NIMCET 2016 — Mathematics PYQ

Magnitude of sum of $\vec{a}$ and $\vec{b}$:
$|\vec{a}+\vec{b}|=\sqrt{a^2+b^2+2ab\cos \theta }$

Magnitude of difference of $\vec{a}$ and $\vec{b}$:
$|\vec{a}-\vec{b}|=\sqrt{a^2+b^2-2ab\cos \theta }$

where $a$, $b$ are magnitude of vectors $\vec{a}$ and $\vec{b}$; and $\theta$ is angle between them.

\begin{aligned}
 & \text{Calculation:} \\
 & \mathrm{Given} \\
 & |{\vec{\mathrm{a}}}|=\alpha,\left|{\vec{\mathrm{b}}}\right|=\alpha\mathrm{~and~}\left|{\vec{\mathrm{a}}}+{\vec{\mathrm{b}}}\right|=\alpha \\
 & \mathrm{As~we~know,} \\
 & \left|\mathrm{\vec{a}}+\mathrm{\vec{b}}\right|=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+2\mathrm{ab}\cos\theta} \\
 & \Rightarrow\alpha=\sqrt{\alpha^{2}+\alpha^{2}+2(\alpha)(\alpha)\cos\theta} \\
 & \Rightarrow\alpha^{2}=2\alpha^{2}+2\alpha^{2}\cos\theta \\
 & \Rightarrow-1=2\cos\theta \\
 & \Rightarrow\cos\theta=-\frac{1}{2} \\
 & \mathrm{Now}, \\
 & \left|\mathrm{\vec{a}}-\mathrm{\vec{b}}\right|=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}-2\mathrm{ab}\cos\theta} \\
 & \Rightarrow\left|\mathrm{\vec{a}}-\mathrm{\vec{b}}\right|=\sqrt{\alpha^{2}+\alpha^{2}-2(\alpha)(\alpha)\cos\theta} \\
 & \because\cos\theta=-\frac{1}{2} \\
 & \Rightarrow\left|\mathrm{\vec{a}}-\mathrm{\vec{b}}\right|=\sqrt{2\alpha^{2}-2\alpha^{2}(\frac{-1}{2})} \\
 & \Rightarrow\left|\mathrm{\vec{a}}-\mathrm{\vec{b}}\right|=\sqrt{2\alpha^{2}+\alpha^{2}} \\
 & \Rightarrow\left|\mathbf{\vec{a}}-\mathbf{\vec{b}}\right|=\sqrt{3}\alpha}
\end{aligned}