Find the angle theta between the vectors vec{ | vedclass

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(A) 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}| |\ve

Vedclass · Accepted Answer

$\cos^{-1}\left(-\frac{1}{3}\right)$

Vedclass · Answer

(A) 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 $\vec{a} \cdot \vec{b}$: $\vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (-1)(1) = 1 - 1 - 1 = -1$ Next,calculate the magnitudes $|\vec{a}|$ and $|\vec{b}|$: $|\vec{a}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$ $|\vec{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$ Now,substitute these values into the formula: $\cos 	heta = \frac{-1}{\sqrt{3} \cdot \sqrt{3}} = \frac{-1}{3}$ Therefore,the angle is $	heta = \cos^{-1}\left(-\frac{1}{3}ight)$.

Find the angle $\theta$ between the vectors $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.

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