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A force, $\overrightarrow{\mathrm{F}}=(4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}) \mathrm{N}$ is acting on a body making an angle $\theta$ with the horizontal. Then the angle ' $\theta$ ' is
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Verified Answer
The correct answer is:
$\cos ^{-1}\left(\frac{2 \sqrt{2}}{5}\right)$
$\vec{F}=4 \hat{j}+3 \hat{j}-5 \hat{k}$
$\begin{aligned} & \overrightarrow{\mathrm{F}}=|\mathrm{F}| \cos \theta \\ & \cos \theta=\frac{\mathrm{F}}{|\mathrm{F}|}=\frac{4}{\sqrt{16+9+25}}=\frac{4}{5 \sqrt{2}}=\frac{2 \sqrt{2}}{5} \\ & \theta=\cos ^{-1}\left(\frac{2 \sqrt{2}}{5}\right)\end{aligned}$
$\begin{aligned} & \overrightarrow{\mathrm{F}}=|\mathrm{F}| \cos \theta \\ & \cos \theta=\frac{\mathrm{F}}{|\mathrm{F}|}=\frac{4}{\sqrt{16+9+25}}=\frac{4}{5 \sqrt{2}}=\frac{2 \sqrt{2}}{5} \\ & \theta=\cos ^{-1}\left(\frac{2 \sqrt{2}}{5}\right)\end{aligned}$
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