question_answer

If the vector \[\vec{b}=3\hat{j}+4\hat{k}\] is written as the sum of a vector \[{{\vec{b}}_{1}},\] parallel to  \[\vec{a}=\hat{i}+\hat{j}\] and a vector \[{{\vec{b}}_{2}}\] perpendicular to \[{{\vec{a}}_{1}}\] then \[{{\vec{b}}_{1}}\times {{\vec{b}}_{2}}\] is equal to         [JEE Online 09-04-2017]

A)  \[6\hat{i}-6\hat{j}+\frac{9}{2}\hat{k}\]

B)  \[-3\hat{i}+3\hat{j}-9\hat{k}\]

C)  \[-6\hat{i}+6\hat{j}-\frac{9}{2}\hat{k}\]              

D)  \[3\hat{i}-3\hat{j}+9\hat{k}\]

Correct Answer: A

Solution :

                \[{{\vec{b}}_{1}}\,=\frac{(\vec{b}.\vec{a})\hat{a}}{1}\]                 \[=\left\{ \frac{(3\hat{j}+4\hat{k}).(\hat{i}+\hat{j})}{\sqrt{2}} \right\}\,\left( \frac{\hat{i}+\hat{j}}{\sqrt{2}} \right)\] \[=\frac{3(\hat{i}+\hat{j})}{\sqrt{2}\times \sqrt{2}}\,=\frac{3(\hat{i}+\hat{j})}{2}\] \[{{\vec{b}}_{1}}+{{\vec{b}}_{2}}=\vec{b}\] \[{{\vec{b}}_{2}}=\vec{b}-{{\vec{b}}_{1}}\] \[=\left( 3\hat{i}+4\hat{k} \right)\,-\frac{3}{2}\,(\hat{i}+\hat{j})\] \[\] \[{{b}_{1}}\times {{b}_{2}}\,=\left| \begin{matrix}    i & j & k  \\    \frac{3}{2} & \frac{3}{2} & 0  \\    -\frac{3}{2} & \frac{3}{2} & 4  \\ \end{matrix} \right|\]