If the vector $\overrightarrow{b}=3\widehat{j}+4\widehat{k}$ is written as the sum of a vector $\overrightarrow{b_1}$, parallel to $\overrightarrow{a}= \widehat{i}+ \widehat{j}$ and a vector ${\overrightarrow{b}}_2,$ perpendicular to $\overrightarrow{a},$ then ${\overrightarrow{b}}_1\times {\overrightarrow{b}}_2$ is equal to :

• Correct answer
  A

  $6\widehat{i}-6\widehat{j}+\frac{9}{2}\widehat{k}$
• B

  $-3\widehat{i}+3\widehat{j}-9\widehat{k}$
• C

  $-6\widehat{i}+6\widehat{j}-\frac{9}{2}\widehat{k}$
• D

  $3\widehat{i}-3\widehat{j}+9\widehat{k}$

Solution

${\overrightarrow{b}}_1=\frac{\left(\overrightarrow{b} \cdot \overrightarrow{a}\right)\widehat{a}}{\sqrt{2}}$

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

$=\frac{3\left(\widehat{i}+ \widehat{j}\right)}{\sqrt{2}\times \sqrt{2}}=\frac{3\left(\widehat{i}+ \widehat{j}\right)}{2}$

${\overrightarrow{b}}_1+{\overrightarrow{b}}_2= \overrightarrow{b}$

${\overrightarrow{b}}_2= \overrightarrow{b}- {\overrightarrow{b}}_1$

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

${\overrightarrow{b}}_2= -\frac{3}{2}\widehat{i}+\frac{3}{2}\widehat{j}+4\widehat{k}$

${\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ \frac{3}{2}& \frac{3}{2}& 0\\ -\frac{3}{2}& \frac{3}{2}& 4\end{array}\right|$

$$\begin{gathered} \Rightarrow \hat{i}\left(6\right)- \hat{j}\left(6\right)+ \hat{k}\left(\frac{9}{4}+\frac{9}{4}\right) \\ =6\hat{i}-6\hat{j}+\frac{9}{2}\hat{k} \end{gathered}$$

Hence, the correct answer is option: 'A'.