosines, then l+m+n=1l + m + n = 1.

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AnswerA · Only I

  1. I: any non-zero multiple of the ratios is again a set of ratios; the cosines are fixed up to the sign of the direction. True.

  2. II: it is l2+m2+n2=1l^{2} + m^{2} + n^{2} = 1; for the xx-axis reversed, −1+0+0≠1-1 + 0 + 0 \ne 1. False.

Question 12

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A line makes angles α\alpha, β\beta, γ\gamma with the coordinate axes. What is sin⁡2α+sin⁡2β+sin⁡2γ\sin^{2}\alpha + \sin^{2}\beta + \sin^{2}\gamma?

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AnswerC · 22

  1. sin⁡2θ=1−cos⁡2θ\sin^{2}\theta = 1 - \cos^{2}\theta, so the sum is 3−(cos⁡2α+cos⁡2β+cos⁡2γ)3 - (\cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma).

  2. 3−1=23 - 1 = 2.

Worth noting

1 is the sum of the squared cosines, not of the squared sines.

Question 13

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A line has direction cosines 27\frac{2}{7}, 37\frac{3}{7} and nn, with n>0n > 0. What is nn?

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AnswerD · 67\frac{6}{7}

  1. l2+m2+n2=1l^{2} + m^{2} + n^{2} = 1: n2=1−449−949=3649n^{2} = 1 - \frac{4}{49} - \frac{9}{49} = \frac{36}{49}.

  2. n=67n = \frac{6}{7}.

Worth noting

27\frac{2}{7} uses l+m+n=1l + m + n = 1; it is the squares that add to 1.

Question 14

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A line has direction ratios 1,−2,21, -2, 2. What angle does it make with the xx-axis?

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AnswerA · cos⁡−113\cos^{-1}\frac{1}{3}

  1. 1+4+4=3\sqrt{1 + 4 + 4} = 3, so l=13l = \frac{1}{3}.

  2. The angle with the xx-axis is cos⁡−113\cos^{-1}\frac{1}{3}.

Worth noting

cos⁡−11\cos^{-1} 1 takes the ratio 1 as the cosine without dividing by the length 3.

Question 15

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

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AnswerD · 12,−12,12\frac{1}{2}, -\frac{1}{2}, \frac{1}{\sqrt{2}}

  1. Direction cosines must satisfy l2+m2+n2=1l^{2} + m^{2} + n^{2} = 1.

  2. For 12,−12,12\frac{1}{2}, -\frac{1}{2}, \frac{1}{\sqrt{2}}: 14+14+12=1\frac{1}{4} + \frac{1}{4} + \frac{1}{2} = 1; the other triples give 38\frac{3}{8}, 54\frac{5}{4} and 23\frac{2}{3}.

Worth noting

12,14,14\frac{1}{2}, \frac{1}{4}, \frac{1}{4} adds to 1, but its squares add to 38\frac{3}{8}.

Question 16

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For which kk are the points A(2,−1,1)A(2, -1, 1), B(4,2,3)B(4, 2, 3) and C(8,k,7)C(8, k, 7) collinear?

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AnswerB · 88

  1. The direction ratios of ABAB are 2,3,22, 3, 2; those of ACAC are 6,k+1,66, k + 1, 6.

  2. Collinear means proportional, here with factor 62=3\frac{6}{2} = 3: k+1=9k + 1 = 9, so k=8k = 8.

Worth noting

9 is k+1k + 1, the ratio of ACAC; kk itself is 1 less.

Question 17

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A hall's corner is the origin OO, with the floor edges along the xx- and yy-axes and the vertical edge along the zz-axis; lengths are in metres. A straight rod runs from OO to the point P(6,2,9)P(6, 2, 9).

A bead on the rod is 4.5 m above the floor. How far is it from OO, measured along the rod?

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AnswerC · 5.5 m

  1. A point of the rod at distance ss from OO is at height s×911s \times \frac{9}{11}.

  2. s×911=4.5s \times \frac{9}{11} = 4.5 gives s=5.5s = 5.5 m.

Worth noting

4.5 m is the height; the rod slants, so the distance along it is longer.

Question 18

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A line is perpendicular to the xx-axis, makes an angle of 5π6\frac{5\pi}{6} with the yy-axis and an acute angle with the zz-axis. What are its direction cosines?

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AnswerC · 0,−32,120, -\frac{\sqrt{3}}{2}, \frac{1}{2}

  1. Perpendicular to the xx-axis: l=cos⁡π2=0l = \cos\frac{\pi}{2} = 0; and m=cos⁡5π6=−32m = \cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}.

  2. n2=1−0−34=14n^{2} = 1 - 0 - \frac{3}{4} = \frac{1}{4}, and the acute angle gives n=12n = \frac{1}{2}.

Worth noting

+32+\frac{\sqrt{3}}{2} misses that cos⁡5π6\cos\frac{5\pi}{6} is negative.

Question 19

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A hall's corner is the origin OO, with the floor edges along the xx- and yy-axes and the vertical edge along the zz-axis; lengths are in metres. A straight rod runs from OO to the point P(6,2,9)P(6, 2, 9).

What angle does the rod make with the vertical edge?

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AnswerA · cos⁡−1(911)\cos^{-1}{\left(\frac{9}{11} \right)}

  1. The vertical edge is the zz-axis, and the rod's direction cosine for it is n=911n = \frac{9}{11}.

  2. The angle is cos⁡−1911\cos^{-1}\frac{9}{11}.

Worth noting

sin⁡−1911\sin^{-1}\frac{9}{11} treats 911\frac{9}{11} as a sine; a direction cosine is the cosine of the angle.

Question 20

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Assertion (A): No line makes angles π6\frac{\pi}{6}, π4\frac{\pi}{4} and π3\frac{\pi}{3} with the xx-, yy- and zz-axes.

Reason (R): The direction cosines l,m,nl, m, n of every line satisfy l2+m2+n2=1l^{2} + m^{2} + n^{2} = 1.

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

  1. cos⁡2π6+cos⁡2π4+cos⁡2π3=34+12+14=32≠1\cos^{2}\frac{\pi}{6} + \cos^{2}\frac{\pi}{4} + \cos^{2}\frac{\pi}{3} = \frac{3}{4} + \frac{1}{2} + \frac{1}{4} = \frac{3}{2} \ne 1, so A is true.

  2. R is true, and it is exactly the rule these angles break: R explains A.

Question 21

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A line LL has direction ratios 6,−2,−36, -2, -3. Which of the following statements are true?

I. The direction cosines of LL are 67,−27,−37\frac{6}{7}, -\frac{2}{7}, -\frac{3}{7}, or these three with their signs reversed.

II. LL is perpendicular to the line with direction ratios 1,3,01, 3, 0.

III. LL is parallel to the line with direction ratios 12,4,612, 4, 6.

IV. The direction cosines of LL add up to 1.

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AnswerA · I and II only

  1. 36+4+9=7\sqrt{36 + 4 + 9} = 7, so I is true; 6(1)+(−2)(3)+(−3)(0)=06(1) + (-2)(3) + (-3)(0) = 0, so II is true.

  2. 12,4,612, 4, 6 is not a multiple of 6,−2,−36, -2, -3, so III is false; 6−2−37=17\frac{6 - 2 - 3}{7} = \frac{1}{7}, so IV is false.

Worth noting

III fails on the signs: twice 6,−2,−36, -2, -3 is 12,−4,−612, -4, -6.

Question 22

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A line has direction ratios 2,k,62, k, 6 with k>0k > 0, and the cosine of the angle it makes with the xx-axis is 27\frac{2}{7}. What is kk?

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AnswerA · 33

  1. l=24+k2+36=27l = \frac{2}{\sqrt{4 + k^{2} + 36}} = \frac{2}{7}, so 40+k2=7\sqrt{40 + k^{2}} = 7.

  2. k2=49−40=9k^{2} = 49 - 40 = 9, and k>0k > 0 gives k=3k = 3.

Worth noting

9 is k2k^{2}; kk is its square root.

Question 23

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A line makes angles α\alpha, β\beta, γ\gamma with the coordinate axes. What is cos⁡2α+cos⁡2β+cos⁡2γ\cos 2\alpha + \cos 2\beta + \cos 2\gamma?

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AnswerA · −1-1

  1. cos⁡2θ=2cos⁡2θ−1\cos 2\theta = 2\cos^{2}\theta - 1, so the sum is 2(cos⁡2α+cos⁡2β+cos⁡2γ)−32(\cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma) - 3.

  2. 2×1−3=−12 \times 1 - 3 = -1.

Worth noting

1 adds the three cos⁡2\cos^{2} terms and forgets the −1-1 in each cos⁡2θ\cos 2\theta.

Question 24

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A line makes angles α\alpha, β\beta, γ\gamma, δ\delta with the four body diagonals of a cube. What is cos⁡2α+cos⁡2β+cos⁡2γ+cos⁡2δ\cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma + \cos^{2}\delta?

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AnswerC · 43\frac{4}{3}

  1. With the edges along the axes, the body diagonals have direction ratios 1,1,11, 1, 1, −1,1,1-1, 1, 1, 1,−1,11, -1, 1 and 1,1,−11, 1, -1, each of length 3\sqrt{3}.

  2. For a line with direction cosines l,m,nl, m, n: cos⁡α=∣l+m+n∣3\cos \alpha = \frac{\lvert l + m + n \rvert}{\sqrt{3}}, and so on for the others.

  3. The four squares add to 4(l2+m2+n2)3=43\frac{4(l^{2} + m^{2} + n^{2})}{3} = \frac{4}{3}.

Worth noting

1 is the identity for the three axes, not for the four diagonals.

Question 25

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A hall's corner is the origin OO, with the floor edges along the xx- and yy-axes and the vertical edge along the zz-axis; lengths are in metres. A straight rod runs from OO to the point P(6,2,9)P(6, 2, 9).

A second rod runs from PP to a point RR on the xx-axis and is perpendicular to the first rod. How far is RR from OO?

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AnswerD · 1216\frac{121}{6} m

  1. With R(r,0,0)R(r, 0, 0), PRPR has direction ratios r−6,−2,−9r - 6, -2, -9, and it must be perpendicular to 6,2,96, 2, 9.

  2. 6(r−6)−4−81=06(r - 6) - 4 - 81 = 0, so 6r=1216r = 121 and OR=1216OR = \frac{121}{6} m.

Worth noting

6 m stops at the xx-coordinate of PP; then PRPR is not perpendicular to OPOP.

Question 26

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A point PP is at a distance of 8 from the origin OO. The line OPOP makes angles 2π3\frac{2\pi}{3} and 3π4\frac{3\pi}{4} with the xx- and yy-axes and an acute angle with the zz-axis. What are the coordinates of PP?

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AnswerD · (−4,−42,4)(-4, -4\sqrt{2}, 4)

  1. l=cos⁡2π3=−12l = \cos\frac{2\pi}{3} = -\frac{1}{2} and m=cos⁡3π4=−12m = \cos\frac{3\pi}{4} = -\frac{1}{\sqrt{2}}, so n2=1−14−12=14n^{2} = 1 - \frac{1}{4} - \frac{1}{2} = \frac{1}{4}; the acute angle gives n=12n = \frac{1}{2}.

  2. P=(8l,8m,8n)=(−4,−42,4)P = (8l, 8m, 8n) = (-4, -4\sqrt{2}, 4).

Worth noting

(4,42,4)(4, 4\sqrt{2}, 4) misses that the cosines of 2π3\frac{2\pi}{3} and 3π4\frac{3\pi}{4} are negative.

Question 27

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Triangle ABCABC has vertices A(2,−1,3)A(2, -1, 3), B(5,1,−2)B(5, 1, -2) and C(1,5,−8)C(1, 5, -8). What are the direction cosines of its median, taken from AA towards the midpoint of BCBC?

