Question

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

Answer 1

vectorA*vectorB = costheta

Answer 2

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

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

$A.B = |A||B|cos \theta \\ \\ (i + j + k).( - 2i - 2j - 2k) = \sqrt{3} \times 2 \sqrt{3} \times \cos( \theta) \\ \\ - 2 - 2 - 2 = 6 \cos( \theta) \\ \\ - 6 = 6\cos( \theta) \\ \\ \cos( \theta) = \frac{ - 6}{6} = - 1 \\ \\ \cos( \theta) = \cos(180^{o} ) \\ \\ \theta = 180^{o}$


Angle between A and B is 180°.