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:

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Aicharya · Accepted Answer

$\begin{aligned} & \overrightarrow{\mathrm{n}}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}} \\ & \overrightarrow{\mathrm{n}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 4 & -11 & 5 \\ 3 & -6 & 1\end{array}\right|=19 \hat{\mathrm{i}}+11 \hat{\mathrm{j}}+9 \hat{\mathrm{k}} \\ & \text { S.d. }=\text { projection of } \overrightarrow{\mathrm{AB}} \text { on } \overrightarrow{\mathrm{n}} \\ & =\left|\frac{\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{n}}}{|\overrightarrow{\mathrm{n}}|}\right|=\left|\frac{(2 \hat{\mathrm{i}}+16 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}) \cdot(19 \hat{\mathrm{i}}+11 \hat{\mathrm{j}}+9 \hat{\mathrm{k}})}{\sqrt{361+121+81}}\right| \\ & =\frac{38+176-27}{\sqrt{563}} \\ & \text { S.d. }=\frac{187}{\sqrt{563}}\end{aligned}$

The shortest distance between the lines $\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}$ and…

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