Question

Find the angle between the vectors A=i+j+k and B = 2i-2j-2k.

Answer 1

$A=i+j+k \\ \\ |A| = \sqrt{ {1}^{2} + {1}^{2} +{1}^{2} } = \sqrt{3} \\ \\ B=2i-2j-2k\\ \\ |B|=\sqrt{ {-2}^{2} + {2}^{2} +{2}^{2} } = \sqrt{12} = 2 \sqrt{3}$

$\text{let the angle between A and B is } \theta.$

$A.B = |A||B|cos \theta \\ \\ (i + j + k).( 2i - 2j - 2k) = \sqrt{3} \times 2 \sqrt{3} \times \cos( \theta) \\ \\ 2 - 2 - 2 = 6 \cos( \theta) \\ \\ - 2= 6\cos( \theta) \\ \\ \cos( \theta) = \frac{ - 2}{6} = \frac{-1}{3} \\ \\ \cos( \theta) = \cos(109.47^{o} ) \\ \\ \theta = 109.47^{o}$

Angle between A and B is 109.47°.