Physics - Vectors Question with Solution | TestHub

Given, $A+B=C$ and $\mathrm{B}=\mathrm{C}-\mathrm{A}$ Since, $\mathbf{C}\perp \mathbf{A}$, therefore $B^2=C^2+A^2$ Also, $|\mathbf{C}|=|\mathbf{A}|$, therefore $B^2=2A^2\Rightarrow B=\sqrt{2}A$ Now, $A^2+B^2+2AB\cos \theta =C^2=A^2$ $\therefore \cos \theta =-\frac{1}{\sqrt{2}}$ This gives $\theta =\frac{3\pi }{4}\mathrm{rad}$