Mathematics - Vector Question with Solution | TestHub

$\begin{array}{rl}& \frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}\&\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}\\ & \text{ S.D }=\frac{\left|\left({\overline{\mathrm{a}}}_2\cdot {\overline{\mathrm{a}}}_1\right)\cdot \left({\overline{\mathrm{b}}}_1\cdot {\overline{\mathrm{b}}}_2\right)\right|}{\left|{\overline{\mathrm{b}}}_1\times {\overline{\mathrm{b}}}_2\right|}\\ & a_1=3,-15,9\\ & {\mathrm{b}}_1=2,-7,5\\ & {\mathrm{a}}_2=-1,1,9\\ & {\mathrm{b}}_2=2,1,-3\\ & a_2-a_1=-4,16,0\\ & {\overline{\mathrm{b}}}_1\times {\overline{\mathrm{b}}}_2=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 2& -7& 5\\ 2& 1& -3\end{array}\right|=\hat{\mathrm{i}}(16)-\hat{\mathrm{j}}(-16)+\hat{\mathrm{k}}(16)\\ & 16(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\\ & \left|{\overline{\mathrm{b}}}_1\times {\overline{\mathrm{b}}}_2\right|=16\sqrt{3}\\ & \therefore \left({\overline{\mathrm{a}}}_2-{\overline{\mathrm{a}}}_1\right)\cdot \left({\overline{\mathrm{b}}}_1-{\overline{\mathrm{b}}}_2\right)=16[-4+16]=(16)(12)\\ & \text{ S.D. }=\frac{(16)(12)}{16\sqrt{3}}=4\sqrt{3}\\ & \end{array}$