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AnswerB · 19,49,−89\frac{1}{9}, \frac{4}{9}, -\frac{8}{9}

  1. The midpoint of BCBC is D(5+12,1+52,−2−82)=(3,3,−5)D\left(\frac{5 + 1}{2}, \frac{1 + 5}{2}, \frac{-2 - 8}{2}\right) = (3, 3, -5).

  2. ADAD has direction ratios 1,4,−81, 4, -8 and length 1+16+64=9\sqrt{1 + 16 + 64} = 9: direction cosines 19,49,−89\frac{1}{9}, \frac{4}{9}, -\frac{8}{9}.

Worth noting

−19,−49,89-\frac{1}{9}, -\frac{4}{9}, \frac{8}{9} points from the midpoint back to AA.

Question 28

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A line makes the same angle θ\theta with the xx- and zz-axes and an angle β\beta with the yy-axis, where sin⁡2β=3sin⁡2θ\sin^{2}\beta = 3\sin^{2}\theta. What is cos⁡2θ\cos^{2}\theta?

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AnswerC · 35\frac{3}{5}

  1. cos⁡2β=1−sin⁡2β=1−3(1−cos⁡2θ)=3cos⁡2θ−2\cos^{2}\beta = 1 - \sin^{2}\beta = 1 - 3(1 - \cos^{2}\theta) = 3\cos^{2}\theta - 2.

  2. The squared direction cosines add to 1: 2cos⁡2θ+cos⁡2β=5cos⁡2θ−2=12\cos^{2}\theta + \cos^{2}\beta = 5\cos^{2}\theta - 2 = 1, so cos⁡2θ=35\cos^{2}\theta = \frac{3}{5}.

Worth noting

34\frac{3}{4} counts θ\theta once, though the line makes that angle with two axes.

The equation of a line

31 questions

Question 29

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What is the Cartesian equation of the line through (5,−2,4)(5, -2, 4) parallel to 3i^+2j^−8k^3\hat{i} + 2\hat{j} - 8\hat{k}?

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AnswerC · x−53=y+22=z−4−8\frac{x - 5}{3} = \frac{y + 2}{2} = \frac{z - 4}{-8}

  1. Through (x1,y1,z1)(x_1, y_1, z_1) with direction ratios a,b,ca, b, c: x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}.

  2. Here x−53=y+22=z−4−8\frac{x - 5}{3} = \frac{y + 2}{2} = \frac{z - 4}{-8}.

Worth noting

Writing x+5x + 5 uses the point's coordinates with the wrong sign.

Question 30

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What is the vector equation of the line through the point (2,−1,4)(2, -1, 4) parallel to i^+j^−2k^\hat{i} + \hat{j} - 2\hat{k}?

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AnswerB · r⃗=(2i^−j^+4k^)+λ(i^+j^−2k^)\vec{r} = (2\hat{i} - \hat{j} + 4\hat{k}) + \lambda(\hat{i} + \hat{j} - 2\hat{k})

  1. A line through a⃗\vec{a} parallel to b⃗\vec{b} is r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda\vec{b}.

  2. Here a⃗=2i^−j^+4k^\vec{a} = 2\hat{i} - \hat{j} + 4\hat{k} and b⃗=i^+j^−2k^\vec{b} = \hat{i} + \hat{j} - 2\hat{k}.

Worth noting

Swapping a⃗\vec{a} and b⃗\vec{b} gives a line through the wrong point, in the wrong direction.

Question 31

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What is the vector equation of the line through the origin and the point (2,−1,3)(2, -1, 3)?

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AnswerA · r⃗=λ(2i^−j^+3k^)\vec{r} = \lambda(2\hat{i} - \hat{j} + 3\hat{k})

  1. The line passes through 0⃗\vec{0} with direction 2i^−j^+3k^−0⃗2\hat{i} - \hat{j} + 3\hat{k} - \vec{0}.

  2. r⃗=0⃗+λ(2i^−j^+3k^)\vec{r} = \vec{0} + \lambda(2\hat{i} - \hat{j} + 3\hat{k}).

Worth noting

r⃗=2i^−j^+3k^\vec{r} = 2\hat{i} - \hat{j} + 3\hat{k} has no parameter, so it is a single point.

Question 32

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What is the Cartesian form of the line r⃗=(2i^−j^+4k^)+λ(i^+2j^−k^)\vec{r} = (2\hat{i} - \hat{j} + 4\hat{k}) + \lambda(\hat{i} + 2\hat{j} - \hat{k})?

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AnswerD · x−21=y+12=z−4−1\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 4}{-1}

  1. The point is (2,−1,4)(2, -1, 4) and the direction ratios are 1,2,−11, 2, -1.

  2. x−21=y+12=z−4−1\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 4}{-1}.

Worth noting

Swapping the point and the direction puts 1,2,−11, 2, -1 in the numerators.

Question 33

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A line passes through A(2,−1,3)A(2, -1, 3) and B(4,3,−1)B(4, 3, -1).

Which of these are direction ratios of the line?

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AnswerA · 2,4,−42, 4, -4

  1. Subtract the coordinates of AA from those of BB.

  2. 4−2,3−(−1),−1−3=2,4,−44 - 2, 3 - (-1), -1 - 3 = 2, 4, -4.

Worth noting

2,2,−42, 2, -4 takes 3−(−1)3 - (-1) as 2.

Question 34

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Which pair of equations describes the line through (3,−1,2)(3, -1, 2) parallel to the yy-axis?

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AnswerC · x=3x = 3 and z=2z = 2

  1. Parallel to the yy-axis means direction ratios 0,1,00, 1, 0: x−30=y+11=z−20\frac{x - 3}{0} = \frac{y + 1}{1} = \frac{z - 2}{0}.

  2. The zero denominators mean xx and zz never change along the line: x=3x = 3 and z=2z = 2, with yy free.

Worth noting

y=−1y = -1 and z=2z = 2 fixes yy, the one coordinate that changes along a line parallel to the yy-axis.

Question 35

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What are the direction ratios of the line x−23=y+14\frac{x - 2}{3} = \frac{y + 1}{4}, z=5z = 5?

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AnswerD · 3,4,03, 4, 0

  1. z=5z = 5 at every point, so zz does not change along the line: its direction ratio is 0.

  2. The other two are the denominators, so the ratios are 3,4,03, 4, 0.

Worth noting

5 is the fixed value of zz, not a direction ratio.

Question 36

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A point P(5,1,4)P(5, 1, 4) and the line L:x−12=y2=z−3−1L: \frac{x - 1}{2} = \frac{y}{2} = \frac{z - 3}{-1} are given. MM is the point of LL with parameter λ\lambda, and FF is the foot of the perpendicular from PP to LL.

What are the coordinates of MM?

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AnswerA · (1+2λ,2λ,3−λ)(1 + 2\lambda, 2\lambda, 3 - \lambda)

  1. Set each fraction equal to λ\lambda: x−1=2λx - 1 = 2\lambda, y=2λy = 2\lambda, z−3=−λz - 3 = -\lambda.

  2. So M=(1+2λ,2λ,3−λ)M = (1 + 2\lambda, 2\lambda, 3 - \lambda).

Worth noting

(2+λ,2,−1+3λ)(2 + \lambda, 2, -1 + 3\lambda) swaps the point (1,0,3)(1, 0, 3) and the direction 2,2,−12, 2, -1.

Question 37

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What is the vector equation of the line through A(3,−2,1)A(3, -2, 1) and B(1,0,4)B(1, 0, 4)?

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AnswerC · r⃗=(3i^−2j^+k^)+λ(−2i^+2j^+3k^)\vec{r} = (3\hat{i} - 2\hat{j} + \hat{k}) + \lambda(-2\hat{i} + 2\hat{j} + 3\hat{k})

  1. Its direction is b⃗−a⃗=(i^+4k^)−(3i^−2j^+k^)=−2i^+2j^+3k^\vec{b} - \vec{a} = (\hat{i} + 4\hat{k}) - (3\hat{i} - 2\hat{j} + \hat{k}) = -2\hat{i} + 2\hat{j} + 3\hat{k}.

  2. Through AA: r⃗=(3i^−2j^+k^)+λ(−2i^+2j^+3k^)\vec{r} = (3\hat{i} - 2\hat{j} + \hat{k}) + \lambda(-2\hat{i} + 2\hat{j} + 3\hat{k}).

Worth noting

i^+4k^\hat{i} + 4\hat{k} is the position vector of BB, not the direction from AA to BB.

Question 38

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Match each line in List I with a set of its direction ratios in List II.

List IList II
A. x−12=y+3−1=z4\frac{x - 1}{2} = \frac{y + 3}{-1} = \frac{z}{4}I. 3,−2,13, -2, 1
B. r⃗=(i^−k^)+λ(3j^+k^)\vec{r} = (\hat{i} - \hat{k}) + \lambda(3\hat{j} + \hat{k})II. 2,−1,42, -1, 4
C. The line through (1,2,0)(1, 2, 0) and (4,0,1)(4, 0, 1)III. −2,3,1-2, 3, 1
D. 1−x2=y3=z+11\frac{1 - x}{2} = \frac{y}{3} = \frac{z + 1}{1}IV. 0,3,10, 3, 1
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AnswerC · A-II, B-IV, C-I, D-III

  1. A: the denominators, 2,−1,42, -1, 4; B: the vector multiplying λ\lambda, 0,3,10, 3, 1.

  2. C: 4−1,0−2,1−0=3,−2,14 - 1, 0 - 2, 1 - 0 = 3, -2, 1; D: 1−x2=x−1−2\frac{1 - x}{2} = \frac{x - 1}{-2}, so −2,3,1-2, 3, 1.

Question 39

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What is the Cartesian equation of the line through (3,−2,−5)(3, -2, -5) and (3,−2,6)(3, -2, 6)?

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AnswerD · x−30=y+20=z+511\frac{x - 3}{0} = \frac{y + 2}{0} = \frac{z + 5}{11}

  1. Direction ratios B−A=0,0,11B - A = 0, 0, 11.

  2. So x−30=y+20=z+511\frac{x - 3}{0} = \frac{y + 2}{0} = \frac{z + 5}{11}; the zero denominators mean x=3x = 3, y=−2y = -2 all along the line.

Worth noting

Adding the points' coordinates gives no direction; the direction is their difference.

Question 40

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Which point lies on the line x−12=y−2−1=z+13\frac{x - 1}{2} = \frac{y - 2}{-1} = \frac{z + 1}{3}?

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AnswerC · (5,0,5)(5, 0, 5)

  1. A point on the line is (1+2λ,2−λ,−1+3λ)(1 + 2\lambda, 2 - \lambda, -1 + 3\lambda).

  2. λ=2\lambda = 2 gives (5,0,5)(5, 0, 5); the others fit no single λ\lambda.

Worth noting

(2,−1,3)(2, -1, 3) is the direction, not a point of the line.

Question 41

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What is the vector form of the line x−53=y+47=z−62\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2}?

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AnswerB · r⃗=(5i^−4j^+6k^)+λ(3i^+7j^+2k^)\vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda(3\hat{i} + 7\hat{j} + 2\hat{k})

  1. The line passes through (5,−4,6)(5, -4, 6): note y+4=y−(−4)y + 4 = y - (-4).

  2. Its direction is 3i^+7j^+2k^3\hat{i} + 7\hat{j} + 2\hat{k}.

Worth noting

y+4y + 4 means y1=−4y_1 = -4, not 4.

Question 42

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Which point lies on the line r⃗=(i^−j^+2k^)+λ(2i^+j^−k^)\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \lambda(2\hat{i} + \hat{j} - \hat{k})?

