Question

Let $\overrightarrow{a}=2i+2j+k$ and $\overrightarrow{b}$ be another vector such that $\overrightarrow{a}.\overrightarrow{b}=14$ and $\overrightarrow{a}\times \overrightarrow{b}=3i+j-8k$, then the vector $\overrightarrow{b}$ is equal to

• A. $5i+j+2k$
• B. $5i-j-2k$
• C. $5i+j-2k$
• D. $3i+j+4k$

Answer 1

The correct option is A $5i+j+2k$

Let $\overrightarrow{b}=x\hat{i}+y\hat{j}+z\hat{k}$

$\overrightarrow{a}.\overrightarrow{b}=2x+2y+z=14$ ....(i)

$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 2& 1\\ x& y& z\end{array}\right|=(2z-y)\hat{i}-(2z-x)\hat{j}+(2y-2x)\hat{k}$

$\Rightarrow (2z-y)\hat{i}-(2z-x)\hat{j}+(2y-2x)\hat{k}=3\hat{i}+\hat{j}-8\hat{k}$

$2z-y=3$ . ....(ii)

$2z-x=-1$ ...(iii)

$y-x=-4$ ...(iv)

Solving (i), (ii), (iii), (iv), we get

$x=5,y=1,z=2$

$\Rightarrow \overrightarrow{b}=5\hat{i}+\hat{j}+2\hat{k}$