Find the angle between the two vectors: vec{a | vedclass

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Question

(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

$\cos^{-1}\left(\frac{26}{\sqrt{1330}}\right)$

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: $\vec{a} \cdot \vec{b} = (3)(5) + (2)(3) + (5)(1) = 15 + 6 + 5 = 26$.Next,calculate the magnitudes of the vectors:$|\vec{a}| = \sqrt{3^2 + 2^2 + 5^2} = \sqrt{9 + 4 + 25} = \sqrt{38}$.$|\vec{b}| = \sqrt{5^2 + 3^2 + 1^2} = \sqrt{25 + 9 + 1} = \sqrt{35}$.Now,substitute these values into the formula:$\cos 	heta = \frac{26}{\sqrt{38} \cdot \sqrt{35}} = \frac{26}{\sqrt{1330}}$.Therefore,$	heta = \cos^{-1}\left(\frac{26}{\sqrt{1330}}ight)$.

Find the angle between the two vectors: $\vec{a}=3 \hat{i}+2 \hat{j}+5 \hat{k}$ and $\vec{b}=5 \hat{i}+3 \hat{j}+\hat{k}$.

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