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,

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

(B) $\hat{u}=\frac{\hat{a}-\hat{b}}{|\hat{a}-\hat{b}|}$ and $M=(1+\hat{a} \cdot \hat{b})^{\frac{1}{2}}$

(C) $\hat{u}=\frac{\hat{a}+\hat{b}}{|\hat{a}+\hat{b}|}$ and $M=(1+2 \hat{a} \cdot \hat{b})^{\frac{1}{2}}$

(D) $\hat{u}=\frac{\hat{a}-\hat{b}}{|\hat{a}-\hat{b}|}$ and $M=(1+2 \hat{a} \cdot \hat{b})^{\frac{1}{2}}$

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Vector Algebra

Let two non-collinear unit vectors

\overset{a}{^}

and

\hat{b}

form an acute angle. A point

P

moves so that at any time

t

the position vector

O P

(where

O

is the origin) is given by

\overset{a}{^} cos t + \hat{b} sin t

. When

P

is farthest from origin

O

, let

M

be the length of

O P

and

\overset{u}{^}

be the unit vector along

O P

. Then,

(A) $\overset{u}{^} = \frac{a ^ + b ^}{∣ a ^ + b ^ ∣}$ and $M = (1 + \overset{a}{^} \cdot \hat{b})^{\frac{1}{2}}$ ✓ Correct
(B) $\overset{u}{^} = \frac{a ^ - b ^}{∣ a ^ - b ^ ∣}$ and $M = (1 + \overset{a}{^} \cdot \hat{b})^{\frac{1}{2}}$
(C) $\overset{u}{^} = \frac{a ^ + b ^}{∣ a ^ + b ^ ∣}$ and $M = (1 + 2 \overset{a}{^} \cdot \hat{b})^{\frac{1}{2}}$
(D) $\overset{u}{^} = \frac{a ^ - b ^}{∣ a ^ - b ^ ∣}$ and $M = (1 + 2 \overset{a}{^} \cdot \hat{b})^{\frac{1}{2}}$

Answer: (A)

The magnitude is

∣ O P ∣ = cos^{2} t + sin^{2} t + 2 cos t sin t (\overset{a}{^} \cdot \hat{b}) = 1 + sin (2 t) (\overset{a}{^} \cdot \hat{b})

. This is maximum when

sin (2 t) = 1

, i.e.,

t = π /4

, giving

M = 1 + \overset{a}{^} \cdot \hat{b}

. At

t = π /4

O P = \frac{a ^ + b ^}{2}

, so

\overset{u}{^} = \frac{a ^ + b ^}{∣ a ^ + b ^ ∣}

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Single correct option