# If vector a and vector b are two non zero vectors such that find the angle between bar a and bar b

Kshitiz Sharma
26 Points
5 years ago
Let $\theta$ be the angle between vectors A and B.
Let magnitude of non zero vecters A and B be a and b.
Magnitude of Vector A + B = $\sqrt{a^{2} + b^{2} + 2*a*b*\cos (\theta )}$
Magnitude of Vector A – B = $\sqrt{a^{2} + b^{2} - 2*a*b*\cos (\theta )}$.
Using Given Equation we are able to get
a = 2*b. using this in
$\left | A + B \right |$ = $\frac{\left | A - B \right |}{4}$.
We get on simplification
$4*b^{2}+b^{2}+4*b^{2}\cos(\theta)=\frac{4*b^{2}+b^{2}-4*b^{2}\cos(\theta)}{4}$
$4(5+4\cos(\theta)) = 5 - 4\cos(\theta)$
$\frac{4}{1} = \frac{5 - 4\cos(\theta) }{5+4\cos(\theta)}$
Now applying Componendo and Dividendo we get
$\frac{4-1}{4+1} = \frac{5 - 4\cos(\theta)- 5 - 4\cos(\theta) }{5+4\cos(\theta)+5 - 4\cos(\theta)}$
$\frac{3}{5} = -\frac{8*\cos(\theta)}{10}$
$\cos(\theta) = -3/4$
$\theta =\cos^{-1} -3/4.$
Solved as $\theta$ is the angle between vector A and vector B.

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