Let two non-collinear unit vectors a and 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 a t+ b t . When P is farthest from origin O , let M be

Correct answer: A. u = a + b | a + b | and M=(1+ a · b )^ 1 / 2 O P= a t+ b t aligned & |O P|= ( a · a ^2 t+ b · b ^2 t+2 a · b t t) & |O P|= 1+2 a · b · t t & |O P|= 1+ a · b · 2 t & |O P|_ = 1+ a · b at 2 t=1 t=(π)/(4) & O P( at t=(π)/(4))= 1 2 ( a + b ) & aligned Unit vector along O P at (t=(π)/(4))= a + b | a + b |

JEE Advanced 2008Paper 2

Exam: JEE Advanced
Year: 2008
Paper: Paper 2
Subject: Mathematics
Chapter: Vector Algebra

JEE Advanced 2008 · Paper 2 · Mathematics · Vector Algebra

Let two non-collinear unit vectors a and b form an acute angle. A...

Let two non-collinear unit vectors

\mathbf{a}

and

\hat{\mathbf{b}}

form an acute angle. A point

P

moves so that at any time

t

the position vector

\mathbf{O P}

(where,

O

is the origin) is given by

\mathbf{a} \cos t+\hat{\mathbf{b}} \sin t

. When

P

is farthest from origin

O

, let

M

be the length of

\mathbf{O P}

and

\hat{\mathbf{u}}

be the unit vector along

\mathbf{O P}

. Then,

Options

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

Correct answer

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

Step-by-step solution

\mathbf{O P}=\hat{\mathbf{a}} \cos t+\hat{\mathbf{b}} \sin t

\begin{aligned} & \Rightarrow \quad|\mathbf{O P}|=\sqrt{\left(\hat{\mathbf{a}} \cdot \hat{\mathbf{a}} \cos ^2 t+\hat{\mathbf{b}} \cdot \hat{\mathbf{b}} \sin ^2 t+2 \hat{\mathbf{a}} \cdot \hat{\mathbf{b}} \sin t \cos t\right)} \\ & \Rightarrow \quad|\mathbf{O P}|=\sqrt{1+2 \hat{\mathbf{a}} \cdot \hat{\mathbf{b}} \cdot \sin t \cos t} \\ & \Rightarrow \quad|\mathbf{O P}|=\sqrt{1+\hat{\mathbf{a}} \cdot \hat{\mathbf{b}} \cdot \sin 2 t} \\ & \Rightarrow \quad|\mathbf{O P}|_{\max }=\sqrt{1+\hat{\mathbf{a}} \cdot \hat{\mathbf{b}}} \text{ at } \sin 2 t=1 \Rightarrow t=\frac{\pi}{4} \\ & \Rightarrow \mathbf{O P}\left(\text{ at } t=\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}(\hat{\mathbf{a}}+\hat{\mathbf{b}}) \\ & \end{aligned}

\therefore

Unit vector along \mathbf{O P} at \left(t=\frac{\pi}{4}\right)=\frac{\hat{\mathbf{a}}+\hat{\mathbf{b}}}{|\hat{\mathbf{a}}+\hat{\mathbf{b}}|}

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