Question

Find the projection of $\overrightarrow{A}=2\hat{i}-\hat{j}+\hat{k}$ along the vector $\overrightarrow{B}=\hat{i}+\hat{j}+\hat{k}$.

• A. $\frac{1}{\sqrt{3}}$
• B. $\frac{2}{\sqrt{3}}$
• C. $\frac{4}{\sqrt{3}}$
• D. $0$

Answer 1

The correct option is B $\frac{2}{\sqrt{3}}$

Given, $\overrightarrow{A}=2\hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{B}=\hat{i}+\hat{j}+\hat{k}$
We know that the projection of $\overrightarrow{A}$ along $\overrightarrow{B}$ is given by $\overrightarrow{A}.\hat{B}$, where $\hat{B}$ is the unit vector of $\overrightarrow{B}$.
So, we have
$\hat{B}=\frac{\overrightarrow{B}}{|\overrightarrow{B}|}=\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{1^2+1^2+1^2}}=\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$
Hence, projection is given by
$\overrightarrow{A}.\hat{B}=(2\hat{i}-\hat{j}+\hat{k}).\left(\frac{1}{\sqrt{3}}\right(\hat{i}+\hat{j}+\hat{k}\left)\right)$
$\Rightarrow \frac{2-1+1}{\sqrt{3}}=\frac{2}{\sqrt{3}}$