Question:
The shortest distance between the lines
$\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and
$\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4} \mathrm{is}:$
Correct Option: , 3
Solution:
$\overrightarrow{A B}=6 \hat{i}+15 \hat{j}+3 \hat{k}$
$\vec{p}=\hat{i}+4 \hat{j}+22 \hat{k}$
$\vec{q}=\hat{i}+\hat{j}+7 \hat{k}$
$\vec{p} \times \vec{q}=\left|\begin{array}{rrr}i & j & k \\ 1 & 4 & 22 \\ 1 & 1 & 7\end{array}\right|=6 \hat{i}+15 \hat{j}-3 \hat{k}$
Shortest distance between the lines is
$=\frac{|\overrightarrow{A B} \cdot(\vec{p} \times \vec{q})|}{|\vec{p} \times \vec{q}|}=\frac{|36+225+9|}{\sqrt{36+225+9}}=3 \sqrt{30}$
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