Find the angle between the vectors 2 hat{i}-h | vedclass

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(A) Let $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = 3\hat{i} + 4\hat{j} - \hat{k}$.The angle $	heta$ between two vectors $\vec{a}$ and $\

Vedclass · Accepted Answer

$\cos^{-1}\left(\frac{1}{2\sqrt{39}}\right)$

Vedclass · Answer

(A) Let $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = 3\hat{i} + 4\hat{j} - \hat{k}$.The angle $	heta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by the formula $\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.First,calculate the dot product: $\vec{a} \cdot \vec{b} = (2)(3) + (-1)(4) + (1)(-1) = 6 - 4 - 1 = 1$.Next,calculate the magnitudes: $|\vec{a}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$ and $|\vec{b}| = \sqrt{3^2 + 4^2 + (-1)^2} = \sqrt{9 + 16 + 1} = \sqrt{26}$.Substitute these values into the formula: $\cos 	heta = \frac{1}{\sqrt{6} \cdot \sqrt{26}} = \frac{1}{\sqrt{156}} = \frac{1}{\sqrt{4 	imes 39}} = \frac{1}{2\sqrt{39}}$.Therefore,$	heta = \cos^{-1}\left(\frac{1}{2\sqrt{39}}ight)$.

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

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