Q. Let two non-collinear unit vectors $\widehat{\mathbf{a}}$ and $\widehat{\mathbf{b}}$ from and acute angle. A point $P$ moves so that at any time $t$ the position vector $\overrightarrow{\mathbf{OP}}$ (where $O$ is the origin) is given by $\widehat{\mathbf{a}}\cos t + \widehat{\mathbf{b}}\sin t$. When $P$ is farthest form origin $O$, let $M$ be the length of $\overrightarrow{\mathbf{OP}}$ and $\widehat{\mathbf{u}}$ be the unit vector along $\overrightarrow{\mathbf{OP}}$ Then,
Explanation
(a)
Given, $\overrightarrow{\mathbf{OP}} = \widehat{\mathbf{a}}\cos{t + \widehat{\mathbf{b}}}\sin t$
$\Longrightarrow \ \left| \overrightarrow{\mathbf{OP}} \right| = \sqrt{{(\widehat{\mathbf{a}} \bullet \widehat{\mathbf{a}}\cos}^{2}t + \widehat{\mathbf{b}} \bullet \widehat{\mathbf{b}}\sin^{2}t + 2\widehat{\mathbf{a}} \bullet \widehat{\mathbf{b}}\sin t\cos t}$
$\Longrightarrow \ \left| \overrightarrow{\mathbf{OP}} \right| = \sqrt{1 + \widehat{\mathbf{a}} \bullet \widehat{\mathbf{b}}\sin{2t}}$
$\Longrightarrow \ \left| \overrightarrow{\mathbf{OP}} \right|_{maxx} = \sqrt{1 + \widehat{\mathbf{a}} \bullet \widehat{\mathbf{b}}}$
$\left\lbrack Max\ (\sin{2t) = 1 \Longrightarrow t = \frac{\pi}{4}} \right\rbrack$
$\Longrightarrow \ \overrightarrow{\mathbf{OP}}\left( at\ t = \frac{\pi}{4} \right) = \frac{1}{\sqrt{2\ }}\left( \widehat{\mathbf{a}} + \widehat{\mathbf{b}} \right)$
$\ \therefore Unit\ vector\ along\ \overrightarrow{\mathbf{OP}}\ at\ \left( t = \frac{\pi}{4} \right) = \frac{\widehat{\mathbf{a}} + \widehat{\mathbf{b}}}{\left| \widehat{\mathbf{a}} + \widehat{\mathbf{b}} \right|}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.