Question
Find the angle between the vectors $2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $3\hat{\text{i}}+4\hat{\text{j}}-\hat{\text{k}}.$

Answer

Let $\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=3\hat{\text{i}}+4\hat{\text{j}}-\hat{\text{k}}$
We know that, angle between two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ is given by
$\cos\theta=\frac{\vec{\text{a}}\cdot\vec{\text{b}}}{|\vec{\text{a}}||\vec{\text{b}}|}$
$=\frac{(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})(3\hat{\text{i}}+4\hat{\text{j}}-\hat{\text{k}})}{\sqrt{4+1+1}\sqrt{9+16+1}}$
$=\frac{6-4-1}{\sqrt{6}\sqrt{26}}=\frac{1}{2\sqrt{39}}$
$\therefore\theta=\cos^{-1}\Big(\frac{1}{2\sqrt{39}}\Big)$

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