The angle between the vectors (hat i + hat j) | vedclass

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Question

(c) $(\hat i + \hat j).(\hat j + \hat k) = 0 + 0 + 1 + 0 = 1$

$\cos 	heta = \frac{{\vec A.\vec B}}{{|\vec A||\vec B|}} = \frac{1}{{\sqrt 2 	imes

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(c) $(\hat i + \hat j).(\hat j + \hat k) = 0 + 0 + 1 + 0 = 1$

$\cos 	heta = \frac{{\vec A.\vec B}}{{|\vec A||\vec B|}} = \frac{1}{{\sqrt 2 	imes \sqrt 2 }} = \frac{1}{2}$ 

$	heta = 60^\circ $

The angle between the vectors $(\hat i + \hat j)$ and $(\hat j + \hat k)$ is ....... $^o$

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