Mathematics - Vector Question with Solution | TestHub

$\begin{array}{rl}& {\mathrm{L}}_1:\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-0}{2}\\ & \text{ Let }\mathrm{Q}(\lambda +1,\lambda +2,\lambda )\\ & \overrightarrow{\mathrm{PQ}}=(\lambda -4,\lambda -1,\lambda +3)\\ & \overrightarrow{\mathrm{PQ}}\cdot \overrightarrow{\mathrm{m}}=0\end{array}$

$\begin{array}{rl}\Rightarrow & \lambda -4+\lambda +1,\lambda +3=0\\ \Rightarrow & 3\lambda =0\\ & \lambda =0\end{array}$
$\begin{array}{rl}& L_2:\frac{x-2}{1}=\frac{y-0}{1}=\frac{z-1}{2}\\ & \text{ Let }\mathrm{R}(\mu +2,\mu ,\mu +1)\overrightarrow{\mathrm{PR}}=(\mu -3,\mu -1,\mu +4)\\ & \overrightarrow{\mathrm{PR}}\cdot \overrightarrow{\mathrm{n}}=0\\ & \mu -3+\mu -1+\mu +4=0\\ & \ne \mu =0\end{array}$
$\begin{array}{rl}& \text{ Area of }△\mathrm{PQR}(\mathrm{A})=\frac{1}{2}|\overrightarrow{\mathrm{PQ}}\times \overrightarrow{\mathrm{PR}}|\\ & \mathrm{A}=\frac{1}{2}|(-4\hat{\mathrm{i}}+\hat{\mathrm{j}}+3\hat{\mathrm{k}})\times (-3\hat{\mathrm{i}}+\hat{\mathrm{j}}+4\hat{\mathrm{k}})|\\ & \mathrm{A}=\frac{1}{2}|7(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})|\\ & \left|\begin{array}{crr}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ -4& 1& 3\\ -3& -1& 4\end{array}\right|\\ & =7\hat{\mathrm{i}}+7\hat{\mathrm{j}}+7\hat{\mathrm{k}}\\ & 4{\mathrm{A}}^2=49\times 3=147\end{array}$