Find the angle between two vectors vec A = 2h | vedclass

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(B) The angle $	heta$ between two vectors $\vec A$ and $\vec B$ is given by the formula $\cos 	heta = \frac{\vec A \cdot \vec B}{|\vec A| |\vec B|}$

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(B) 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: $\vec A \cdot \vec B = (2)(1) + (1)(0) + (-1)(-1) = 2 + 0 + 1 = 3$.Next,calculate the magnitudes: $|\vec A| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$ and $|\vec B| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{1 + 0 + 1} = \sqrt{2}$.Now,substitute these values into the formula: $\cos 	heta = \frac{3}{\sqrt{6} \cdot \sqrt{2}} = \frac{3}{\sqrt{12}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.Since $\cos 	heta = \frac{\sqrt{3}}{2}$,the angle $	heta = 30^o$.

Find the angle between two vectors $\vec A = 2\hat i + \hat j - \hat k$ and $\vec B = \hat i - \hat k$ in degrees.

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