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 $(\overrightarrow{a}+\overrightarrow{b}).(-7\hat{i}+2\hat{j}+3\hat{k})$ is

• A. 8
• B. 3
• C. 3
• D. 4

Answer 1

The correct option is A 8

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

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

$\Rightarrow (\overrightarrow{a}+\overrightarrow{b}=\lambda (2\overrightarrow{i}+3\overrightarrow{j}+4\overrightarrow{k}\left)\right)$

Given, $|\overrightarrow{a}+\overrightarrow{b}|=\sqrt{29}$

$\Rightarrow \lambda =1$

So, $\overrightarrow{a}+\overrightarrow{b}=2\overrightarrow{i}+3\overrightarrow{j}+4\overrightarrow{k}$

Now, $(\overrightarrow{a}+\overrightarrow{b})(-7\hat{i}+2\hat{j}+3\hat{k})=4$