Question

The projection of $\overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k}$ on the vector $\overrightarrow{b}=\hat{i}+2\hat{j}-\hat{k}$ is?

• A. $\sqrt{\frac{2}{3}}$
• B. $\frac{2}{\sqrt{21}}$
• C. $\sqrt{\frac{3}{2}}$
• D. $\frac{5}{\sqrt{21}}$

Answer 1

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

The given vectors are $\overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k}$ and $\overrightarrow{b}=\hat{i}+2\hat{j}-\hat{k}$

Let us first find the dot product of the given vectors as shown below:

$\overrightarrow{a}\cdot \overrightarrow{b}=(1\times 1)+(2\times 2)+(3\times -1)=1+4-3=5-3=2$

Now, we find the magnitude of the vector $\overrightarrow{b}$ as follows:

$\left|\overrightarrow{b}\right|=\sqrt{1^2+2^2+(-1{)}^2}=\sqrt{1+4+1}=\sqrt{6}$

We know that the projection of $\overrightarrow{a}$ on $\overrightarrow{b}$ is $\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{\left|\overrightarrow{b}\right|}$, therefore, we have:

$\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{\left|\overrightarrow{b}\right|}=\frac{2}{\sqrt{6}}=\frac{2}{\sqrt{2\times 3}}=\frac{2}{\sqrt{2}\times \sqrt{3}}=\frac{\sqrt{2}}{\sqrt{3}}=\sqrt{\frac{2}{3}}$

Hence, the projection of the vector $\overrightarrow{a}$ on $\overrightarrow{b}$ is $\sqrt{\frac{2}{3}}$.