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. $0$
• B. $3$
• C. $4$
• D. $8$

Answer 1

Answer: (C)

$4$

We have,

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

$\begin{array}{c}\overrightarrow{c}=2\overline{i}+3\overline{j}+4\overline{k}\\ \overrightarrow{a}\times \overrightarrow{c}=\overrightarrow{c}\times \overrightarrow{b}\\ \overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{b}\times \overrightarrow{c}=0\end{array}$

$\left(\overrightarrow{a}+\overrightarrow{b}\right)\times \overrightarrow{c}=0$

$\overrightarrow{a}+\overrightarrow{b}=\lambda \overrightarrow{c}$

$\sqrt{29}=\lambda \sqrt{29}$

$\begin{array}{c}\lambda =1\\ \overrightarrow{a}+\overrightarrow{b}=\overrightarrow{c}\end{array}$

$\left(2\overline{i}+3\overline{j}+4\overline{k}\right)\left(-7\overline{i}+2\overline{j}+3\overline{k}\right)$

$=-14+6+12$

$=4$

Then,

Option $C$ is correct answer.