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Three-Dimensional Geometry · Question 110 of 120

Question 110

HardOriginal question

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})?

Options for question 110

Show the answer and explanation
CorrectNot quiteAnswer C

Correct answer: C

Explanation

  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.

Concept
The shortest distance between skew lines
Typical time
120 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.