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Q.
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
Work, Energy and Power
Solution:
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 $\vec{d}$.
$\therefore cos\, \theta= \frac{ \vec{F}. \vec{d}}{\left| \vec{F}\right|\left| \vec{d}\right|}=\frac{\left(3\hat{i}+4\hat{j}-5\hat{k}\right)\cdot\left(5\hat{i}+4\hat{j}+3\hat{k}\right)}{\sqrt{50}\,\sqrt{50}}$
$= \frac{15+16-15}{50}=\frac{16}{50}=0.32$
or $\theta = cos^{-1}\,\left(0.32\right)$