Let two non-collinear unit vectors $\mathbf{a}$ and $\hat{\mathbf{b}}$ form an acute angle. A point $P$…

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 $\hat{\mathbf{a}} \cos t+\hat{\mathbf{b}} \sin t$. 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,
  1. $\hat{\mathbf{u}}=\frac{\hat{\mathbf{a}}+\hat{\mathbf{b}}}{|\hat{\mathbf{a}}+\hat{\mathbf{b}}|}$ and $M=(1+\hat{\mathbf{a}} \cdot \hat{\mathbf{b}})^{1 / 2}$
  2. $\hat{\mathbf{u}}=\frac{\hat{\mathbf{a}}-\hat{\mathbf{b}}}{|\hat{\mathbf{a}}-\hat{\mathbf{b}}|}$ and $M=(1+\hat{\mathbf{a}} \cdot \hat{\mathbf{b}})^{1 / 2}$
  3. $\hat{\mathbf{u}}=\frac{\hat{\mathbf{a}}+\hat{\mathbf{b}}}{|\hat{\mathbf{a}}+\hat{\mathbf{b}}|}$ and $M=(1+2 \hat{\mathbf{a}} \cdot \hat{\mathbf{b}})^{1 / 2}$
  4. $\hat{\mathbf{u}}=\frac{\hat{\mathbf{a}}-\hat{\mathbf{b}}}{|\hat{\mathbf { a }}-\hat{\mathbf{b}}|}$ and $M=(1+2 \hat{\mathbf{a}} \cdot \hat{\mathbf{b}})^{1 / 2}$

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 $\left(t=\frac{\pi}{4}\right)=\frac{\hat{\mathbf{a}}+\hat{\mathbf{b}}}{|\hat{\mathbf{a}}+\hat{\mathbf{b}}|}$

Asked in: JEE Advanced 2008 (Paper 2)

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