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Three-Dimensional Geometry · Question 115 of 120

Question 115

HardOriginal question

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}.

Options for question 115

Show the answer and explanation
CorrectNot quiteAnswer A

Correct answer: A

Explanation

  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.

Concept
Facts about skew lines
Typical time
150 s
Your time
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The rule this testsSkew lines and the shortest distance
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Ten questions from Three-Dimensional Geometry, at your own pace.

Practise this chapter →

Ten questions from Three-Dimensional Geometry, at your own pace.