overrightarrow A = 2hat i + 4hat j + 4hat k a | vedclass

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(C) The angle $	heta$ between two vectors $\overrightarrow A$ and $\overrightarrow B$ is given by the dot product formula: $\cos 	heta = \frac{\over

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(C) The angle $	heta$ between two vectors $\overrightarrow A$ and $\overrightarrow B$ is given by the dot product formula: $\cos 	heta = \frac{\overrightarrow A \cdot \overrightarrow B}{|\overrightarrow A| |\overrightarrow B|}$.First,calculate the dot product $\overrightarrow A \cdot \overrightarrow B = (2)(4) + (4)(2) + (4)(-4) = 8 + 8 - 16 = 0$.Since the dot product is $0$,we have $\cos 	heta = 0$.Therefore,$	heta = \cos^{-1}(0) = 90^\circ$.

$\overrightarrow A = 2\hat i + 4\hat j + 4\hat k$ and $\overrightarrow B = 4\hat i + 2\hat j - 4\hat k$ are two vectors. The angle between them will be ........ $^o$.

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