MCQ
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 $\vec a$ , then ${\vec {b_1}} \times {\vec {b_2}}$ is equal to
  • A
    $ - 3\hat i + 3\hat j - 9\hat k$
  • $  6\hat i - 6\hat j + \frac{9}{2}\hat k$
  • C
    $ - 6\hat i + 6\hat j - \frac{9}{2}\hat k$
  • D
    $3\hat i - 3\hat j + 9\hat k$

Answer

Correct option: B.
$  6\hat i - 6\hat j + \frac{9}{2}\hat k$
b
$\overrightarrow{\mathrm{b}_{1}}=\frac{(\overrightarrow{\mathrm{b}_{1}} \cdot \overrightarrow{\mathrm{a}}) \hat a}{1}$

$=\left\{\frac{(3 \hat{j}+4 \hat{k}) \cdot(\hat{i}+\hat{j})}{\sqrt{2}}\right\}\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right)$

$=-\frac{3(\hat{i}+\hat{j})}{\sqrt{2} \times \sqrt{2}}=\frac{3(\hat{i}+\hat{j})}{2}$

$\overrightarrow{\mathrm{b}_{1}}+\overrightarrow{\mathrm{b}_{2}}=\overrightarrow{\mathrm{b}}$

$\overrightarrow{b_{2}}=\vec{b}-\overrightarrow{b_{1}}$

$ = \left( {\left. {3\hat j + 4\hat k - \frac{3}{2}} \right)(\hat i + \hat j)} \right.$

$\boxed{\overrightarrow {{b_2}}  =  - \frac{3}{2}\widehat {\text{i}} + \frac{3}{2}\widehat {\text{j}} + 4\widehat {\text{k}}}$

$\overrightarrow {{b_1}}  \times \overrightarrow {{b_2}}  = \begin{array}{*{20}{c}}
{\widehat i}&{\widehat j}&{\widehat k}\\
{\frac{3}{2}}&{\frac{3}{2}}&0\\
{ - \frac{3}{2}}&{\frac{3}{2}}&4
\end{array}$

$\overrightarrow {{b_1}}  \times \overrightarrow {{b_2}}  = \widehat {\rm{i}}(6) - \widehat {\rm{j}}({\rm{6}}) + \widehat {\rm{k}}\left( { - \frac{9}{4} + \frac{9}{4}} \right)$

$\Rightarrow 6 \hat{i}-6 \hat{j}+\frac{9}{2} \hat{k}$

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