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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Mathematics
Chapter
Three Dimensional Geometry

BITSAT 2023 · Paper 2 · Mathematics · Three Dimensional Geometry

The shortest distance between the lines x=y+2=6 z-6 and x+1=2 y=-12...

The shortest distance between the lines x=y+2=6z−6\mathrm{x}=\mathrm{y}+2=6 z-6 and x+1=2y=−12zx+1=2 y=-12 z is

Options

  1. A12\frac{1}{2}
  2. B
    2
  3. C
    1
  4. D32\frac{3}{2}

Correct answer

B.
2

Step-by-step solution

The lines are x6=y+26=z−11\frac{x}{6}=\frac{y+2}{6}=\frac{z-1}{1}and x+112=y6=z−1\frac{x+1}{12}=\frac{y}{6}=\frac{z}{-1}Here,a⃗1=−2j^+k^,b1+6i^+6j^+k^,a⃗2=−i^,b⃗2=12i^+6j^−k^b⃗1×b⃗2=∣i^j^k661126−1∣=−12i^+18j^−36k^ Shortest distance =∣(a⃗2−a⃗1)⋅(b⃗1−b⃗2)∣∣b⃗1×b⃗2∣\begin{aligned} \\ & \vec{a}_1=-2 \hat{j}+\hat{k}, b_1+6 \hat{i}+6 \hat{j}+\hat{k}, \vec{a}_2=-\hat{i}, \\ \\ & \vec{b}_2=12 \hat{i}+6 \hat{j}-\hat{k} \\ \\ & \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc} \\ \hat{i} & \hat{j} & k \\ \\ 6 & 6 & 1 \\ \\ 12 & 6 & -1 \\ \end{array}\right|=-12 \hat{i}+18 \hat{j}-36 \hat{k} \\ \\ & \text{ Shortest distance }=\frac{\left|\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1-\vec{b}_2\right)\right|}{\left|\vec{b}_1 \times \vec{b}_2\right|} \\ \end{aligned}=∣(−i^+2j^−k^)⋅(−12i+18j^−36k^)∣(−12)2+(18)2+(−36)2=∣+12+36+36∣1764=8442=2\begin{aligned} & =\frac{|(-\hat{i}+2 \hat{j}-\hat{k}) \cdot(-12 i+18 \hat{j}-36 \hat{k})|}{\sqrt{(-12)^2+(18)^2+(-36)^2}} \\ & =\frac{|+12+36+36|}{\sqrt{1764}}=\frac{84}{42}=2\end{aligned}$
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