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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Aicharya · Accepted Answer

Given lines are: $\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $\frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{0}$ $	herefore \quad$ Required distance $=\mid \frac{\left|\begin{array}{lll} 3 & 0 & 4 \ 3 & 1 & 2 \ 1 & 2 & 0 \end{array}ight|}{\sqrt{(6-1)^2+(0-2)^2+(0-4)^2}}$ $\begin{aligned} & =\left|\frac{3(0-4)+0+4(6-1)}{\sqrt{25+4+16}}ight| \ & =\left|\frac{8}{\sqrt{45}}ight| \ & =\frac{8}{3 \sqrt{5}}\end{aligned}$

The shortest distance (in units) between the lines $\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and…

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