Let $\mathrm{L}_1$ $\frac{x+1}{3}=\frac{y+2}{2}=\frac{z+1}{1}$ and $L_2: \frac{x-2}{2}=\frac{y+2}{1}=\frac{z-3}{3}$ be the given lines. Then the unit vector perpendicular to $L_1$ and $L_2$ is

Question

Aicharya · Accepted Answer

Lines $\mathrm{L}_1$ and $\mathrm{L}_2$ are parallel to the vectors $\begin{aligned} & \overline{\mathrm{b}}_1=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \ & \overline{\mathrm{~b}}_2=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}} 	ext { respectively. } \end{aligned}$ $	herefore \quad$ The unit vector perpendicular to both $\mathrm{L}_1$ and $\mathrm{L}_2$ is $\hat{n}=\frac{\bar{b}_1 	imes \bar{b}_2}{\left|\bar{b}_1 	imes \bar{b}_2ight|}$ Now, $\begin{aligned} \overline{\mathrm{b}}_1 	imes \overline{\mathrm{b}}_2 & =\left|\begin{array}{lll}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \ 3 & 2 & 1 \ 2 & 1 & 3\end{array}ight| \ & =5 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-\hat{\mathrm{k}}\end{aligned}$ $	herefore \quad \hat{\mathrm{n}}=\frac{5 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-\hat{\mathrm{k}}}{5 \sqrt{3}}$

Let $\mathrm{L}_1$ $\frac{x+1}{3}=\frac{y+2}{2}=\frac{z+1}{1}$ and $L_2:…

Solution