Question
Find angle between $\vec A = 3\hat i - \hat j + 4\hat k$ and $Z-$ axis
✓
Answer
$\cos \theta=\frac{\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}}{\mathrm{AB}}=\frac{(3 \hat{\mathrm{i}}-\hat{j}+4 \hat{\mathrm{k}}) \cdot(\hat{\mathrm{k}})}{\sqrt{(3)^{2}+(-1)^{2}+(4)^{2}} \sqrt{(1)^{2}}}=\frac{4}{\sqrt{26}}$
Base $=4,$ hypotenuse $=\sqrt{26} \cdot$ perpendicular $=\sqrt{10}$
$\tan \theta=\frac{\sqrt{10}}{4}, \quad \theta=\tan ^{-1}\left(\frac{\sqrt{10}}{4}\right)$
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