Mathematics › Line and Plane › Line in Space
Consider the lines $\begin{aligned} & \mathrm{L}_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{\mathrm{z}+1}{2} \\ &…
Consider the lines
$\begin{aligned}
& \mathrm{L}_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{\mathrm{z}+1}{2} \\
& \mathrm{~L}_2: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}
\end{aligned}$
then the unit vector perpendicular to both $\mathrm{L}_1$ and $\mathrm{L}_2$ is
$\frac{-\hat{\mathrm{i}}+7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{5 \sqrt{3}}$ $\frac{-\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{5 \sqrt{3}}$ $\frac{+\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{5 \sqrt{3}}$ $\frac{\hat{\mathrm{i}}+7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{5 \sqrt{3}}$ Solution
Lines $L_1$ and $L_2$ are parallel to the vectors
$\overline{\mathrm{b}}_1=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}} \text { and } \overline{\mathrm{b}}_2=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
respectively.
$\therefore$ The unit vector perpendicular to both $\mathrm{L}_1$ and $\mathrm{L}_2$ is $\hat{\mathrm{n}}=\frac{\overline{\mathrm{b}}_1 \times \overline{\mathrm{b}}_2}{\left|\overline{\mathrm{b}}_1 \times \overline{\mathrm{b}}_2\right|}$
Now, $\bar{b}_1 \times \bar{b}_2=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 1 & 2 & 3\end{array}\right|=-\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$
$\therefore \quad \hat{\mathrm{n}}=\frac{1}{5 \sqrt{3}}(-\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})$
Asked in: MHT CET 2023 (14 May Shift 1)
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