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AnswerB · (5,1,0)(5, 1, 0)

  1. A point on the line is (1+2λ,−1+λ,2−λ)(1 + 2\lambda, -1 + \lambda, 2 - \lambda).

  2. λ=2\lambda = 2 gives (5,1,0)(5, 1, 0); each other option fails one coordinate.

Worth noting

(2,1,−1)(2, 1, -1) is the direction vector, not a point of the line.

Question 43

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The point (p,4,−3)(p, 4, -3) lies on the line x−12=y−21=z+1−1\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z + 1}{-1}. What is pp?

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AnswerB · 55

  1. From yy: 4−21=2\frac{4 - 2}{1} = 2; zz agrees, −3+1−1=2\frac{-3 + 1}{-1} = 2.

  2. p−12=2\frac{p - 1}{2} = 2, so p=5p = 5.

Worth noting

9 uses y=4y = 4 itself as the common ratio instead of 4−21\frac{4 - 2}{1}.

Question 44

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What is the Cartesian equation of the line through (1,−2,3)(1, -2, 3) parallel to the line x−42=y+1−3=z−26\frac{x - 4}{2} = \frac{y + 1}{-3} = \frac{z - 2}{6}?

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AnswerC · x−12=y+2−3=z−36\frac{x - 1}{2} = \frac{y + 2}{-3} = \frac{z - 3}{6}

  1. A parallel line has the same direction ratios, 2,−3,62, -3, 6.

  2. Through (1,−2,3)(1, -2, 3): x−12=y+2−3=z−36\frac{x - 1}{2} = \frac{y + 2}{-3} = \frac{z - 3}{6}.

Worth noting

Taking 4,−1,24, -1, 2 as the direction uses the given line's point, not its direction.

Question 45

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A line passes through A(2,−1,3)A(2, -1, 3) and B(4,3,−1)B(4, 3, -1).

What are its direction cosines, taken in the direction from AA to BB?

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AnswerD · 13,23,−23\frac{1}{3}, \frac{2}{3}, -\frac{2}{3}

  1. 4+16+16=6\sqrt{4 + 16 + 16} = 6.

  2. 26,46,−46=13,23,−23\frac{2}{6}, \frac{4}{6}, -\frac{4}{6} = \frac{1}{3}, \frac{2}{3}, -\frac{2}{3}.

Worth noting

15,25,−25\frac{1}{5}, \frac{2}{5}, -\frac{2}{5} divides by 2+4+4=102 + 4 + 4 = 10, not by the length 6.

Question 46

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A line passes through A(2,−1,3)A(2, -1, 3) and B(4,3,−1)B(4, 3, -1).

Which point lies on the line ABAB?

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AnswerD · (6,7,−5)(6, 7, -5)

  1. Points of the line: (2+2t,−1+4t,3−4t)(2 + 2t, -1 + 4t, 3 - 4t).

  2. t=2t = 2 gives (6,7,−5)(6, 7, -5); each other option fails one coordinate.

Worth noting

(6,7,5)(6, 7, 5) gets xx and yy right for t=2t = 2 but adds 4t4t to zz instead of taking it away.

Question 47

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A line passes through A(2,−1,3)A(2, -1, 3) and B(4,3,−1)B(4, 3, -1).

What is the Cartesian equation of the line?

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AnswerA · x−21=y+12=z−3−2\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 3}{-2}

  1. Through A(2,−1,3)A(2, -1, 3), with ratios 2,4,−42, 4, -4 or, halved, 1,2,−21, 2, -2.

  2. x−21=y+12=z−3−2\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 3}{-2}.

Worth noting

Denominators 4,3,−14, 3, -1 use the coordinates of BB as the direction.

Question 48

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Which point of the line x−22=y+3−2=z−11\frac{x - 2}{2} = \frac{y + 3}{-2} = \frac{z - 1}{1} is at a distance of 6 from (2,−3,1)(2, -3, 1) and has z>1z > 1?

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AnswerB · (6,−7,3)(6, -7, 3)

  1. A point of the line is (2+2λ,−3−2λ,1+λ)(2 + 2\lambda, -3 - 2\lambda, 1 + \lambda), at distance 4+4+1 ∣λ∣=3∣λ∣\sqrt{4 + 4 + 1}\,\lvert \lambda \rvert = 3\lvert \lambda \rvert from (2,−3,1)(2, -3, 1).

  2. 3∣λ∣=63\lvert \lambda \rvert = 6 gives λ=±2\lambda = \pm 2, and z>1z > 1 needs λ=2\lambda = 2: the point (6,−7,3)(6, -7, 3).

Worth noting

(14,−15,7)(14, -15, 7) takes λ=6\lambda = 6, but each unit of λ\lambda moves 3 units along the line, not 1.

Question 49

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At which point does the line through A(4,−1,3)A(4, -1, 3) and B(2,3,−1)B(2, 3, -1) have xx-coordinate 0?

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AnswerA · (0,7,−5)(0, 7, -5)

  1. The direction B−AB - A is −2,4,−4-2, 4, -4, so the points of the line are (4−2t,−1+4t,3−4t)(4 - 2t, -1 + 4t, 3 - 4t).

  2. 4−2t=04 - 2t = 0 gives t=2t = 2: the point is (0,7,−5)(0, 7, -5).

Worth noting

(0,3,−1)(0, 3, -1) only replaces the xx-coordinate of BB by 0; that point is not on the line.

Question 50

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Read the two statements and choose the correct option.

I. The line r⃗=(i^+2j^)+λ(2i^−k^)\vec{r} = (\hat{i} + 2\hat{j}) + \lambda(2\hat{i} - \hat{k}) passes through (5,2,2)(5, 2, 2).

II. The Cartesian form of r⃗=(2i^−k^)+λ(j^+3k^)\vec{r} = (2\hat{i} - \hat{k}) + \lambda(\hat{j} + 3\hat{k}) is x−20=y1=z+13\frac{x - 2}{0} = \frac{y}{1} = \frac{z + 1}{3}.

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AnswerB · Only II

  1. I: x=1+2λ=5x = 1 + 2\lambda = 5 needs λ=2\lambda = 2, which gives z=−2z = -2, not 2. False.

  2. II: the point is (2,0,−1)(2, 0, -1) and the direction ratios are 0,1,30, 1, 3. True.

Question 51

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A point P(5,1,4)P(5, 1, 4) and the line L:x−12=y2=z−3−1L: \frac{x - 1}{2} = \frac{y}{2} = \frac{z - 3}{-1} are given. MM is the point of LL with parameter λ\lambda, and FF is the foot of the perpendicular from PP to LL.

For which λ\lambda is PMPM perpendicular to LL?

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AnswerB · 11

  1. PMPM has direction ratios 2λ−4,2λ−1,−1−λ2\lambda - 4, 2\lambda - 1, -1 - \lambda, and it must be perpendicular to 2,2,−12, 2, -1.

  2. 2(2λ−4)+2(2λ−1)−(−1−λ)=9λ−9=02(2\lambda - 4) + 2(2\lambda - 1) - (-1 - \lambda) = 9\lambda - 9 = 0, so λ=1\lambda = 1.

Worth noting

9 is (p⃗−a⃗)⋅b⃗(\vec{p} - \vec{a}) \cdot \vec{b}, which still has to be divided by ∣b⃗∣2=9\lvert \vec{b} \rvert^{2} = 9.

Question 52

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A point P(5,1,4)P(5, 1, 4) and the line L:x−12=y2=z−3−1L: \frac{x - 1}{2} = \frac{y}{2} = \frac{z - 3}{-1} are given. MM is the point of LL with parameter λ\lambda, and FF is the foot of the perpendicular from PP to LL.

What are the coordinates of FF?

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AnswerC · (3,2,2)(3, 2, 2)

  1. Put λ=1\lambda = 1 in MM: F=(1+2,2,3−1)=(3,2,2)F = (1 + 2, 2, 3 - 1) = (3, 2, 2).

  2. Check: PFPF has direction ratios −2,1,−2-2, 1, -2, and 2(−2)+2(1)−1(−2)=02(-2) + 2(1) - 1(-2) = 0.

Worth noting

(19,18,−6)(19, 18, -6) puts λ=9\lambda = 9, the dot product before it is divided by ∣b⃗∣2\lvert \vec{b} \rvert^{2}.

Question 53

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For the line L:x−36=y+1−2=z3L: \frac{x - 3}{6} = \frac{y + 1}{-2} = \frac{z}{3}, which of the following statements are true?

I. LL passes through (−3,1,−3)(-3, 1, -3).

II. The vector form of LL is r⃗=(3i^−j^)+λ(6i^−2j^+3k^)\vec{r} = (3\hat{i} - \hat{j}) + \lambda(6\hat{i} - 2\hat{j} + 3\hat{k}).

III. LL is parallel to 12i^+4j^+6k^12\hat{i} + 4\hat{j} + 6\hat{k}.

IV. The point of LL with z=6z = 6 is (15,−5,6)(15, -5, 6).

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AnswerA · I, II and IV only

  1. λ=−1\lambda = -1 gives (−3,1,−3)(-3, 1, -3), so I is true; II reads the point (3,−1,0)(3, -1, 0) and the ratios 6,−2,36, -2, 3, so it is true.

  2. 2(6i^−2j^+3k^)=12i^−4j^+6k^2(6\hat{i} - 2\hat{j} + 3\hat{k}) = 12\hat{i} - 4\hat{j} + 6\hat{k}, so III is false; z=3λ=6z = 3\lambda = 6 gives λ=2\lambda = 2 and (15,−5,6)(15, -5, 6), so IV is true.

Worth noting

Leaving out IV misses that λ=2\lambda = 2 gives exactly (15,−5,6)(15, -5, 6).

Question 54

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Which is the Cartesian equation of the line through (4,−2,1)(4, -2, 1) that makes equal acute angles with the coordinate axes?

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AnswerD · x−41=y+21=z−11\frac{x - 4}{1} = \frac{y + 2}{1} = \frac{z - 1}{1}

  1. Equal acute angles give direction cosines 13,13,13\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, so the direction ratios are 1,1,11, 1, 1.

  2. Through (4,−2,1)(4, -2, 1): x−41=y+21=z−11\frac{x - 4}{1} = \frac{y + 2}{1} = \frac{z - 1}{1}.

Worth noting

x−44=y+2−2=z−11\frac{x - 4}{4} = \frac{y + 2}{-2} = \frac{z - 1}{1} uses the point's coordinates as the direction.

Question 55

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What are the direction ratios of the line x−21=2y+14=3−z2\frac{x - 2}{1} = \frac{2y + 1}{4} = \frac{3 - z}{2}?

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AnswerC · 1,2,−21, 2, -2

  1. Make each coefficient of xx, yy, zz equal to 1: 2y+14=y+122\frac{2y + 1}{4} = \frac{y + \frac{1}{2}}{2} and 3−z2=z−3−2\frac{3 - z}{2} = \frac{z - 3}{-2}.

  2. So x−21=y+122=z−3−2\frac{x - 2}{1} = \frac{y + \frac{1}{2}}{2} = \frac{z - 3}{-2}: ratios 1,2,−21, 2, -2.

Worth noting

1,4,21, 4, 2 reads the denominators as they stand, before yy and zz have coefficient 1.

Question 56

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Which is the vector equation of the line through (−1,2,1)(-1, 2, 1) perpendicular to both x−31=y+12=z−2−1\frac{x - 3}{1} = \frac{y + 1}{2} = \frac{z - 2}{-1} and x2=y−41=z+11\frac{x}{2} = \frac{y - 4}{1} = \frac{z + 1}{1}?

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AnswerA · r⃗=(−i^+2j^+k^)+λ(i^−j^−k^)\vec{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} - \hat{k})

  1. A direction perpendicular to both is b⃗1×b⃗2=(i^+2j^−k^)×(2i^+j^+k^)=3i^−3j^−3k^\vec{b}_1 \times \vec{b}_2 = (\hat{i} + 2\hat{j} - \hat{k}) \times (2\hat{i} + \hat{j} + \hat{k}) = 3\hat{i} - 3\hat{j} - 3\hat{k}, or i^−j^−k^\hat{i} - \hat{j} - \hat{k}.

