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  4. Given that vecA+ vecB= vecR and A2+B2=R2. The angle between A and B is

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Q. Given that $\vec{A}+\vec{B}=\vec{R}$ and $A^{2}+B^{2}=R^{2}$. The angle between $A$ and $B$ is

BITSATBITSAT 2009

Solution:

$cos \theta = \frac{R^{2}- A^{2}-B^{2}}{2 AB} = \frac{R^{2}-R^{2}}{2 AB} =0 $
$ \therefore \theta = \pi/2$