Question

If $\overrightarrow{A}$ = 2 $\hat{i}$ + 3$\hat{j}$ - $\hat{k}$ and $\overrightarrow{B}$ = -$\hat{i}$ + 3$\hat{j}$ + 4$\hat{k}$ then find the projection of

$\overrightarrow{A}$ on $\overrightarrow{B}$

• A.
• B.
• C. 25
• D. None of these

Answer 1

The correct option is A

$\overrightarrow{A}$ = 2 $\hat{i}$ + 3$\hat{j}$ - $\hat{k}$

$\overrightarrow{B}$ = -$\hat{i}$ + 3$\hat{j}$ + 4$\hat{k}$

Component of $\overrightarrow{A}$ along $\overrightarrow{B}$ is |A|cos$\theta$ is the angle between vector $\overrightarrow{A}$ & $\overrightarrow{B}$

We know $\overrightarrow{A}$.$\overrightarrow{B}$ = |A||B| cos$\theta$ ------------(1)

We need |A| cos$\theta$

Dividing $e{q}^n$ (1) by |B|

We get $\frac{\overrightarrow{A}.\overrightarrow{B}}{|B|}=|A|cos\theta$

$\overrightarrow{A}$.$\overrightarrow{B}$ = $(2i+3j-\hat{k})(-\hat{i}+3\hat{j}+4\hat{k})$

= (-2 + 9 - 4) = 3

|B| =$\sqrt{(-1{)}^2+9+16}$

|A|cos$\theta =\frac{3}{\sqrt{26}}$