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BITSAT 2023Paper 1
Exam
BITSAT
Year
2023
Paper
Paper 1
Subject
Mathematics
Chapter
Three Dimensional Geometry

BITSAT 2023 · Paper 1 · Mathematics · Three Dimensional Geometry

The shortest distance between the lines...

The shortest distance between the lines x−32=y−23=z−1−1\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1} and x+32=y−61=z−53\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3} is :

Options

  1. A185\frac{18}{\sqrt{5}}
  2. B2235\frac{22}{3 \sqrt{5}}
  3. C4635\frac{46}{3 \sqrt{5}}
  4. D636 \sqrt{3}

Correct answer

A. 185\frac{18}{\sqrt{5}}

Step-by-step solution

Give lines are x−32=y−23=z−1−1\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1} and x+32=y−61=z−53\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}
Lines passes through the points
a⃗1=(3,2,1) and a⃗2=(−3,6,5), \vec{a}_1=(3,2,1) \text{ and } \vec{a}_2=(-3,6,5) \text{, }
b⃗1=2i^+3j^−k^\vec{b}_1=2 \hat{i}+3 \hat{j}-\hat{k}
b⃗1=2i^+j^−3k^,a⃗2−a⃗1=6i^−4j−4k^\vec{b}_1=2 \hat{i}+\hat{j}-3 \hat{k}, \vec{a}_2-\vec{a}_1=6 \hat{i}-4 j-4 \hat{k}
Shortest distance =∣(a⃗2−a⃗1)(b⃗1×b⃗2)∣∣(b⃗1×b⃗2)∣=\frac{\left|\left(\vec{a}_2-\vec{a}_1\right)\left(\vec{b}_1 \times \vec{b}_2\right)\right|}{\left|\left(\vec{b}_1 \times \vec{b}_2\right)\right|}
\vec{b}_1 \times \vec{b}_2=\left|i^j^k^23−1213\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 2 & 1 & 3\end{array}\right|=10 \hat{i}-8 \hat{j}-4 \hat{k}
\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)=60+32+16=108<br/>∣b⃗1×b⃗2∣=100+64+16=180<br />\left|\vec{b}_1 \times \vec{b}_2\right|=\sqrt{100+64+16}=\sqrt{180}
S.D=108180=10865=185S . D=\frac{108}{\sqrt{180}}=\frac{108}{6 \sqrt{5}}=\frac{18}{\sqrt{5}}
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