Question

If a body placed at the origin is acted upon by a force $\overrightarrow{F}$ = $(\hat{i}+\hat{j}+\sqrt{2}\hat{k})$, then the following statements are correct?
1. Magnitude of $\overrightarrow{F}$ is $(2+\sqrt{2})$.
2. Magnitude of $\overrightarrow{F}$ is 2.
3. $\overrightarrow{F}$ makes an angle of ${45}^0$ with the Z-axis.
4. $\overrightarrow{F}$ makes an angle of ${30}^0$ with the Z-axis.
Select the correct answer using the codes given below.

• A. 1 and 3
• B. 2 and 3
• C. 1 and 4
• D. 2 and 4

Answer 1

The correct option is B 2 and 3

Here,

$\overrightarrow{F}=\hat{i}+\hat{j}+\sqrt{2}\hat{k}$

$|\overrightarrow{F}|=\sqrt{1^2+1^2+{\sqrt{2}}^2}$

$|\overrightarrow{F}|=2$

Thus$\hat{F}=\frac{\underset{F}{\to }}{|\overrightarrow{F}|}$

$\hat{F}=\frac{1}{2}\hat{i}+\frac{1}{2}\hat{j}+\frac{\sqrt{2}}{2}\hat{k}$ --(1)

Let $\alpha ,\beta and\gamma$ are angled by force F from x,y and z-axis respectively. Then

$\hat{F}=cos\alpha \hat{i}+cos\beta \hat{j}+cos\gamma \hat{k}$ -- (2)

Comparing equations (1) and (2) , we get

$cos\alpha =\frac{1}{2},so\alpha ={60}^0$

$cos\beta =\frac{1}{2},so\beta ={60}^0$

$cos\gamma =\frac{1}{\sqrt{2}},so\gamma ={45}^0$

Thus statement 2 and 3 are correct.

Correct option is C.