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

Abhijeet sinha , 7 Years ago
Grade 12th pass
anser 1 Answers
Kshitiz Sharma

Last Activity: 7 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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