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

Answers (1)

Given:

$\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{b} =3 \hat{i}+4 \hat{j}-\hat{k}$

Assume $\theta$ is angle between $\vec{a}$ and $\vec{b}$ .

$\\ \cos \theta =\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \\ \\ =\frac{(2 \hat{i}-\hat{j}+\hat{k}) \cdot(3 \hat{i}+4 \hat{j}-\hat{k})}{\sqrt{4+1+1} \sqrt{9+16+1}} \\\\ =\frac{6-4-1}{\sqrt{6} \sqrt{26}}=\frac{1}{2 \sqrt{39}} \\\\ \theta =\cos ^{-1} \frac{1}{2 \sqrt{39}}$

Posted by

Ravindra Pindel

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