  2. Through (−1,2,1)(-1, 2, 1): r⃗=(−i^+2j^+k^)+λ(i^−j^−k^)\vec{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} - \hat{k}).

Worth noting

i^+j^−k^\hat{i} + \hat{j} - \hat{k} forgets the minus sign in the j^\hat{j} term of the cross product.

Question 57

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Which of these equations describes the same line as x−22=y−13=z+3−1\frac{x - 2}{2} = \frac{y - 1}{3} = \frac{z + 3}{-1}?

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AnswerB · x−6−4=y−7−6=z+52\frac{x - 6}{-4} = \frac{y - 7}{-6} = \frac{z + 5}{2}

  1. Two equations give the same line when the directions are parallel and a point is shared: −4,−6,2=−2(2,3,−1)-4, -6, 2 = -2(2, 3, -1).

  2. And (6,7,−5)(6, 7, -5) is on the given line: 6−22=7−13=−5+3−1=2\frac{6 - 2}{2} = \frac{7 - 1}{3} = \frac{-5 + 3}{-1} = 2.

Worth noting

x+22=y+13=z−3−1\frac{x + 2}{2} = \frac{y + 1}{3} = \frac{z - 3}{-1} has the right direction but passes through (−2,−1,3)(-2, -1, 3), which is not on the line.

Question 58

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A point P(5,1,4)P(5, 1, 4) and the line L:x−12=y2=z−3−1L: \frac{x - 1}{2} = \frac{y}{2} = \frac{z - 3}{-1} are given. MM is the point of LL with parameter λ\lambda, and FF is the foot of the perpendicular from PP to LL.

What is the image of PP in LL?

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AnswerD · (1,3,0)(1, 3, 0)

  1. FF is the midpoint of PP and its image P′P', so P′=2F−PP' = 2F - P.

  2. P′=(6−5,4−1,4−4)=(1,3,0)P' = (6 - 5, 4 - 1, 4 - 4) = (1, 3, 0).

Worth noting

(3,2,2)(3, 2, 2) is the foot FF, only halfway to the image.

Question 59

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What is the vector equation of the line 2x−4=3y+3=6−z2x - 4 = 3y + 3 = 6 - z?

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AnswerD · r⃗=(2i^−j^+6k^)+λ(3i^+2j^−6k^)\vec{r} = (2\hat{i} - \hat{j} + 6\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 6\hat{k})

  1. Take out the coefficients: 2(x−2)=3(y+1)=−(z−6)2(x - 2) = 3(y + 1) = -(z - 6); dividing by 6, x−23=y+12=z−6−6\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 6}{-6}.

  2. So the line passes through (2,−1,6)(2, -1, 6) with direction 3i^+2j^−6k^3\hat{i} + 2\hat{j} - 6\hat{k}: r⃗=(2i^−j^+6k^)+λ(3i^+2j^−6k^)\vec{r} = (2\hat{i} - \hat{j} + 6\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 6\hat{k}).

Worth noting

2i^+3j^−k^2\hat{i} + 3\hat{j} - \hat{k} takes the coefficients of xx, yy, zz as the direction; the direction ratios are their reciprocals.

The angle between two lines

26 questions

Question 60

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Read the two statements and choose the correct option.

I. Lines with direction ratios 1,2,31, 2, 3 and −2,−4,−6-2, -4, -6 are parallel.

II. Lines with direction ratios 1,2,31, 2, 3 and 3,0,−13, 0, -1 are perpendicular.

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AnswerC · Both I and II

  1. I: −2,−4,−6=−2(1,2,3)-2, -4, -6 = -2(1, 2, 3), proportional. True.

  2. II: 1(3)+2(0)+3(−1)=01(3) + 2(0) + 3(-1) = 0. True.

Question 61

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Which of these lines is perpendicular to the line x−21=y+32=z−1−2\frac{x - 2}{1} = \frac{y + 3}{2} = \frac{z - 1}{-2}?

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AnswerA · x2=y−10=z+21\frac{x}{2} = \frac{y - 1}{0} = \frac{z + 2}{1}

  1. Perpendicular lines have a1a2+b1b2+c1c2=0a_1a_2 + b_1b_2 + c_1c_2 = 0; the given ratios are 1,2,−21, 2, -2.

  2. For 2,0,12, 0, 1: 2+0−2=02 + 0 - 2 = 0; the other lines give 99, −4-4 and 11.

Worth noting

x+11=y2=z−3−2\frac{x + 1}{1} = \frac{y}{2} = \frac{z - 3}{-2} has the same ratios as the given line: it is parallel, not perpendicular.

Question 62

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What is the cosine of the angle between the lines with direction cosines 27,37,67\frac{2}{7}, \frac{3}{7}, \frac{6}{7} and 67,−27,37\frac{6}{7}, -\frac{2}{7}, \frac{3}{7}?

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AnswerC · 2449\frac{24}{49}

  1. With direction cosines, cos⁡θ=∣l1l2+m1m2+n1n2∣\cos \theta = \lvert l_1l_2 + m_1m_2 + n_1n_2 \rvert: there is nothing to divide by.

  2. 12−6+1849=2449\frac{12 - 6 + 18}{49} = \frac{24}{49}.

Worth noting

3649\frac{36}{49} drops the minus sign of −27-\frac{2}{7}.

Question 63

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What is the angle between the lines r⃗=(i^+k^)+λ(2i^−3j^+k^)\vec{r} = (\hat{i} + \hat{k}) + \lambda(2\hat{i} - 3\hat{j} + \hat{k}) and r⃗=(3j^−k^)+μ(−4i^+6j^−2k^)\vec{r} = (3\hat{j} - \hat{k}) + \mu(-4\hat{i} + 6\hat{j} - 2\hat{k})?

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AnswerA · 00

  1. −4i^+6j^−2k^=−2(2i^−3j^+k^)-4\hat{i} + 6\hat{j} - 2\hat{k} = -2(2\hat{i} - 3\hat{j} + \hat{k}): the directions are proportional.

  2. cos⁡θ=∣−8−18−2∣1456=2828=1\cos \theta = \frac{\lvert -8 - 18 - 2 \rvert}{\sqrt{14}\sqrt{56}} = \frac{28}{28} = 1, so θ=0\theta = 0: the lines are parallel.

Worth noting

π\pi treats the opposite direction vectors as an angle; between lines the angle comes from ∣cos⁡θ∣\lvert \cos \theta \rvert, so it is 0.

Question 64

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What is the cosine of the angle between the lines with direction ratios 2,2,12, 2, 1 and 4,1,84, 1, 8?

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AnswerC · 23\frac{2}{3}

  1. cos⁡θ=∣a1a2+b1b2+c1c2∣a12+b12+c12a22+b22+c22\cos \theta = \frac{\lvert a_1a_2 + b_1b_2 + c_1c_2 \rvert}{\sqrt{a_1^{2} + b_1^{2} + c_1^{2}}\sqrt{a_2^{2} + b_2^{2} + c_2^{2}}}.

  2. 8+2+83×9=1827=23\frac{8 + 2 + 8}{3 \times 9} = \frac{18}{27} = \frac{2}{3}.

Worth noting

1881\frac{18}{81} divides by 9×99 \times 9, using 9, the square of the first magnitude, in place of 3.

Question 65

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For which kk are the lines with direction ratios 2,−1,32, -1, 3 and 4,k,64, k, 6 parallel?

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AnswerB · −2-2

  1. Parallel lines have proportional direction ratios: 42=k−1=63\frac{4}{2} = \frac{k}{-1} = \frac{6}{3}.

  2. k=−2k = -2.

Worth noting

2 is the ratio between the lines, not kk.

Question 66

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What is the angle between the lines r⃗=(3i^+j^)+λ(i^+2j^+2k^)\vec{r} = (3\hat{i} + \hat{j}) + \lambda(\hat{i} + 2\hat{j} + 2\hat{k}) and r⃗=(j^−2k^)+μ(3i^+2j^+6k^)\vec{r} = (\hat{j} - 2\hat{k}) + \mu(3\hat{i} + 2\hat{j} + 6\hat{k})?

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AnswerD · cos⁡−1(1921)\cos^{-1}{\left(\frac{19}{21} \right)}

  1. Only the directions matter: cos⁡θ=∣3+4+12∣3×7\cos \theta = \frac{\lvert 3 + 4 + 12 \rvert}{3 \times 7}.

  2. θ=cos⁡−1(1921)\theta = \cos^{-1}\left(\frac{19}{21}\right).

Worth noting

The points 3i^+j^3\hat{i} + \hat{j} and j^−2k^\hat{j} - 2\hat{k} play no part in the angle.

Question 67

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For which kk are the lines with direction ratios k,2,−1k, 2, -1 and 3,−1,43, -1, 4 perpendicular?

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AnswerD · 22

  1. Perpendicular: 3k+2(−1)+(−1)(4)=03k + 2(-1) + (-1)(4) = 0.

  2. 3k=63k = 6, so k=2k = 2.

Worth noting

23\frac{2}{3} leaves out the last product, (−1)(4)(-1)(4).

Question 68

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For which kk are the lines x−32=y+1k=z−1\frac{x - 3}{2} = \frac{y + 1}{k} = \frac{z}{-1} and x+13=y2=z−24\frac{x + 1}{3} = \frac{y}{2} = \frac{z - 2}{4} perpendicular?

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AnswerB · −1-1

  1. Direction ratios 2,k,−12, k, -1 and 3,2,43, 2, 4: 6+2k−4=06 + 2k - 4 = 0.

  2. 2k=−22k = -2, so k=−1k = -1.

Worth noting

−5-5 takes (−1)(4)(-1)(4) as +4+4.

Question 69

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What is the angle between the lines x−11=y+12=z2\frac{x - 1}{1} = \frac{y + 1}{2} = \frac{z}{2} and x+22=y−3−2=z−11\frac{x + 2}{2} = \frac{y - 3}{-2} = \frac{z - 1}{1}?

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AnswerD · π2\frac{\pi}{2}

  1. Direction ratios 1,2,21, 2, 2 and 2,−2,12, -2, 1: 1(2)+2(−2)+2(1)=01(2) + 2(-2) + 2(1) = 0.

  2. cos⁡θ=0\cos \theta = 0, so the lines are perpendicular: π2\frac{\pi}{2}.

Worth noting

0 would need the direction ratios in proportion; their dot product is 0, the opposite case.

Question 70

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What is the cosine of the acute angle between the lines with direction ratios 3,−6,23, -6, 2 and 2,2,12, 2, 1?

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AnswerB · 421\frac{4}{21}

  1. 3(2)+(−6)(2)+2(1)=−43(2) + (-6)(2) + 2(1) = -4; the lengths are 7 and 3.

  2. For the acute angle, cos⁡θ=∣−4∣7×3=421\cos \theta = \frac{\lvert -4 \rvert}{7 \times 3} = \frac{4}{21}.

Worth noting

−421-\frac{4}{21} belongs to the obtuse angle; the acute one takes the absolute value.

Question 71

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What is the sine of the angle between the lines with direction ratios 1,−2,21, -2, 2 and 4,0,34, 0, 3?

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AnswerD · 53\frac{\sqrt{5}}{3}

  1. cos⁡θ=∣4+0+6∣3×5=23\cos \theta = \frac{\lvert 4 + 0 + 6 \rvert}{3 \times 5} = \frac{2}{3}.

  2. sin⁡θ=1−49=53\sin \theta = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}.

Worth noting

33=1−23\frac{\sqrt{3}}{3} = \sqrt{1 - \frac{2}{3}} forgets to square the cosine.

