Question

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

• A. $co{s}^{-1}(0.16)$
• B. $co{s}^{-1}(0.32)$
• C. $co{s}^{-1}(0.24)$
• D. $co{s}^{-1}(0.64)$

Answer 1

Answer: (B)

Here, $\vec{F}=3\hat{i}+4\hat{j}-5\hat{k}$ unit
$\vec{d}=5\hat{i}+4\hat{j}+3\hat{k}$ unit
$\therefore \left|\vec{F}\right|=\sqrt{{\left(3\right)}^2+{\left(4\right)}^2+{\left(-5\right)}^2}=\sqrt{50}$ unit
$\left|\vec{d}\right|=\sqrt{{\left(5\right)}^2+{\left(4\right)}^2+{\left(3\right)}^2}=\sqrt{50}$ unit
Let $\theta$ be angle between $\vec{F}$ and $\vec{d}$.
$\therefore cos\theta =\frac{\vec{F}.\vec{d}}{|\vec{F}||\vec{d}|}=\frac{\left(3\hat{i}+4\hat{j}-5\hat{k}\right)\cdot \left(5\hat{i}+4\hat{j}+3\hat{k}\right)}{\sqrt{50}\sqrt{50}}$
$=\frac{15+16-15}{50}=\frac{16}{50}=0.32$
or $\theta =co{s}^{-1}\left(0.32\right)$