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The shortest distance (in units) between the lines $\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $\overline{\mathrm{r}}=(2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+2 \hat{\mathrm{j}})$ is
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Verified Answer
The correct answer is:
$\frac{8}{3 \sqrt{5}}$
Given lines are: $\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $\frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{0}$
$\therefore \quad$ Required distance
$$
=\mid \frac{\left|\begin{array}{lll}
3 & 0 & 4 \\
3 & 1 & 2 \\
1 & 2 & 0
\end{array}\right|}{\sqrt{(6-1)^2+(0-2)^2+(0-4)^2}}
$$
$\begin{aligned} & =\left|\frac{3(0-4)+0+4(6-1)}{\sqrt{25+4+16}}\right| \\ & =\left|\frac{8}{\sqrt{45}}\right| \\ & =\frac{8}{3 \sqrt{5}}\end{aligned}$
$\therefore \quad$ Required distance
$$
=\mid \frac{\left|\begin{array}{lll}
3 & 0 & 4 \\
3 & 1 & 2 \\
1 & 2 & 0
\end{array}\right|}{\sqrt{(6-1)^2+(0-2)^2+(0-4)^2}}
$$
$\begin{aligned} & =\left|\frac{3(0-4)+0+4(6-1)}{\sqrt{25+4+16}}\right| \\ & =\left|\frac{8}{\sqrt{45}}\right| \\ & =\frac{8}{3 \sqrt{5}}\end{aligned}$
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