Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

Let $L_1:\vec{r}=\vec{a}+t\vec{b}$ $\overrightarrow{\mathrm{b}}=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 2& 3& 1\\ 1& 2& 1\end{array}\right|=\hat{\mathrm{i}}(1)-\hat{\mathrm{j}}(1)+\hat{\mathrm{k}}(1)$ Dr's of $L_1:<1,-1,1>$ $\begin{array}{rl}& L_1:\frac{x+1}{1}=\frac{y-0}{-1}=\frac{z-6}{1}\\ & \&L_2:\frac{x-2}{1}=\frac{y+1}{-1}=\frac{z-3}{1}=\lambda \\ & M:2x+y-2z-6=0\end{array}$ Let point on ${\mathrm{L}}_2(\lambda +2,-\lambda -1,\lambda +3)$ $\begin{array}{rl}& 2(\lambda +2)-\lambda -1-2\lambda -6-6=0\\ & 2\lambda +4-3\lambda -13=0\\ & \lambda =-9\\ & \therefore Q(-7,8,-6)\end{array}$
Line PR: $\frac{x-2}{2}=\frac{y+1}{1}=\frac{z-3}{-2}=\mu$ $\mathrm{R}(2\mu +2,\mu -1,-2\mu +3)$ Put in plane $\begin{array}{rl}& 2(2\mu +2)+\mu -1-2(-2\mu +3)-6=0\\ & 4\mu +4+\mu -1+4\mu -6-6=0\\ & 9\mu -9=0\Rightarrow \mu =1\\ & \mathrm{R}(4,0,1)\\ & \mathrm{P}(2,-1,3)\&\mathrm{Q}(-7,8,-6)\\ & \mathrm{PQ}=\sqrt{81+81+81}=9\sqrt{3}\\ & \mathrm{Q}(-7,8,-6)\&\mathrm{R}(4,0,1)\end{array}$ $\mathrm{QR}=\sqrt{121+64+49}=\sqrt{234}$ Area ( $△\mathrm{PQR}$ ) $\begin{array}{rl}& =\frac{1}{2}|\overrightarrow{\mathrm{QP}}\times \overrightarrow{\mathrm{QR}}|\\ & =\frac{1}{2}\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 9& -9& 9\\ 11& -8& 7\end{array}\right|\\ & =\frac{3}{2}\sqrt{234}\\ & \overrightarrow{\mathrm{PQ}}=-9\hat{\mathrm{i}}+9\hat{\mathrm{j}}-9\hat{\mathrm{k}}=-9(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})\\ & \overrightarrow{\mathrm{PR}}=2\hat{\mathrm{i}}+\hat{\mathrm{j}}-2\hat{\mathrm{k}}\end{array}$ $\begin{array}{rl}& \cos \theta =\left|\frac{\overrightarrow{\mathrm{PQ}}\cdot \overrightarrow{\mathrm{PR}}}{\mathrm{PQ}\cdot \mathrm{PR}}\right|\\ & =\frac{9}{9\sqrt{3}\times 3}=\frac{1}{3\sqrt{3}}\\ & \theta ={\cos }^{-1}\left(\frac{1}{3\sqrt{3}}\right)\end{array}$