• # question_answer Let $\vec{\alpha }=3\hat{i}+\hat{j}$and$\vec{\beta }=2\hat{i}-\hat{j}+3\hat{k}.$If$\vec{\beta }={{\vec{\beta }}_{1}}-{{\vec{\beta }}_{2}},$where ${{\vec{\beta }}_{1}}$ is parallel to $\vec{\alpha }$ and ${{\vec{\beta }}_{2}}$ is perpendicular to $\vec{\alpha },$then ${{\vec{\beta }}_{1}}\times {{\vec{\beta }}_{2}}$is equal to:- [JEE Main 9-4-2019 Morning] A) $-3\hat{i}+9\hat{j}+5\hat{k}$                   B) $3\hat{i}-9\hat{j}-5\hat{k}$C) $\frac{1}{2}\left( -3\hat{i}+9\hat{j}+5\hat{k} \right)$      D) $\frac{1}{2}\left( 3\hat{i}-9\hat{j}+5\hat{k} \right)$

$\vec{\alpha }=3\hat{i}+\hat{j}$           $\vec{\beta }=2\hat{i}-\hat{j}+3\hat{k}$           $\vec{\beta }={{\vec{\beta }}_{1}}-{{\vec{\beta }}_{2}}$           ${{\vec{\beta }}_{1}}=\lambda \left( 3\hat{i}+\hat{j} \right),{{\vec{\beta }}_{2}}=\lambda \left( 3\hat{i}+\hat{j} \right)-2\hat{i}+j-3\hat{k}$           ${{\vec{\beta }}_{2}}.\vec{\alpha }=0$           $(3\lambda -2).3+(\lambda +1)=0$           $9\lambda -6+\lambda +1=0$           $\lambda =\frac{1}{2}$           $\Rightarrow$${{\vec{\beta }}_{1}}=\frac{3}{2}\hat{i}+\frac{1}{2}\hat{j}$ $\Rightarrow$${{\vec{\beta }}_{2}}=-\frac{1}{2}\hat{i}+\frac{3}{2}\hat{j}-3\hat{k}$ Now${{\vec{\beta }}_{1}}\times {{\vec{\beta }}_{2}}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ \frac{3}{2} & \frac{1}{2} & 0 \\ -\frac{1}{2} & \frac{3}{2} & -3 \\ \end{matrix} \right|$ $=\hat{i}\left( -\frac{3}{2}-0 \right)-\hat{j}\left( -\frac{9}{2}-0 \right)+\hat{k}\left( \frac{9}{4}+\frac{1}{4} \right)$ $=-\frac{3}{2}\hat{i}+\frac{9}{2}\hat{j}+\frac{5}{2}\hat{k}$ $=\frac{1}{2}\left( -3\hat{i}+9\hat{j}+5\hat{k} \right)$ Aliter: $\vec{\beta }={{\vec{\beta }}_{1}}-{{\vec{\beta }}_{2}}\Rightarrow \vec{\beta }.\hat{\alpha }={{\vec{\beta }}_{1}}.\hat{\alpha }=\left| {{{\vec{\beta }}}_{1}} \right|$ $\Rightarrow$${{\vec{\beta }}_{1}}=\left( \vec{\beta }.\hat{\alpha } \right)\hat{\alpha }$ $\Rightarrow$${{\vec{\beta }}_{2}}=\left( \vec{\beta }.\hat{\alpha } \right)\hat{\alpha }-\vec{\beta }$ $\Rightarrow$${{\vec{\beta }}_{1}}\times {{\vec{\beta }}_{2}}=-\left( \vec{\beta }.\hat{\alpha } \right)\hat{\alpha }\times \vec{\beta }$ $=\frac{-5}{10}\left( 3\hat{i}+\hat{j} \right)\times \left( 2\hat{i}-\hat{j}+3\hat{k} \right)$ $=\frac{1}{2}\left( -3\hat{i}+9\hat{j}+5\hat{k} \right)$