Question 72

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Assertion (A): The lines with direction ratios 3,−1,23, -1, 2 and 1,1,−11, 1, -1 are perpendicular.

Reason (R): Two lines are perpendicular when their direction ratios are proportional.

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AnswerC · A is true, but R is false.

  1. A: 3(1)+(−1)(1)+2(−1)=03(1) + (-1)(1) + 2(-1) = 0, so A is true.

  2. R is false: proportional ratios make lines parallel; perpendicular lines need a1a2+b1b2+c1c2=0a_1a_2 + b_1b_2 + c_1c_2 = 0.

Question 73

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For which pp and qq are the lines x−12=y+1p=z−23\frac{x - 1}{2} = \frac{y + 1}{p} = \frac{z - 2}{3} and x+3q=y−2−6=z9\frac{x + 3}{q} = \frac{y - 2}{-6} = \frac{z}{9} parallel?

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AnswerA · p=−2,q=6p = -2, q = 6

  1. Parallel lines have proportional direction ratios: q2=−6p=93=3\frac{q}{2} = \frac{-6}{p} = \frac{9}{3} = 3.

  2. So q=6q = 6 and p=−2p = -2.

Worth noting

p=−18p = -18 multiplies −6-6 by 3 where it should divide.

Question 74

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Read the two statements and choose the correct option.

I. The lines with direction ratios 1,2,−21, 2, -2 and 2,−2,12, -2, 1 are perpendicular.

II. The angle between r⃗=a⃗1+λb⃗1\vec{r} = \vec{a}_1 + \lambda\vec{b}_1 and r⃗=a⃗2+μb⃗2\vec{r} = \vec{a}_2 + \mu\vec{b}_2 depends on a⃗1\vec{a}_1 and a⃗2\vec{a}_2.

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AnswerD · Neither I nor II

  1. I: 1(2)+2(−2)+(−2)(1)=−4≠01(2) + 2(-2) + (-2)(1) = -4 \ne 0. False.

  2. II: the angle uses only b⃗1\vec{b}_1 and b⃗2\vec{b}_2; the points play no part. False.

Question 75

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What is the angle between the lines with direction ratios a,b,ca, b, c and b−c,c−a,a−bb - c, c - a, a - b, where a,b,ca, b, c are not all equal?

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AnswerD · π2\frac{\pi}{2}

  1. a(b−c)+b(c−a)+c(a−b)=ab−ac+bc−ab+ca−bc=0a(b - c) + b(c - a) + c(a - b) = ab - ac + bc - ab + ca - bc = 0.

  2. So cos⁡θ=0\cos \theta = 0 and θ=π2\theta = \frac{\pi}{2}, whatever a,b,ca, b, c are.

Worth noting

0 would need the two sets of ratios in proportion; their dot product is 0 instead.

Question 76

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Lines L1L_1, L2L_2, L3L_3 have direction ratios 2,−1,22, -1, 2; 2,2,−12, 2, -1; and −1,2,2-1, 2, 2, respectively. Which of the following statements are true?

I. L1L_1 is perpendicular to L2L_2.

II. L1L_1, L2L_2 and L3L_3 are mutually perpendicular.

III. L3L_3 is parallel to the line with direction ratios 1,−2,−21, -2, -2.

IV. L1L_1 is parallel to the line with direction ratios 4,−2,−44, -2, -4.

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AnswerA · I, II and III only

  1. For L1,L2L_1, L_2: 4−2−2=04 - 2 - 2 = 0; for L1,L3L_1, L_3: −2−2+4=0-2 - 2 + 4 = 0; for L2,L3L_2, L_3: −2+4−2=0-2 + 4 - 2 = 0. So I and II are true.

  2. −1,2,2=−(1,−2,−2)-1, 2, 2 = -(1, -2, -2), so III is true; twice 2,−1,22, -1, 2 is 4,−2,44, -2, 4, not 4,−2,−44, -2, -4, so IV is false.

Worth noting

IV fails on one sign: twice 2,−1,22, -1, 2 is 4,−2,44, -2, 4.

Question 77

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What is the angle between the lines x+13=4−y2=z6\frac{x + 1}{3} = \frac{4 - y}{2} = \frac{z}{6} and x−22=y+36=z−13\frac{x - 2}{2} = \frac{y + 3}{6} = \frac{z - 1}{3}?

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AnswerD · cos⁡−1(1249)\cos^{-1}{\left(\frac{12}{49} \right)}

  1. 4−y2=y−4−2\frac{4 - y}{2} = \frac{y - 4}{-2}, so the first line has direction ratios 3,−2,63, -2, 6 and the second 2,6,32, 6, 3; both have length 7.

  2. cos⁡θ=∣6−12+18∣7×7=1249\cos \theta = \frac{\lvert 6 - 12 + 18 \rvert}{7 \times 7} = \frac{12}{49}, so θ=cos⁡−11249\theta = \cos^{-1}\frac{12}{49}.

Worth noting

cos⁡−13649\cos^{-1}\frac{36}{49} reads 4−y2\frac{4 - y}{2} as direction ratio 2; it is −2-2.

Question 78

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What is the cosine of the angle between the line through A(1,3,−2)A(1, 3, -2) and B(3,1,−1)B(3, 1, -1) and the line through C(−1,0,2)C(-1, 0, 2) and D(5,2,5)D(5, 2, 5)?

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AnswerC · 1121\frac{11}{21}

  1. Direction ratios: ABAB gives 2,−2,12, -2, 1 (length 3) and CDCD gives 6,2,36, 2, 3 (length 7).

  2. cos⁡θ=∣12−4+3∣3×7=1121\cos \theta = \frac{\lvert 12 - 4 + 3 \rvert}{3 \times 7} = \frac{11}{21}.

Worth noting

1921\frac{19}{21} takes 1−31 - 3 as +2+2 in the direction of ABAB.

Question 79

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A line L1L_1 makes equal acute angles with the coordinate axes, and a line L2L_2 has direction ratios 6,−2,36, -2, 3. What is the cosine of the angle between L1L_1 and L2L_2?

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AnswerB · 33\frac{\sqrt{3}}{3}

  1. L1L_1 has direction ratios 1,1,11, 1, 1, of length 3\sqrt{3}; L2L_2 has length 36+4+9=7\sqrt{36 + 4 + 9} = 7.

  2. cos⁡θ=∣6−2+3∣73=13=33\cos \theta = \frac{\lvert 6 - 2 + 3 \rvert}{7\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}.

Worth noting

13\frac{1}{3} takes each direction cosine of L1L_1 as 13\frac{1}{3} instead of 13\frac{1}{\sqrt{3}}.

Question 80

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For which kk are the lines x−1−3=y−22k=z−32\frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2} and x−13k=y−11=z−6−5\frac{x - 1}{3k} = \frac{y - 1}{1} = \frac{z - 6}{-5} perpendicular?

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AnswerA · −107- \frac{10}{7}

  1. Perpendicular: a1a2+b1b2+c1c2=0a_1a_2 + b_1b_2 + c_1c_2 = 0, so −9k+2k−10=0-9k + 2k - 10 = 0.

  2. −7k=10-7k = 10: k=−107k = -\frac{10}{7}.

Worth noting

107\frac{10}{7} loses the sign when solving −7k=10-7k = 10.

Question 81

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What is the angle between the lines r⃗=(i^−k^)+λ(i^+j^+2k^)\vec{r} = (\hat{i} - \hat{k}) + \lambda(\hat{i} + \hat{j} + 2\hat{k}) and r⃗=3j^+μ(2i^−j^+k^)\vec{r} = 3\hat{j} + \mu(2\hat{i} - \hat{j} + \hat{k})?

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AnswerC · π3\frac{\pi}{3}

  1. Only the directions matter: (1)(2)+(1)(−1)+(2)(1)=3(1)(2) + (1)(-1) + (2)(1) = 3, and each has length 6\sqrt{6}.

  2. cos⁡θ=36=12\cos \theta = \frac{3}{6} = \frac{1}{2}, so θ=π3\theta = \frac{\pi}{3}.

Worth noting

π6\frac{\pi}{6} reads 12\frac{1}{2} as a sine; the dot product gives the cosine.

Question 82

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The direction cosines l,m,nl, m, n of two lines satisfy 2l+2m−3n=02l + 2m - 3n = 0 and mn+nl−3lm=0mn + nl - 3lm = 0. What is the cosine of the angle between the lines?

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AnswerD · 89\frac{8}{9}

  1. From the first, n=2(l+m)3n = \frac{2(l + m)}{3}; then mn+nl=n(l+m)=3lmmn + nl = n(l + m) = 3lm gives 2(l+m)2=9lm2(l + m)^{2} = 9lm, that is (2l−m)(l−2m)=0(2l - m)(l - 2m) = 0.

  2. m=2lm = 2l gives direction ratios 1,2,21, 2, 2; l=2ml = 2m gives 2,1,22, 1, 2.

  3. cos⁡θ=2+2+43×3=89\cos \theta = \frac{2 + 2 + 4}{3 \times 3} = \frac{8}{9}.

Worth noting

49\frac{4}{9} leaves out the product of the zz-ratios, 2×22 \times 2.

Question 83

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What is the acute angle between two body diagonals of a cube?

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AnswerC · cos⁡−1(13)\cos^{-1}{\left(\frac{1}{3} \right)}

  1. With side aa and edges along the axes, two body diagonals run from (0,0,0)(0, 0, 0) to (a,a,a)(a, a, a) and from (a,0,0)(a, 0, 0) to (0,a,a)(0, a, a): direction ratios 1,1,11, 1, 1 and −1,1,1-1, 1, 1.

  2. cos⁡θ=∣−1+1+1∣33=13\cos \theta = \frac{\lvert -1 + 1 + 1 \rvert}{\sqrt{3}\sqrt{3}} = \frac{1}{3}, so θ=cos⁡−113\theta = \cos^{-1}\frac{1}{3}.

Worth noting

π2\frac{\pi}{2} holds for the diagonals of a square face, not for the body diagonals of a cube.

Question 84

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For which kk is the angle between the lines with direction ratios 2,−1,22, -1, 2 and 1,1,k1, 1, k equal to π4\frac{\pi}{4}?

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AnswerB · 44

  1. cos⁡π4=∣2−1+2k∣32+k2=12\cos\frac{\pi}{4} = \frac{\lvert 2 - 1 + 2k \rvert}{3\sqrt{2 + k^{2}}} = \frac{1}{\sqrt{2}}, so 2(1+2k)2=9(2+k2)2(1 + 2k)^{2} = 9(2 + k^{2}).

  2. 8k2+8k+2=9k2+188k^{2} + 8k + 2 = 9k^{2} + 18 gives k2−8k+16=0k^{2} - 8k + 16 = 0, that is (k−4)2=0(k - 4)^{2} = 0, so k=4k = 4.

Worth noting

−4-4 comes from a sign slip in −8k-8k, which turns the equation into (k+4)2=0(k + 4)^{2} = 0.

Question 85

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What is the angle between the lines x=2y=−zx = 2y = -z and x=y=2zx = y = 2z?

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AnswerB · cos⁡−1(49)\cos^{-1}{\left(\frac{4}{9} \right)}

  1. Put each equal to tt: the first line is x=tx = t, y=t2y = \frac{t}{2}, z=−tz = -t, with direction ratios 1,12,−11, \frac{1}{2}, -1, or 2,1,−22, 1, -2.

  2. Likewise the second has 1,1,121, 1, \frac{1}{2}, or 2,2,12, 2, 1.

  3. cos⁡θ=∣4+2−2∣3×3=49\cos \theta = \frac{\lvert 4 + 2 - 2 \rvert}{3 \times 3} = \frac{4}{9}, so θ=cos⁡−149\theta = \cos^{-1}\frac{4}{9}.

