Question

Component of vector $\overrightarrow{A}=3\hat{i}+2\hat{j}+\hat{k}$ perpendicular to vector $\overrightarrow{B}=\hat{i}-2\hat{j}$ is

• A. $\frac{16}{5}\hat{i}+\frac{8}{5}\hat{j}+\hat{k}$
• B. $5\hat{i}+2\hat{j}+\hat{k}$
• C. $\hat{k}$
• D. $\frac{-3}{5}\hat{i}+\frac{2}{5}\hat{j}-\hat{k}$

Answer 1

The correct option is A $\frac{16}{5}\hat{i}+\frac{8}{5}\hat{j}+\hat{k}$


$\overrightarrow{C}=\overrightarrow{A}-Acos\theta \hat{B}=\overrightarrow{A}-\frac{\overrightarrow{A}.\overrightarrow{B}}{B}\times \frac{\overrightarrow{B}}{B}=\overrightarrow{A}-\frac{\overrightarrow{A}.\overrightarrow{B}}{B^2}\overrightarrow{B}$
$\overrightarrow{A}.\overrightarrow{B}=3-4=-1$, $B=\sqrt{5}$
$\overrightarrow{C}=3\hat{i}+2\hat{j}+\hat{k}-\frac{(-1)}{5}(\hat{i}-2\hat{j})=3\hat{i}+2\hat{j}+k+\frac{\hat{i}}{5}-\frac{2}{5}\hat{j}=\frac{16}{5}\hat{i}+\frac{8}{5}\hat{j}+k$