The angle between the two vectors (2hat{i} + | vedclass

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(D) Let $\vec{A} = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{B} = -3\hat{i} + 0\hat{j} + 6\hat{k}$.The angle $	heta$ between two vectors is given by $

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$90$

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(D) Let $\vec{A} = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{B} = -3\hat{i} + 0\hat{j} + 6\hat{k}$.The angle $	heta$ between two vectors is given by $\cos	heta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|}$.Calculate the dot product: $\vec{A} \cdot \vec{B} = (2)(-3) + (3)(0) + (1)(6) = -6 + 0 + 6 = 0$.Since the dot product is $0$,$\cos	heta = 0$,which implies $	heta = 90^\circ$.

The angle between the two vectors $(2\hat{i} + 3\hat{j} + \hat{k})$ and $(-3\hat{i} + 6\hat{k})$ is ...... $^\circ$.

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