Worth noting

cos⁡−116\cos^{-1}\frac{1}{6} uses the coefficients 1,2,−11, 2, -1 and 1,1,21, 1, 2 as direction ratios; the ratios are their reciprocals.

Skew lines and the shortest distance

35 questions

Question 86

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What is the shortest distance between the xx-axis and the line through (0,0,3)(0, 0, 3) parallel to the yy-axis?

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AnswerC · 33

  1. b⃗1×b⃗2=i^×j^=k^\vec{b}_1 \times \vec{b}_2 = \hat{i} \times \hat{j} = \hat{k}, of length 1, and a⃗2−a⃗1=3k^\vec{a}_2 - \vec{a}_1 = 3\hat{k}.

  2. d=∣3k^⋅k^∣=3d = \lvert 3\hat{k} \cdot \hat{k} \rvert = 3: the common perpendicular runs along the zz-axis from (0,0,0)(0, 0, 0) to (0,0,3)(0, 0, 3).

Worth noting

0 assumes perpendicular lines meet; these two are perpendicular but skew.

Question 87

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What is the distance between the xx-axis and the line through (0,3,4)(0, 3, 4) parallel to it?

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AnswerB · 55

  1. Each point (t,3,4)(t, 3, 4) of the line is nearest to (t,0,0)(t, 0, 0) on the xx-axis.

  2. The gap is 32+42=5\sqrt{3^{2} + 4^{2}} = 5 all along, so the distance is 5.

Worth noting

7 adds 3 and 4; the gap is the hypotenuse, 9+16\sqrt{9 + 16}.

Question 88

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What is the shortest distance between the lines x−21=y+12=z−43\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 4}{3} and x−23=y+1−1=z−42\frac{x - 2}{3} = \frac{y + 1}{-1} = \frac{z - 4}{2}?

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AnswerA · 00

  1. Both lines pass through (2,−1,4)(2, -1, 4): every numerator is 0 there.

  2. Lines with a common point meet, so the shortest distance is 0.

Worth noting

14\sqrt{14} is the distance between (1,2,3)(1, 2, 3) and (3,−1,2)(3, -1, 2), the denominators read as points.

Question 89

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The lines r⃗=a⃗1+λb⃗\vec{r} = \vec{a}_1 + \lambda\vec{b} and r⃗=a⃗2+μb⃗\vec{r} = \vec{a}_2 + \mu\vec{b} are parallel and distinct. Which expression gives the distance between them?

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AnswerD · ∣b⃗×(a⃗2−a⃗1)∣∣b⃗∣\frac{\lvert \vec{b} \times (\vec{a}_2 - \vec{a}_1) \rvert}{\lvert \vec{b} \rvert}

  1. For parallel lines b⃗×b⃗=0⃗\vec{b} \times \vec{b} = \vec{0}, so the skew-lines formula would divide by 0.

  2. The gap is the part of a⃗2−a⃗1\vec{a}_2 - \vec{a}_1 across b⃗\vec{b}: d=∣b⃗×(a⃗2−a⃗1)∣∣b⃗∣d = \frac{\lvert \vec{b} \times (\vec{a}_2 - \vec{a}_1) \rvert}{\lvert \vec{b} \rvert}.

Worth noting

The skew-lines formula needs b⃗1×b⃗2≠0⃗\vec{b}_1 \times \vec{b}_2 \ne \vec{0}; with one direction b⃗\vec{b} it becomes 00\frac{0}{0}.

Question 90

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Two lines in space are neither parallel nor intersecting. What are they called?

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AnswerD · Skew lines

  1. Lines that do not meet and are not parallel cannot lie in one plane.

  2. Such lines are skew.

Worth noting

Perpendicular lines can still meet, as the axes do.

Question 91

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Which of the following statements are true?

I. The shortest distance between two skew lines is along the line perpendicular to both.

II. If (a⃗2−a⃗1)⋅(b⃗1×b⃗2)=0(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 0 for non-parallel lines, the lines intersect.

III. Two parallel lines are skew.

IV. Two skew lines always lie in one plane.

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AnswerA · I and II only

  1. I and II are true: the formula measures along b⃗1×b⃗2\vec{b}_1 \times \vec{b}_2, and a zero numerator means distance 0.

  2. III and IV are false: parallel lines lie in a plane, and skew lines never do.

Question 92

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What is the shortest distance between r⃗=(i^+2j^+3k^)+λ(i^−j^+2k^)\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + 2\hat{k}) and r⃗=(i^+4k^)+μ(i^+j^+k^)\vec{r} = (\hat{i} + 4\hat{k}) + \mu(\hat{i} + \hat{j} + \hat{k})?

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AnswerA · 00

  1. b1⃗×b2⃗=−3i^+j^+2k^\vec{b_1} \times \vec{b_2} = -3\hat{i} + \hat{j} + 2\hat{k} and a2⃗−a1⃗=−2j^+k^\vec{a_2} - \vec{a_1} = -2\hat{j} + \hat{k}; their dot product is 00.

  2. So the distance is 0: the lines meet, at (2,1,5)(2, 1, 5) where λ=μ=1\lambda = \mu = 1.

Worth noting

5\sqrt{5} is ∣a2⃗−a1⃗∣\lvert \vec{a_2} - \vec{a_1} \rvert, the gap between the two given points, which need not be the shortest.

Question 93

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How are the lines x1=y2=z3\frac{x}{1} = \frac{y}{2} = \frac{z}{3} and x−12=y4=z6\frac{x - 1}{2} = \frac{y}{4} = \frac{z}{6} placed?

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AnswerA · Parallel and distinct

  1. The direction ratios 2,4,62, 4, 6 are twice 1,2,31, 2, 3, so the lines are parallel.

  2. (1,0,0)(1, 0, 0) is on the second line but not the first (11≠02\frac{1}{1} \ne \frac{0}{2}), so they are distinct.

Worth noting

Skew lines are never parallel; parallel lines always lie in one plane.

Question 94

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Assertion (A): Skew lines do not lie in any one plane.

Reason (R): Two lines in space that do not meet must be parallel.

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AnswerC · A is true, but R is false.

  1. A is true: that is what skew means.

  2. R is false: skew lines do not meet and are not parallel either.

Question 95

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Read the two statements and choose the correct option.

I. Two lines with direction cosines l1,m1,n1l_1, m_1, n_1 and l2,m2,n2l_2, m_2, n_2 are perpendicular exactly when l1l2+m1m2+n1n2=0l_1l_2 + m_1m_2 + n_1n_2 = 0.

II. The shortest distance between skew lines is ∣(a2⃗−a1⃗)⋅(b1⃗×b2⃗)∣\lvert (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) \rvert.

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AnswerA · Only I

  1. I: cos⁡θ=l1l2+m1m2+n1n2\cos \theta = l_1l_2 + m_1m_2 + n_1n_2, which is 0 for 90∘90^\circ. True.

  2. II: it must still be divided by ∣b1⃗×b2⃗∣\lvert \vec{b_1} \times \vec{b_2} \rvert. False.

Question 96

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Two lines are L1:r⃗=(i^+2k^)+λ(i^+2j^−k^)L_1: \vec{r} = (\hat{i} + 2\hat{k}) + \lambda(\hat{i} + 2\hat{j} - \hat{k}) and L2:r⃗=(j^−k^)+μ(2i^+j^+k^)L_2: \vec{r} = (\hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + \hat{k}).

What is b1⃗×b2⃗\vec{b_1} \times \vec{b_2}?

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AnswerD · 3i^−3j^−3k^3\hat{i} - 3\hat{j} - 3\hat{k}

  1. i^(2⋅1−(−1)⋅1)−j^(1⋅1−(−1)⋅2)+k^(1⋅1−2⋅2)\hat{i}(2 \cdot 1 - (-1) \cdot 1) - \hat{j}(1 \cdot 1 - (-1) \cdot 2) + \hat{k}(1 \cdot 1 - 2 \cdot 2).

  2. =3i^−3j^−3k^= 3\hat{i} - 3\hat{j} - 3\hat{k}.

Worth noting

−3i^+3j^+3k^-3\hat{i} + 3\hat{j} + 3\hat{k} is b2⃗×b1⃗\vec{b_2} \times \vec{b_1}, the product in the other order.

Question 97

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Two lines are L1:r⃗=(i^+2k^)+λ(i^+2j^−k^)L_1: \vec{r} = (\hat{i} + 2\hat{k}) + \lambda(\hat{i} + 2\hat{j} - \hat{k}) and L2:r⃗=(j^−k^)+μ(2i^+j^+k^)L_2: \vec{r} = (\hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + \hat{k}).

What is (a2⃗−a1⃗)⋅(b1⃗×b2⃗)(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})?

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AnswerB · 33

  1. a2⃗−a1⃗=−i^+j^−3k^\vec{a_2} - \vec{a_1} = -\hat{i} + \hat{j} - 3\hat{k}.

  2. (−1)(3)+(1)(−3)+(−3)(−3)=−3−3+9=3(-1)(3) + (1)(-3) + (-3)(-3) = -3 - 3 + 9 = 3.

Worth noting

15 adds the three products without their signs.

Question 98

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Two lines are L1:r⃗=(i^+2k^)+λ(i^+2j^−k^)L_1: \vec{r} = (\hat{i} + 2\hat{k}) + \lambda(\hat{i} + 2\hat{j} - \hat{k}) and L2:r⃗=(j^−k^)+μ(2i^+j^+k^)L_2: \vec{r} = (\hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + \hat{k}).

What is the angle between L1L_1 and L2L_2?

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AnswerC · π3\frac{\pi}{3}

  1. b1⃗⋅b2⃗=2+2−1=3\vec{b_1} \cdot \vec{b_2} = 2 + 2 - 1 = 3, and each has length 6\sqrt{6}.

  2. cos⁡θ=36=12\cos \theta = \frac{3}{6} = \frac{1}{2}, so θ=π3\theta = \frac{\pi}{3}.

Worth noting

π2\frac{\pi}{2} assumes skew lines are perpendicular; they can meet at any angle in direction.

Question 99

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The lines r⃗=(i^+3j^−6k^)+λ(i^−j^+2k^)\vec{r} = (\hat{i} + 3\hat{j} - 6\hat{k}) + \lambda(\hat{i} - \hat{j} + 2\hat{k}) and r⃗=(i^−2k^)+μ(2i^+j^)\vec{r} = (\hat{i} - 2\hat{k}) + \mu(2\hat{i} + \hat{j}) intersect. At which point?

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AnswerB · (3,1,−2)(3, 1, -2)

  1. Equate the components: 1+λ=1+2μ1 + \lambda = 1 + 2\mu, 3−λ=μ3 - \lambda = \mu and −6+2λ=−2-6 + 2\lambda = -2.

  2. The last gives λ=2\lambda = 2, then μ=1\mu = 1, and the first checks: 3=33 = 3. The point is (3,1,−2)(3, 1, -2).

Worth noting

(5,2,−2)(5, 2, -2) puts λ=2\lambda = 2 into the second line, whose parameter is μ\mu.

Question 100

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Assertion (A): The lines L1:r⃗=(i^+2j^)+λ(i^−j^+k^)L_1: \vec{r} = (\hat{i} + 2\hat{j}) + \lambda(\hat{i} - \hat{j} + \hat{k}) and L2:r⃗=(2i^−k^)+μ(i^+2j^)L_2: \vec{r} = (2\hat{i} - \hat{k}) + \mu(\hat{i} + 2\hat{j}) are skew.

Reason (R): The direction ratios of L1L_1 and L2L_2 are not proportional.

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AnswerB · Both A and R are true, but R is not the correct explanation of A.

