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 $\overrightarrow{OP}$ (where $O$ is the origin) is given as $\overrightarrow{OP}=\left(t+\frac{1}{t}\right)\vec{a}+\left(t-\frac{1}{t}\right)\vec{b}$ 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 $|\overrightarrow{OP}{|}^2={\left(t+\frac{1}{t}\right)}^2|\vec{a}{|}^2+{\left(t-\frac{1}{t}\right)}^2|\vec{b}{|}^2$
$+2\left(t^2-\frac{1}{t^2}\right)|\vec{a}\Vert \vec{b}|\cos \left(\frac{2\pi }{3}\right)$
$|\overrightarrow{OP}{|}^2=9{\left(t+\frac{1}{t}\right)}^2+4{\left(t-\frac{1}{t}\right)}^2+2\left(t^2-\frac{1}{t^2}\right)3\cdot 2\cdot \left(\frac{-1}{2}\right)$
$=9\left(t^2+\frac{1}{t^2}+2\right)+4\left(t^2+\frac{1}{t^2}-2\right)-6\left(t^2-\frac{1}{t^2}\right)$
$=7t^2+\frac{19}{t^2}+10$
$\Rightarrow |\overrightarrow{OP}{|}^2\ge 2\cdot \sqrt{7t^2\cdot \frac{19}{t^2}}+10(\because$ A.M. $\ge$ G.M. $)$
$\therefore$ Minimum value of $|\overrightarrow{OP}|=\sqrt{10+2\sqrt{133}}$