Question

Let two non-collinear unit vectors $\hat{a}$ and $\hat{b}$ form an acute angle. A point $P$ moves so that at any time $t$ the position vector $OP$ (where $O$ is the origin) is given by $\hat{a}\cos t+\hat{b}\sin t$. When $P$ is farthest from origin $O$, let $M$ be the length of $OP$ and $\hat{u}$ be the unit vector along vector $OP$. Then:

• A. $u=\frac{\hat{a}+\hat{b}}{\left|\hat{a}+\hat{b}\right|}$ and $M={\left(1+\hat{a}·\hat{b}\right)}^{\frac{1}{2}}$
• B. $u=\frac{\hat{a}-\hat{b}}{\left|\hat{a}-\hat{b}\right|}$ and $M={\left(1+\hat{a}·\hat{b}\right)}^{\frac{1}{2}}$
• C. $u=\frac{\hat{a}+\hat{b}}{\left|\hat{a}+\hat{b}\right|}$ and $M={\left(1+2\hat{a}·\hat{b}\right)}^{\frac{1}{2}}$
• D. $u=\frac{\hat{a}-\hat{b}}{\left|\hat{a}-\hat{b}\right|}$ and $M={\left(1+2\hat{a}·\hat{b}\right)}^{\frac{1}{2}}$

Answer 1

The correct option is A

$u=\frac{\hat{a}+\hat{b}}{\left|\hat{a}+\hat{b}\right|}$ and $M={\left(1+\hat{a}·\hat{b}\right)}^{\frac{1}{2}}$

Explanation for the correct option:

Finding the unit vector:

Given the non-collinear unit vectors $\hat{a}$ and $\hat{b}$ which form an acute angle.

The position vector $OP$ is given as, $\hat{a}\cos t+\hat{b}\sin t$ .

$M$ is the maximum length of $OP$ and $\hat{u}$ is the unit vector $OP$.

Finding the value of $M$,

$M$ is the maximum length of $OP$,

$\Rightarrow M=\left|\overrightarrow{OP}\right|$

$\begin{array}{rcl}M& =& \sqrt{{\left(\hat{a}\cos t+\hat{b}\sin t\right)}^2}\\ & =& \sqrt{{\left(\hat{a}\cos t\right)}^2+{\left(\hat{b}\sin t\right)}^2+2\left(\hat{a}\cos t\right)\left(\hat{b}\sin t\right)}\\ & =& \sqrt{{\cos }^{2}t+{\sin }^{2}t+\hat{a}·\hat{b}\left(2\cos t\sin t\right)}\\ & =& \sqrt{1+\hat{a}·\hat{b}\left(\sin 2t\right)}\end{array}$

We know that the maximum value of $\sin 2t=1$.

$\begin{array}{rcl}& \Rightarrow & \sin 2t=1\\ 2t& =& {\sin }^{-1}\left(1\right)\\ 2t& =& \frac{\pi }{2}\\ t& =& \frac{\pi }{4}\end{array}$

Then $\sin 2t$ can be replaced as $1$.

$\begin{array}{rcl}& \Rightarrow & M=\sqrt{1+\hat{a}·\hat{b}\left(1\right)}\\ & =& \sqrt{1+\hat{a}·\hat{b}}\\ & =& {\left(1+\hat{a}·\hat{b}\right)}^{\frac{1}{2}}\end{array}$

Therefore, the maximum length of $OP$ is $M={\left(1+\hat{a}·\hat{b}\right)}^{\frac{1}{2}}$.

Finding the unit vector of $OP$, we have

$\begin{array}{rcl}& \Rightarrow & \hat{u}=\frac{OP}{\left|OP\right|}\\ & =& \frac{\hat{a}\cos t+\hat{b}\sin t}{\left|\hat{a}\cos t+\hat{b}\sin t\right|}\end{array}$

Substituting $t$ as $\frac{\pi }{4}$ and we know that $\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}$ and $\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$ then,

$\begin{array}{rcl}\hat{u}& =& \frac{\hat{a}\cos {\displaystyle \frac{\pi }{4}}+\hat{b}\sin {\displaystyle \frac{\pi }{4}}}{\left|\hat{a}\cos \frac{\pi }{4}+\hat{b}\sin \frac{\pi }{4}\right|}\\ & =& \frac{\hat{a}\left({\displaystyle \frac{{\displaystyle 1}}{\sqrt{2}}}\right)+\hat{b}\left({\displaystyle \frac{1}{\sqrt{2}}}\right)}{\left|\hat{a}\left(\frac{{\displaystyle 1}}{\sqrt{2}}\right)+\hat{b}\left(\frac{1}{\sqrt{2}}\right)\right|}\\ & =& \frac{\left({\displaystyle \frac{1}{\sqrt{2}}}\right)\left(\hat{a}+\hat{b}\right)}{\left({\displaystyle \frac{1}{\sqrt{2}}}\right)\left|\left(\hat{a}+\hat{b}\right)\right|}\\ & =& \frac{\hat{a}+\hat{b}}{\left|\left(\hat{a}+\hat{b}\right)\right|}\end{array}$

Therefore, the unit vector of $OP$ is $\hat{u}=\frac{\hat{a}+\hat{b}}{\left|\hat{a}+\hat{b}\right|}$

Hence, the correct answer is option (A).