Question

The component of vector $2\hat{i}-3\hat{j}+2\hat{k}$ perpendicualr to $\hat{i}+\hat{j}+\hat{k}$ is:

• A. $\frac{5}{3}(\hat{i}-2\hat{j}+\hat{k})$
• B. $\frac{1}{3}(\hat{i}+\hat{j}-2\hat{k})$
• C. $\frac{(7\hat{i}-10\hat{j}+7\hat{k})}{3}$
• D. $\frac{(5\hat{i}-8\hat{j}+5\hat{k})}{3}$

Answer 1

The correct option is A $\frac{5}{3}(\hat{i}-2\hat{j}+\hat{k})$

Let, two vectors be $\overrightarrow{a}=2\hat{i}-3\hat{j}+2\hat{k},\overrightarrow{b}=\hat{i}+\hat{j}+\hat{k}$
We know the component of vector of $\overrightarrow{a}$ perpendicular to $\overrightarrow{b}$ can be given by:
$\overrightarrow{c}=\overrightarrow{a}-\left(\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{b}{|}^2}\right)\overrightarrow{b}\cdots \left(i\right)$
Now,
$\overrightarrow{a}.\overrightarrow{b}=(2\hat{i}-3\hat{j}+2\hat{k}).(\hat{i}+\hat{j}+\hat{k})$
$\Rightarrow \overrightarrow{a}.\overrightarrow{b}=2-3+2=1$
Now putting all these values in equation $\left(i\right),$ we will get:
$\overrightarrow{c}=(2\hat{i}-3\hat{j}+2\hat{k})-\frac{1}{(\sqrt{3}{)}^2}\times (\hat{i}+\hat{j}+\hat{k})$
$=(2\hat{i}-3\hat{j}+2\hat{k})-\frac{1}{3}\times (\hat{i}+\hat{j}+\hat{k})$
$=(2-\frac{1}{3})\hat{i}+(-3-\frac{1}{3})\hat{j}+(2-\frac{1}{3})\hat{k}$
$=\frac{5}{3}\hat{i}-\frac{10}{3}\hat{j}+\frac{5}{3}\hat{k}$
$\Rightarrow \overrightarrow{c}=\frac{5}{3}(\hat{i}-2\hat{j}+\hat{k})$