Question

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are vectors such that $|\overrightarrow{a}+\overrightarrow{b}|=\sqrt{29}$ and $\overrightarrow{a}\times (2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times \overrightarrow{b}$, then a possible value of $(\overrightarrow{a}+\overrightarrow{b})\cdot (-7\hat{i}+2\hat{j}+3\hat{k})$ is

• A. $0$
• B. $3$
• C. $4$
• D. $8$

Answer 1

The correct option is C $4$

$\overrightarrow{a}\times (2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times \overrightarrow{b}$

$\Rightarrow \overrightarrow{a}\times (2\hat{i}+3\hat{j}+4\hat{k})=-\overrightarrow{b}\times (2\hat{i}+3\hat{j}+4\hat{k})$

$\Rightarrow (\overrightarrow{a}+\overrightarrow{b})\times (2\hat{i}+3\hat{j}+4\hat{k})=0$

$\Rightarrow$ Unit vector in the direction of $\overrightarrow{a}+\overrightarrow{b}$ is $\pm \frac{2\hat{i}+3\hat{j}+4\hat{k}}{\sqrt{29}}$

Also given that $\mid \overrightarrow{a}+\overrightarrow{b}\mid =\sqrt{29}$

$\therefore \overrightarrow{a}+\overrightarrow{b}=\sqrt{29}\times [\pm \frac{(2\hat{i}+3\hat{j}+4\hat{k})}{\sqrt{29}}]$
$=\pm (2\hat{i}+3\hat{j}+4\hat{k})$

Hence $(\overrightarrow{a}+\overrightarrow{b})\cdot (-7\hat{i}+2\hat{j}+2\hat{k})$

$=\pm (-14+6+12)=\pm 4$