Q69 / 100 · BITSAT 2023 (Memory Based Paper 1) · Mathematics
The shortest distance between the lines $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}$ and $\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$ is :
- Correct answerA$\frac{18}{\sqrt{5}}$
- B$\frac{22}{3 \sqrt{5}}$
- C$\frac{46}{3 \sqrt{5}}$
- D$6 \sqrt{3}$
Solution
Give lines are $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}$ and $\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$
Lines passes through the points
$\vec{a}_1=(3,2,1) \text { and } \vec{a}_2=(-3,6,5) \text {, }$
$\vec{b}_1=2 \hat{i}+3 \hat{j}-\hat{k}$
$\vec{b}_1=2 \hat{i}+\hat{j}-3 \hat{k}, \vec{a}_2-\vec{a}_1=6 \hat{i}-4 j-4 \hat{k}$
Shortest distance $=\frac{\left|\left(\vec{a}_2-\vec{a}_1\right)\left(\vec{b}_1 \times \vec{b}_2\right)\right|}{\left|\left(\vec{b}_1 \times \vec{b}_2\right)\right|}$
$\vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 2 & 1 & 3\end{array}\right|=10 \hat{i}-8 \hat{j}-4 \hat{k}$
$\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)=60+32+16=108$
$\left|\vec{b}_1 \times \vec{b}_2\right|=\sqrt{100+64+16}=\sqrt{180}$
$S . D=\frac{108}{\sqrt{180}}=\frac{108}{6 \sqrt{5}}=\frac{18}{\sqrt{5}}$
Lines passes through the points
$\vec{a}_1=(3,2,1) \text { and } \vec{a}_2=(-3,6,5) \text {, }$
$\vec{b}_1=2 \hat{i}+3 \hat{j}-\hat{k}$
$\vec{b}_1=2 \hat{i}+\hat{j}-3 \hat{k}, \vec{a}_2-\vec{a}_1=6 \hat{i}-4 j-4 \hat{k}$
Shortest distance $=\frac{\left|\left(\vec{a}_2-\vec{a}_1\right)\left(\vec{b}_1 \times \vec{b}_2\right)\right|}{\left|\left(\vec{b}_1 \times \vec{b}_2\right)\right|}$
$\vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 2 & 1 & 3\end{array}\right|=10 \hat{i}-8 \hat{j}-4 \hat{k}$
$\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)=60+32+16=108$
$\left|\vec{b}_1 \times \vec{b}_2\right|=\sqrt{100+64+16}=\sqrt{180}$
$S . D=\frac{108}{\sqrt{180}}=\frac{108}{6 \sqrt{5}}=\frac{18}{\sqrt{5}}$
Hence, the correct answer is option: 'A'.