Mathematics - Vector Question with Solution | TestHub

$\begin{array}{rl}& {\mathrm{L}}_1:\frac{\mathrm{x}+11}{1}=\frac{\mathrm{y}+21}{2}=\frac{\mathrm{z}+29}{3}=\mathrm{a}\\ & {\mathrm{L}}_2:\frac{\mathrm{x}+16}{3}=\frac{\mathrm{y}+11}{2}=\frac{\mathrm{z}+4}{\gamma }=\mathrm{b}\end{array}$ $\begin{array}{rl}& x=a-11=3b-16\Rightarrow a-3b=-\mathrm{5...}(1)\\ & y=2a-21=2b-11\Rightarrow 2a-2b=\mathrm{10...}(2)\\ & z=3a-29=br-4\Rightarrow 3a-b\gamma =\mathrm{25...}(3)\end{array}$ from (1) & (2) $\mathrm{a}=10,\mathrm{b}=5$ Now from (3) $3(10)-5\gamma =25\therefore \gamma =1$ $\begin{array}{rl}& {\mathrm{R}}_1\equiv (-1,-1,1)\\ & {\mathrm{OR}}_1=-\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\\ & \overrightarrow{\mathrm{n}}=\left|\begin{array}{lrr}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 1& 2& 3\\ 3& 2& 1\end{array}\right|=-4\hat{\mathrm{i}}-(-8)\hat{\mathrm{j}}-4\hat{\mathrm{k}}\\ & \overrightarrow{\mathrm{n}}=-4\hat{\mathrm{i}}+8\hat{\mathrm{j}}+4\hat{\mathrm{k}}=-4(\hat{\mathrm{i}}-2\hat{\mathrm{j}}+\hat{\mathrm{k}})\\ & \hat{\mathrm{n}}=\pm \frac{4(\hat{\mathrm{i}}-2\hat{\mathrm{j}}+\hat{\mathrm{k}})}{4\sqrt{6}}=\pm \frac{(\hat{\mathrm{i}}-2\hat{\mathrm{j}}+\hat{\mathrm{k}})}{\sqrt{6}}\\ & {\overrightarrow{\mathrm{OR}}}_1\cdot \hat{\mathrm{n}}=\pm (-\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})\left(\frac{\hat{\mathrm{i}}-2\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{6}}\right)=\pm \frac{2}{\sqrt{6}}=\pm \sqrt{\frac{4}{6}}=\pm \sqrt{\frac{2}{3}}\end{array}$