The angle between the two vectors vec A = 3ha | vedclass

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Question

(b) $\cos 	heta = \frac{{\overrightarrow A .\overrightarrow B }}{{|A||B|}} = \frac{{9 + 16 + 25}}{{\sqrt {9 + 16 + 25} \sqrt {9 + 16 + 25} }}$

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Vedclass · Accepted Answer

$0$

Vedclass · Answer

(b) $\cos 	heta = \frac{{\overrightarrow A .\overrightarrow B }}{{|A||B|}} = \frac{{9 + 16 + 25}}{{\sqrt {9 + 16 + 25} \sqrt {9 + 16 + 25} }}$

$=\frac{{50}}{{50}} = 1ightarrow \cos 	heta = 1$

$	herefore 	heta = {\cos ^{ - 1}}(1)$

The angle between the two vectors $\vec A = 3\hat i + 4\hat j + 5\hat k$ and $\vec B = 3\hat i + 4\hat j + 5\hat k$ is ....... $^o$

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