  1. A: (a⃗2−a⃗1)⋅(b⃗1×b⃗2)=(i^−2j^−k^)⋅(−2i^+j^+3k^)=−7≠0(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (\hat{i} - 2\hat{j} - \hat{k}) \cdot (-2\hat{i} + \hat{j} + 3\hat{k}) = -7 \ne 0, and the lines are not parallel, so A is true.

  2. R is true, but it only rules out parallel lines; lines that are not parallel can still meet, so R does not explain A.

Question 101

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What is the perpendicular distance of the point (5,1,5)(5, 1, 5) from the line r⃗=(i^−j^)+λ(2i^−2j^+k^)\vec{r} = (\hat{i} - \hat{j}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k})?

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AnswerB · 66

  1. p⃗−a⃗=4i^+2j^+5k^\vec{p} - \vec{a} = 4\hat{i} + 2\hat{j} + 5\hat{k} and b⃗×(p⃗−a⃗)=−12i^−6j^+12k^\vec{b} \times (\vec{p} - \vec{a}) = -12\hat{i} - 6\hat{j} + 12\hat{k}, of length 18.

  2. d=18∣b⃗∣=183=6d = \frac{18}{\lvert \vec{b} \rvert} = \frac{18}{3} = 6.

Worth noting

18 is ∣b⃗×(p⃗−a⃗)∣\lvert \vec{b} \times (\vec{p} - \vec{a}) \rvert before it is divided by ∣b⃗∣=3\lvert \vec{b} \rvert = 3.

Question 102

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Which describes the lines x−12=y+21=z−3−1\frac{x - 1}{2} = \frac{y + 2}{1} = \frac{z - 3}{-1} and x+11=y−1−1=z−22\frac{x + 1}{1} = \frac{y - 1}{-1} = \frac{z - 2}{2}?

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AnswerA · Skew

  1. 2,1,−12, 1, -1 and 1,−1,21, -1, 2 are not proportional, so the lines are not parallel.

  2. (a⃗2−a⃗1)⋅(b⃗1×b⃗2)=(−2i^+3j^−k^)⋅(i^−5j^−3k^)=−14≠0(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-2\hat{i} + 3\hat{j} - \hat{k}) \cdot (\hat{i} - 5\hat{j} - 3\hat{k}) = -14 \ne 0: they do not meet either, so they are skew.

Worth noting

Lines that are not parallel need not meet: in space they can be skew.

Question 103

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Read the two statements and choose the correct option.

I. The distance between the parallel lines r⃗=λ(i^+j^+k^)\vec{r} = \lambda(\hat{i} + \hat{j} + \hat{k}) and r⃗=2i^+μ(i^+j^+k^)\vec{r} = 2\hat{i} + \mu(\hat{i} + \hat{j} + \hat{k}) is 263\frac{2\sqrt{6}}{3}.

II. The lines r⃗=(2i^+k^)+λ(i^−j^)\vec{r} = (2\hat{i} + \hat{k}) + \lambda(\hat{i} - \hat{j}) and r⃗=(i^+2j^+2k^)+μ(j^+k^)\vec{r} = (\hat{i} + 2\hat{j} + 2\hat{k}) + \mu(\hat{j} + \hat{k}) meet at (1,1,1)(1, 1, 1).

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AnswerC · Both I and II

  1. I: (i^+j^+k^)×2i^=2j^−2k^(\hat{i} + \hat{j} + \hat{k}) \times 2\hat{i} = 2\hat{j} - 2\hat{k}, of length 222\sqrt{2}; divided by 3\sqrt{3} it gives 263\frac{2\sqrt{6}}{3}. True.

  2. II: λ=−1\lambda = -1 gives (1,1,1)(1, 1, 1) on the first line and μ=−1\mu = -1 gives (1,1,1)(1, 1, 1) on the second. True.

Question 104

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Which describes the lines r⃗=(2i^−j^+k^)+λ(i^+2j^−3k^)\vec{r} = (2\hat{i} - \hat{j} + \hat{k}) + \lambda(\hat{i} + 2\hat{j} - 3\hat{k}) and x−4−2=y−3−4=z+56\frac{x - 4}{-2} = \frac{y - 3}{-4} = \frac{z + 5}{6}?

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AnswerB · The same line

  1. −2,−4,6=−2(1,2,−3)-2, -4, 6 = -2(1, 2, -3): the lines are parallel.

  2. λ=2\lambda = 2 on the first line gives (4,3,−5)(4, 3, -5), a point of the second; sharing a point, they are the same line.

Worth noting

Parallel and distinct lines share no point; here (4,3,−5)(4, 3, -5) lies on both.

Question 105

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Match each line in List I with how it lies relative to L:r⃗=λ(2i^+j^)L: \vec{r} = \lambda(2\hat{i} + \hat{j}) in List II.

List IList II
A. r⃗=k^+μ(2i^+j^)\vec{r} = \hat{k} + \mu(2\hat{i} + \hat{j})I. Skew
B. r⃗=k^+μ(i^−2j^)\vec{r} = \hat{k} + \mu(\hat{i} - 2\hat{j})II. The same line
C. r⃗=k^+μ(2i^+j^−k^)\vec{r} = \hat{k} + \mu(2\hat{i} + \hat{j} - \hat{k})III. Parallel and distinct
D. r⃗=(4i^+2j^)+μ(−2i^−j^)\vec{r} = (4\hat{i} + 2\hat{j}) + \mu(-2\hat{i} - \hat{j})IV. Meeting at one point
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AnswerD · A-III, B-I, C-IV, D-II

  1. A has the direction of LL, and (0,0,1)(0, 0, 1) is not on LL: parallel and distinct. D has direction −(2i^+j^)-(2\hat{i} + \hat{j}) and passes through (4,2,0)(4, 2, 0), a point of LL: the same line.

  2. B and C are not parallel to LL. With a⃗2−a⃗1=k^\vec{a}_2 - \vec{a}_1 = \hat{k}, k^⋅(b⃗1×b⃗2)\hat{k} \cdot (\vec{b}_1 \times \vec{b}_2) is −5-5 for B, so B is skew, and 0 for C, which meets LL at (2,1,0)(2, 1, 0).

Question 106

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What is the shortest distance between r⃗=(i^+2j^+3k^)+λ(i^−j^+k^)\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k}) and r⃗=(2i^−j^−k^)+μ(2i^+j^+2k^)\vec{r} = (2\hat{i} - \hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k})?

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AnswerB · 522\frac{5 \sqrt{2}}{2}

  1. b⃗1×b⃗2=−3i^+3k^\vec{b}_1 \times \vec{b}_2 = -3\hat{i} + 3\hat{k}, with magnitude 323 \sqrt{2}; a⃗2−a⃗1=i^−3j^−4k^\vec{a}_2 - \vec{a}_1 = \hat{i} - 3\hat{j} - 4\hat{k}.

  2. d=∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣∣b⃗1×b⃗2∣=1532=522d = \frac{\lvert (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) \rvert}{\lvert \vec{b}_1 \times \vec{b}_2 \rvert} = \frac{15}{3 \sqrt{2}} = \frac{5 \sqrt{2}}{2}.

Worth noting

1515 is the numerator alone; it still has to be divided by ∣b⃗1×b⃗2∣\lvert \vec{b}_1 \times \vec{b}_2 \rvert.

Question 107

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What is the distance between the parallel lines r⃗=(i^+2j^−4k^)+λ(2i^+3j^+6k^)\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) and r⃗=(3i^+3j^−5k^)+μ(2i^+3j^+6k^)\vec{r} = (3\hat{i} + 3\hat{j} - 5\hat{k}) + \mu(2\hat{i} + 3\hat{j} + 6\hat{k})?

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AnswerB · 2937\frac{\sqrt{293}}{7}

  1. a⃗2−a⃗1=2i^+j^−k^\vec{a}_2 - \vec{a}_1 = 2\hat{i} + \hat{j} - \hat{k} and b⃗×(a⃗2−a⃗1)=−9i^+14j^−4k^\vec{b} \times (\vec{a}_2 - \vec{a}_1) = -9\hat{i} + 14\hat{j} - 4\hat{k}.

  2. d=∣b⃗×(a⃗2−a⃗1)∣∣b⃗∣=2937=2937d = \frac{\lvert \vec{b} \times (\vec{a}_2 - \vec{a}_1) \rvert}{\lvert \vec{b} \rvert} = \frac{\sqrt{293}}{7} = \frac{\sqrt{293}}{7}.

Worth noting

6\sqrt{6} is the distance between the two given points, which need not be the closest pair.

Question 108

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Assertion (A): The lines r⃗=i^+λ(i^+j^)\vec{r} = \hat{i} + \lambda(\hat{i} + \hat{j}) and r⃗=j^+μ(i^+k^)\vec{r} = \hat{j} + \mu(\hat{i} + \hat{k}) intersect.

Reason (R): Two lines intersect exactly when the shortest distance between them is 0.

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AnswerD · A is false, but R is true.

  1. A is false: (a⃗2−a⃗1)⋅(b⃗1×b⃗2)=(−i^+j^)⋅(i^−j^−k^)=−2≠0(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-\hat{i} + \hat{j}) \cdot (\hat{i} - \hat{j} - \hat{k}) = -2 \ne 0, so the distance is not 0.

  2. R is true for non-parallel lines, which these are.

Question 109

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What is the distance between the parallel lines r⃗=λ(i^+2j^+2k^)\vec{r} = \lambda(\hat{i} + 2\hat{j} + 2\hat{k}) and r⃗=3i^+μ(i^+2j^+2k^)\vec{r} = 3\hat{i} + \mu(\hat{i} + 2\hat{j} + 2\hat{k})?

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AnswerB · 222 \sqrt{2}

  1. b⃗×(a2⃗−a1⃗)=6j^−6k^\vec{b} \times (\vec{a_2} - \vec{a_1}) = 6\hat{j} - 6\hat{k}, of length 626\sqrt{2}; ∣b⃗∣=3\lvert \vec{b} \rvert = 3.

  2. d=623=22d = \frac{6\sqrt{2}}{3} = 2\sqrt{2}.

Worth noting

3 is the distance between the two given points, measured along the xx-axis, not across the lines.

Question 110

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What is the shortest distance between r⃗=(i^+j^)+λ(i^+k^)\vec{r} = (\hat{i} + \hat{j}) + \lambda(\hat{i} + \hat{k}) and r⃗=(3j^+5k^)+μ(j^+k^)\vec{r} = (3\hat{j} + 5\hat{k}) + \mu(\hat{j} + \hat{k})?

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AnswerC · 433\frac{4 \sqrt{3}}{3}

  1. b1⃗×b2⃗=−i^−j^+k^\vec{b_1} \times \vec{b_2} = -\hat{i} - \hat{j} + \hat{k}, of length 3\sqrt{3}, and a2⃗−a1⃗=−i^+2j^+5k^\vec{a_2} - \vec{a_1} = -\hat{i} + 2\hat{j} + 5\hat{k}.

  2. d=∣1−2+5∣3=43d = \frac{\lvert 1 - 2 + 5 \rvert}{\sqrt{3}} = \frac{4}{\sqrt{3}}.

Worth noting

4 stops at the scalar triple product and never divides by ∣b1⃗×b2⃗∣\lvert \vec{b_1} \times \vec{b_2} \rvert.

Question 111

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What is the distance between the parallel lines x1=y2=z2\frac{x}{1} = \frac{y}{2} = \frac{z}{2} and x−31=y−32=z−32\frac{x - 3}{1} = \frac{y - 3}{2} = \frac{z - 3}{2}?

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AnswerB · 2\sqrt{2}

  1. b⃗×(a2⃗−a1⃗)=3j^−3k^\vec{b} \times (\vec{a_2} - \vec{a_1}) = 3\hat{j} - 3\hat{k}, of length 323\sqrt{2}, and ∣b⃗∣=3\lvert \vec{b} \rvert = 3.

