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Three-Dimensional Geometry · Question 56 of 120

Question 56

HardOriginal question

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}?

Options for question 56

Show the answer and explanation
CorrectNot quiteAnswer A

Correct answer: A

Explanation

  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.

Concept
A line perpendicular to two lines
Typical time
120 s
Your time
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The rule this testsThe equation of a line
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Ten questions from Three-Dimensional Geometry, at your own pace.

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