Question

Let two non-collinear vectors $\vec{a}$ and $\vec{b}$ inclined at an angle $\frac{2 \pi}{3}$ be such that $|\vec{a}|=3$ and $|\vec{b}|=2$. If a point $P$ moves so that at any time $t$ its position vector $\overrightarrow{O P}$ (where $O$ is the origin) is given as $\overrightarrow{O P}=\left(t+\frac{1}{t}\right) \vec{a}+\left(t-\frac{1}{t}\right) \vec{b}$ then least distance of $P$ from the origin is

• A. $\sqrt{2 \sqrt{133}-10}$
• B. $\sqrt{2 \sqrt{133}+10}$
• C. $\sqrt{5+\sqrt{133}}$
• D. none of these

Answer 1

Answer: (B)

We have $|\overrightarrow{O P}|^{2}=\left(t+\frac{1}{t}\right)^{2}|\vec{a}|^{2}+\left(t-\frac{1}{t}\right)^{2}|\vec{b}|^{2}$
$+2\left(t^{2}-\frac{1}{t^{2}}\right)|\vec{a} \| \vec{b}| \cos \left(\frac{2 \pi}{3}\right)$
$|\overrightarrow{O P}|^{2}=9\left(t+\frac{1}{t}\right)^{2}+4\left(t-\frac{1}{t}\right)^{2}+2\left(t^{2}-\frac{1}{t^{2}}\right) 3 \cdot 2 \cdot\left(\frac{-1}{2}\right)$
$=9\left(t^{2}+\frac{1}{t^{2}}+2\right)+4\left(t^{2}+\frac{1}{t^{2}}-2\right)-6\left(t^{2}-\frac{1}{t^{2}}\right)$
$=7 t^{2}+\frac{19}{t^{2}}+10$
$\Rightarrow |\overrightarrow{O P}|^{2} \geq 2 \cdot \sqrt{7 t^{2} \cdot \frac{19}{t^{2}}}+10(\because$ A.M. $\geq$ G.M. $)$
$\therefore$ Minimum value of $|\overrightarrow{OP}| = \sqrt{10 + 2\sqrt{133}}$