The shortest distance between the lines $\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}$ and $\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$ is:
Solution
<p>Given lines are</p>
<p>$\frac{x-3}{4}=\frac{y-(-7)}{-11}=\frac{z-1}{5}$ and</p>
<p>$\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z-(-2)}{1}$</p>
<p>Shortest distance between two lines,</p>
<p>$$\begin{aligned}
& d=\frac{\left|\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)\right|}{\left|\left(\vec{b}_1 \times \vec{b}_2\right)\right|} \\
& \vec{a}_2-\vec{a}_1=2 \hat{i}+16 \hat{j}-3 \hat{k} \text { and } \\
& \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
4 & -11 & 5 \\
3 & -6 & 1
\end{array}\right|=19 \hat{i}+11 \hat{j}+9 \hat{k} \\
& \therefore \quad d=\frac{187}{\sqrt{563}}
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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