  2. d=323=2d = \frac{3\sqrt{2}}{3} = \sqrt{2}.

Worth noting

333\sqrt{3} is the distance between the two given points, (0,0,0)(0, 0, 0) and (3,3,3)(3, 3, 3).

Question 112

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What is the shortest distance between the lines x1=y−11=z−21\frac{x}{1} = \frac{y - 1}{1} = \frac{z - 2}{1} and x−11=y−1=z2\frac{x - 1}{1} = \frac{y}{-1} = \frac{z}{2}?

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AnswerC · 4147\frac{4 \sqrt{14}}{7}

  1. Points (0,1,2)(0, 1, 2), (1,0,0)(1, 0, 0); directions 1,1,11, 1, 1 and 1,−1,21, -1, 2; b1⃗×b2⃗=3i^−j^−2k^\vec{b_1} \times \vec{b_2} = 3\hat{i} - \hat{j} - 2\hat{k}.

  2. (a2⃗−a1⃗)⋅(b1⃗×b2⃗)=3+1+4=8(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 3 + 1 + 4 = 8, so d=814d = \frac{8}{\sqrt{14}}.

Worth noting

47=814\frac{4}{7} = \frac{8}{14} divides by ∣b1⃗×b2⃗∣2\lvert \vec{b_1} \times \vec{b_2} \rvert^{2} instead of its length.

Question 113

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Two lines are L1:r⃗=(i^+2k^)+λ(i^+2j^−k^)L_1: \vec{r} = (\hat{i} + 2\hat{k}) + \lambda(\hat{i} + 2\hat{j} - \hat{k}) and L2:r⃗=(j^−k^)+μ(2i^+j^+k^)L_2: \vec{r} = (\hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + \hat{k}).

What is the shortest distance between L1L_1 and L2L_2?

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AnswerB · 33\frac{\sqrt{3}}{3}

  1. ∣b1⃗×b2⃗∣=9+9+9=33\lvert \vec{b_1} \times \vec{b_2} \rvert = \sqrt{9 + 9 + 9} = 3\sqrt{3}.

  2. d=333=13d = \frac{3}{3\sqrt{3}} = \frac{1}{\sqrt{3}}.

Worth noting

19\frac{1}{9} divides by ∣b1⃗×b2⃗∣2=27\lvert \vec{b_1} \times \vec{b_2} \rvert^{2} = 27, not by its length.

Question 114

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For which kk do the lines x−21=y+12=z−3−1\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 3}{-1} and x−42=y−k1=z−11\frac{x - 4}{2} = \frac{y - k}{1} = \frac{z - 1}{1} intersect?

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AnswerC · 33

  1. The lines are not parallel, so they meet exactly when (a⃗2−a⃗1)⋅(b⃗1×b⃗2)=0(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 0.

  2. a⃗2−a⃗1=2i^+(k+1)j^−2k^\vec{a}_2 - \vec{a}_1 = 2\hat{i} + (k + 1)\hat{j} - 2\hat{k} and b⃗1×b⃗2=3i^−3j^−3k^\vec{b}_1 \times \vec{b}_2 = 3\hat{i} - 3\hat{j} - 3\hat{k}.

  3. 6−3(k+1)+6=9−3k=06 - 3(k + 1) + 6 = 9 - 3k = 0, so k=3k = 3.

Worth noting

−3-3 reads y−ky - k as if the point's yy-coordinate were −k-k.

Question 115

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For L1:r⃗=(i^−k^)+λ(2i^+2j^+k^)L_1: \vec{r} = (\hat{i} - \hat{k}) + \lambda(2\hat{i} + 2\hat{j} + \hat{k}) and L2:r⃗=(2j^+3k^)+μ(i^−2j^+2k^)L_2: \vec{r} = (2\hat{j} + 3\hat{k}) + \mu(\hat{i} - 2\hat{j} + 2\hat{k}), which of the following statements are true?

I. L1L_1 and L2L_2 are perpendicular.

II. L1L_1 and L2L_2 intersect.

III. The shortest distance between L1L_1 and L2L_2 is 4.

IV. b⃗1×b⃗2=6i^+3j^−6k^\vec{b}_1 \times \vec{b}_2 = 6\hat{i} + 3\hat{j} - 6\hat{k}.

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AnswerA · I and III only

  1. b⃗1⋅b⃗2=2−4+2=0\vec{b}_1 \cdot \vec{b}_2 = 2 - 4 + 2 = 0, so I is true; b⃗1×b⃗2=6i^−3j^−6k^\vec{b}_1 \times \vec{b}_2 = 6\hat{i} - 3\hat{j} - 6\hat{k}, so IV is false.

  2. (a⃗2−a⃗1)⋅(b⃗1×b⃗2)=(−i^+2j^+4k^)⋅(6i^−3j^−6k^)=−36≠0(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-\hat{i} + 2\hat{j} + 4\hat{k}) \cdot (6\hat{i} - 3\hat{j} - 6\hat{k}) = -36 \ne 0: the lines do not meet, so II is false, and d=369=4d = \frac{36}{9} = 4, so III is true.

Worth noting

Perpendicular lines need not meet: II fails, as the distance 4 shows.

Question 116

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What is the distance between the lines r⃗=(i^+2j^−k^)+λ(2i^−j^+2k^)\vec{r} = (\hat{i} + 2\hat{j} - \hat{k}) + \lambda(2\hat{i} - \hat{j} + 2\hat{k}) and r⃗=(3i^−j^+k^)+μ(−4i^+2j^−4k^)\vec{r} = (3\hat{i} - \hat{j} + \hat{k}) + \mu(-4\hat{i} + 2\hat{j} - 4\hat{k})?

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AnswerB · 423\frac{4 \sqrt{2}}{3}

  1. −4i^+2j^−4k^=−2(2i^−j^+2k^)-4\hat{i} + 2\hat{j} - 4\hat{k} = -2(2\hat{i} - \hat{j} + 2\hat{k}): the lines are parallel, so use one direction b⃗=2i^−j^+2k^\vec{b} = 2\hat{i} - \hat{j} + 2\hat{k}, with ∣b⃗∣=3\lvert \vec{b} \rvert = 3.

  2. b⃗×(a⃗2−a⃗1)=(2i^−j^+2k^)×(2i^−3j^+2k^)=4i^−4k^\vec{b} \times (\vec{a}_2 - \vec{a}_1) = (2\hat{i} - \hat{j} + 2\hat{k}) \times (2\hat{i} - 3\hat{j} + 2\hat{k}) = 4\hat{i} - 4\hat{k}, of length 424\sqrt{2}.

  3. d=423d = \frac{4\sqrt{2}}{3}.

Worth noting

823\frac{8\sqrt{2}}{3} crosses with −4i^+2j^−4k^-4\hat{i} + 2\hat{j} - 4\hat{k} but divides by the length of 2i^−j^+2k^2\hat{i} - \hat{j} + 2\hat{k}.

Question 117

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What is the shortest distance between the line through A(1,−1,0)A(1, -1, 0) and B(2,1,2)B(2, 1, 2) and the line through C(3,0,1)C(3, 0, 1) and D(5,1,−1)D(5, 1, -1)?

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AnswerB · 11

  1. b⃗1=AB→=i^+2j^+2k^\vec{b}_1 = \overrightarrow{AB} = \hat{i} + 2\hat{j} + 2\hat{k} and b⃗2=CD→=2i^+j^−2k^\vec{b}_2 = \overrightarrow{CD} = 2\hat{i} + \hat{j} - 2\hat{k}; b⃗1×b⃗2=−6i^+6j^−3k^\vec{b}_1 \times \vec{b}_2 = -6\hat{i} + 6\hat{j} - 3\hat{k}, of length 9.

  2. AC→=2i^+j^+k^\overrightarrow{AC} = 2\hat{i} + \hat{j} + \hat{k}, so d=∣−12+6−3∣9=1d = \frac{\lvert -12 + 6 - 3 \rvert}{9} = 1.

Worth noting

9 is the numerator alone; it still has to be divided by ∣b⃗1×b⃗2∣\lvert \vec{b}_1 \times \vec{b}_2 \rvert.

Question 118

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What is the shortest distance between the line r⃗=(3i^−j^+2k^)+λ(2i^+j^−k^)\vec{r} = (3\hat{i} - \hat{j} + 2\hat{k}) + \lambda(2\hat{i} + \hat{j} - \hat{k}) and the zz-axis?

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AnswerB · 5\sqrt{5}

  1. The zz-axis is r⃗=μk^\vec{r} = \mu\hat{k}, so b⃗1×b⃗2=(2i^+j^−k^)×k^=i^−2j^\vec{b}_1 \times \vec{b}_2 = (2\hat{i} + \hat{j} - \hat{k}) \times \hat{k} = \hat{i} - 2\hat{j}, of length 5\sqrt{5}.

  2. a⃗2−a⃗1=−3i^+j^−2k^\vec{a}_2 - \vec{a}_1 = -3\hat{i} + \hat{j} - 2\hat{k}, so d=∣−3−2∣5=55=5d = \frac{\lvert -3 - 2 \rvert}{\sqrt{5}} = \frac{5}{\sqrt{5}} = \sqrt{5}.

Worth noting

5 is the numerator alone; divided by 5\sqrt{5} it gives 5\sqrt{5}.

Question 119

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The shortest distance between the lines r⃗=i^+λ(2i^+2j^+k^)\vec{r} = \hat{i} + \lambda(2\hat{i} + 2\hat{j} + \hat{k}) and r⃗=(i^+pk^)+μ(2i^−j^−2k^)\vec{r} = (\hat{i} + p\hat{k}) + \mu(2\hat{i} - \hat{j} - 2\hat{k}) is 2, where p>0p > 0. What is pp?

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AnswerC · 33

  1. b⃗1×b⃗2=−3i^+6j^−6k^\vec{b}_1 \times \vec{b}_2 = -3\hat{i} + 6\hat{j} - 6\hat{k}, of length 9, and a⃗2−a⃗1=pk^\vec{a}_2 - \vec{a}_1 = p\hat{k}.

  2. d=∣−6p∣9=2p3=2d = \frac{\lvert -6p \rvert}{9} = \frac{2p}{3} = 2, so p=3p = 3.

Worth noting

13\frac{1}{3} sets the numerator 6p6p equal to 2 without dividing by 9.

Question 120

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What is the shortest distance between the lines r⃗=(1+t)i^+(3−2t)j^+2tk^\vec{r} = (1 + t)\hat{i} + (3 - 2t)\hat{j} + 2t\hat{k} and r⃗=(2s−1)i^+2sj^+(s+2)k^\vec{r} = (2s - 1)\hat{i} + 2s\hat{j} + (s + 2)\hat{k}?

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AnswerC · 53\frac{5}{3}

  1. Split each into a point and a direction: r⃗=(i^+3j^)+t(i^−2j^+2k^)\vec{r} = (\hat{i} + 3\hat{j}) + t(\hat{i} - 2\hat{j} + 2\hat{k}) and r⃗=(−i^+2k^)+s(2i^+2j^+k^)\vec{r} = (-\hat{i} + 2\hat{k}) + s(2\hat{i} + 2\hat{j} + \hat{k}).

  2. b⃗1×b⃗2=−6i^+3j^+6k^\vec{b}_1 \times \vec{b}_2 = -6\hat{i} + 3\hat{j} + 6\hat{k}, of length 9, and a⃗2−a⃗1=−2i^−3j^+2k^\vec{a}_2 - \vec{a}_1 = -2\hat{i} - 3\hat{j} + 2\hat{k}.

  3. d=∣12−9+12∣9=159=53d = \frac{\lvert 12 - 9 + 12 \rvert}{9} = \frac{15}{9} = \frac{5}{3}.

Worth noting

57\frac{5}{7} takes the constant parts as the directions and the coefficients of the parameters as the points.

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.