IIT-JEE 2008 · Paper 2 · Mathematics Q7

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MathematicsSingle correct+3 / -1Question 7 of 66

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,

\hat u=\dfrac{\hat a+\hat b}{|\hat a+\hat b|}

and

M=(1+\hat a\cdot\hat b)^{1/2}

\hat u=\dfrac{\hat a-\hat b}{|\hat a-\hat b|}

and

M=(1+\hat a\cdot\hat b)^{1/2}

\hat u=\dfrac{\hat a+\hat b}{|\hat a+\hat b|}

and

M=(1+2\hat a\cdot\hat b)^{1/2}

\hat u=\dfrac{\hat a-\hat b}{|\hat a-\hat b|}

and

M=(1+2\hat a\cdot\hat b)^{1/2}

Answer and worked solution

Official answer: A

Distance.
$|\overrightarrow{OP}|^2=\cos^2t+\sin^2t+2\hat a\cdot\hat b\sin t\cos t=1+(\hat a\cdot\hat b)\sin2t.$
With $\hat a\cdot\hat b>0$ (acute angle), the maximum is at $t=\frac\pi4$ : $M=(1+\hat a\cdot\hat b)^{1/2}$ .

Direction. At $t=\frac\pi4$ , $\overrightarrow{OP}=\frac1{\sqrt2}(\hat a+\hat b)$ , so $\hat u=\dfrac{\hat a+\hat b}{|\hat a+\hat b|}$ .

(A)

Topics:Dot product Maxima and minima

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IIT-JEE 2008 · Paper 2, Mathematics question 7