Let two non-collinear unit vectors â and b̂ form an acute angle. A point P moves so that at any time t the position vector OP⃗ (where O is the origin) is given by âcos t + b̂sin t. When P is farthest from origin O, let M be the length of OP⃗ and û be the unit vector along OP⃗. Then,

Let two non-collinear unit vectors â and b̂ form an acute angle. A point P moves so that at any time t the position vector OP⃗ (where O is the origin) is given by âcos t + b̂sin t. When P is farthest from origin O, let M be the length of OP⃗ and û be the unit vector along OP⃗. Then, â,b̂ unit, non-collinear, acute ⇒ â·b̂>0 OP⃗=âcos t+b̂sin t Find M=|OP⃗|ₘₐₓ, û=OP̂ Options: û=â±b̂/|â±b̂|, M=(1+â·b̂)⁽1/2) or (1+2â·b̂)⁽1/2) 1. |OP⃗|²=OP⃗·OP⃗ 2. maximise over t ⇒ M 3. direction at that t ⇒ û |OP⃗|²=(âcos t+b̂sin t)·(âcos t+b̂sin t) =cos²t+sin²t+2(â·b̂)cos tsin t =1+(â·b̂)sin 2t max|OP⃗|²=1+â·b̂ at sin 2t=1 M=(1+â·b̂)⁽1/2) sin 2t=1 ⇒ t=π/4 OP⃗=âcosπ/4+b̂sinπ/4=1/√2(â+b̂) ⇒ û=(â+b̂)/(|â+b̂|) û=(â+b̂)/(|â+b̂|), M=(1+â·b̂)⁽1/2) ⇒ option (1); (2),(3),(4) mismatch boxedAnswer: option (1)

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 $\overrightarrow{OP}$ (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 $\overrightarrow{OP}$ and $\hat{u}$ be the unit vector along $\overrightarrow{OP}$ . Then,

Answer: A. $\hat{u} = \frac{\hat{a}+\hat{b}}{|\hat{a}+\hat{b}|}$ and $M = (1+\hat{a}\cdot\hat{b})^{1/2}$

Vector Algebra 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

\overrightarrow{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

\overrightarrow{OP}

and

\hat{u}

be the unit vector along

\overrightarrow{OP}

. Then,

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Written solution

$\hat{a},\hat{b} \text{ unit, non-collinear, acute} \Rightarrow \hat{a}\cdot\hat{b}>0$ $\overrightarrow{OP}=\hat{a}\cos t+\hat{b}\sin t$ $\text{Find } M=|\overrightarrow{OP}|_{\max}, \quad \hat{u}=\widehat{OP}$ $\text{Options: } \hat{u}=\dfrac{\hat{a}\pm\hat{b}}{|\hat{a}\pm\hat{b}|}, \quad M=(1+\hat{a}\cdot\hat{b})^{1/2} \text{ or } (1+2\hat{a}\cdot\hat{b})^{1/2}$ $1.\ |\overrightarrow{OP}|^2=\overrightarrow{OP}\cdot\overrightarrow{OP}$ $2.\ \text{maximise over } t \Rightarrow M$ $3.\ \text{direction at that } t \Rightarrow \hat{u}$ $|\overrightarrow{OP}|^2=(\hat a\cos t+\hat b\sin t)\cdot(\hat a\cos t+\hat b\sin t)$ $=\cos^2t+\sin^2t+2(\hat a\cdot\hat b)\cos t\sin t$ $=1+(\hat a\cdot\hat b)\sin 2t$ $\max|\overrightarrow{OP}|^2=1+\hat a\cdot\hat b \ \text{ at } \sin 2t=1$ $M=(1+\hat a\cdot\hat b)^{1/2}$ $\sin 2t=1 \Rightarrow t=\dfrac{\pi}{4}$ $\overrightarrow{OP}=\hat a\cos\dfrac{\pi}{4}+\hat b\sin\dfrac{\pi}{4}=\dfrac{1}{\sqrt2}(\hat a+\hat b)$ $\Rightarrow \hat u=\dfrac{\hat a+\hat b}{|\hat a+\hat b|}$ $\hat u=\dfrac{\hat a+\hat b}{|\hat a+\hat b|}, \quad M=(1+\hat a\cdot\hat b)^{1/2}$ $\Rightarrow \text{option (1)}; \quad (2),(3),(4) \text{ mismatch}$ $\boxed{\text{Answer: option (1)}}$

Written solution

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