Question

The angle between force $\overrightarrow{F}=(3\hat{i}+4\hat{j}-5\hat{k})$ unit and displacement $\overrightarrow{d}=(5\hat{i}+4\hat{j}+3\hat{k})$ unit is

• A. ${\cos }^{-1}\left(0.16\right)$
• B. ${\cos }^{-1}\left(0.32\right)$
• C. ${\cos }^{-1}\left(0.24\right)$
• D. ${\cos }^{-1}\left(0.64\right)$

Answer 1

The correct option is B ${\cos }^{-1}\left(0.32\right)$

Here, $\overrightarrow{F}=3\hat{i}+4\hat{j}-5\hat{k}$ unit

$\overrightarrow{d}=5\hat{i}+4\hat{j}+3\hat{k}$unit

$\therefore |\overrightarrow{F}|=\sqrt{(3{)}^2+(4{)}^2+(-5{)}^2}=\sqrt{50}units$

$|\overrightarrow{d}|=\sqrt{(5{)}^2+(4{)}^2+(3{)}^2}=\sqrt{50}units$

Let $\theta$ be angle between $\overrightarrow{F}$ and $\overrightarrow{d}$

$\therefore \cos \theta =\frac{\overrightarrow{F}\cdot \overrightarrow{d}}{|\overrightarrow{F}||\overrightarrow{d}|}$

$=\frac{(3\hat{i}+4\hat{j}-5\hat{k})\cdot (5\hat{i}+4\hat{j}+3\hat{k})}{\sqrt{50}\sqrt{50}}$

$=\frac{15+16-15}{50}$

$=\frac{16}{50}=0.32$

or $\theta ={\cos }^{-1}\left(0.32\right)$.