Question

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

Answer 1

It is given that $\overrightarrow{a}=2\hat{i}+3\hat{j}+2\hat{k}$ and $\overrightarrow{b}=\hat{i}+2\hat{j}+\hat{k}$

As we know that projection of vector $\overrightarrow{a}$ on $\overrightarrow{b}=\frac{1}{\left|\begin{array}{c}\overrightarrow{b}\end{array}\right|}(\overrightarrow{a}.\overrightarrow{b})$
$(\overrightarrow{a}.\overrightarrow{b})=(2\hat{i}+3\hat{j}+2\hat{k}).(\hat{i}+2\hat{j}+\hat{k})$
$(\overrightarrow{a}.\overrightarrow{b})=2+6+2$
$(\overrightarrow{a}.\overrightarrow{b})=10$
Magnitude of $\overrightarrow{b}=\sqrt{1^2+2^2+1^2}$
$\left|\begin{array}{c}\overrightarrow{b}\end{array}\right|=\sqrt{1+4+1}\sqrt{6}$
Projection of vector $\overrightarrow{a}$ on $\overrightarrow{b}=\frac{(\overrightarrow{a}.\overrightarrow{b})}{\left|\begin{array}{c}\overrightarrow{b}\end{array}\right|}$
Projection of vector $\overrightarrow{a}$ on $\overrightarrow{b}=\frac{10}{\sqrt{6}}$
$=\frac{10}{\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}}=\frac{10}{6}\times \sqrt{6}$
$=\frac{5}{3}\times \sqrt{6}$

Final answer:
Hence, the projection of vector $\overrightarrow{a}$ on $\overrightarrow{b}=\frac{5}{3}\times \sqrt{6}$