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. $4$
• C. $3$
• D. $8$

Answer 1

Answer: (B)

$4$

$\begin{array}{c}\left|\overrightarrow{a}+\overrightarrow{b}\right|=\sqrt{29}\\ \overrightarrow{a}\times \left(2\hat{i}+3\hat{j}+4\hat{k}\right)=\left(2\hat{i}+3\hat{j}+4\hat{k}\right)\times \overrightarrow{b}\\ Also,\\ \left|2\hat{i}+3\hat{j}+4\hat{k}\right|=\sqrt{2^2+3^2+4^2}=\sqrt{29}\end{array}$

Therefore,

$\overrightarrow{a}$ and $\overrightarrow{b}$ both can be $\frac{2\hat{i}+3\hat{j}+4\hat{k}}{2}$ because this will satisfy

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

Then,

$\overrightarrow{a}=\frac{2\hat{i}+3\hat{j}+4\hat{k}}{2}$

$\overrightarrow{b}=\frac{2\hat{i}+3\hat{j}+4\hat{k}}{2}$

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

Now,

$\begin{array}{c}\left(\overrightarrow{a}+\overrightarrow{b}\right)\cdot \left(-7\hat{i}+2\hat{j}+3\hat{k}\right)\\ =\left(2\hat{i}+3\hat{j}+4\hat{k}\right)\cdot \left(-7\hat{i}+2\hat{j}+3\hat{k}\right)\\ =-14+6+12\end{array}=4$

Therefore, possible value is $4$

Hence option $B$ is correct.