Vectors 1 Question 9
10. Let two non-collinear unit vectors $\hat{\mathbf{a}}$ and $\hat{\mathbf{b}}$ form an acute angle. A point $P$ moves, so that at any time $t$ the position vector $\overrightarrow{\mathbf{O P}}$ (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 $\overrightarrow{\mathbf{O P}}$ and $\hat{\mathbf{u}}$ be the unit vector along $\overrightarrow{\mathbf{O P}}$. Then,
(2008, 3M)
(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) $\hat{\mathbf{u}}=\frac{\hat{\mathbf{a}}+\mathbf{b}}{|\hat{\mathbf{a}}+\hat{\mathbf{b}}|}$ and $M=(1+2 \hat{\mathbf{a}} \cdot \hat{\mathbf{b}})^{1 / 2}$
(d) $\hat{\mathbf{u}}=\frac{\hat{\mathbf{a}}-\hat{\mathbf{b}}}{|\hat{\mathbf{a}}-\hat{\mathbf{b}}|}$ and $M=(1+2 \hat{\mathbf{a}} \cdot \hat{\mathbf{b}})^{1 / 2}$
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Answer:
Correct Answer: 10. Orthocentre 23. $(5 \sqrt{2})$
Solution:
- Given, $\overrightarrow{\mathbf{O P}}=\hat{\mathbf{a}} \cos t+\hat{\mathbf{b}} \sin t$
$\Rightarrow|\overrightarrow{\mathbf{O P}}|=\sqrt{(\hat{\mathbf{a}} \cdot \hat{\mathbf{a}}) \cos ^{2} t+(\hat{\mathbf{b}} \cdot \hat{\mathbf{b}}) \sin ^{2} t+2 \hat{\mathbf{a}} \cdot \hat{\mathbf{b}} \sin t \cos t}$
$\Rightarrow|\overrightarrow{\mathbf{O P}}|=\sqrt{1+\hat{\mathbf{a}} \cdot \hat{\mathbf{b}} \sin 2 t}$
$\Rightarrow|\overrightarrow{\mathbf{O P}}| _{\max }=M=\sqrt{1+\hat{\mathbf{a}} \cdot \hat{\mathbf{b}}}$ at $\sin 2 t=1 \Rightarrow t=\frac{\pi}{4}$
At $t=\frac{\pi}{4}, \overrightarrow{\mathbf{O P}}=\frac{1}{\sqrt{2}}(\hat{\mathbf{a}}+\hat{\mathbf{b}})$
Unit vector along $\overrightarrow{\mathbf{O P}}$ at $\left(t=\frac{\pi}{4}\right)=\frac{\hat{\mathbf{a}}+\hat{\mathbf{b}}}{|\hat{\mathbf{a}}+\hat{\mathbf{b}}|}$
