Question

find the angle between two vector of magnitude 4 unit and 6 unit so that their resultant vector is 8 unit​

Answer 1

$\theta = cos^{ - 1} ( \frac{1}{3} )$

Explanation:

$Given, \: \: |A| = 4, \: |B| = 6, \: and \: \: |AB| = 8 \\ we \: \: know \: \: \boxed{|AB| =|A|.|B|.cos \theta } \\ Now, \: \: \: 8 = 4 \times 6 \times cos \theta \\ \: \: \: \: \: \: \: \: \: \: 8 = 24 \times cos \theta \\ cos \theta = \frac{8}{24} = \frac{1}{3} \\ \: \: \: \: \: \: \: \: \theta = cos^{ - 1} ( \frac{1}{3} )$