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

MCQ

Let two non-collinear unit vectors $\hat{\mathrm{a}}$ and $\hat{\mathrm{b}}$ form an acute angle. A point $\mathrm{P}$ moves so that at any time $\mathrm{t}$ the position vector $\overline{\mathrm{OP}}$ (where $\mathrm{O}$ is the origin) is given by $\hat{\mathrm{a}} \cos t+\hat{b} \sin t$. When $\mathrm{P}$ is farthest from origin $O$, let $M$ be the length of $\overline{\mathrm{OP}}$ and $\mathrm{u}$ be the unit vector along $\overline{\mathrm{OP}}$. Then,

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

✓

Answer

Correct option: A.

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

a
$ |\overline{\mathrm{OP}}|=|\hat{\mathrm{a}} \cos \mathrm{t}+\hat{\mathrm{b}} \sin t| $

$ =\left(\cos ^2 \mathrm{t}+\sin ^2 t+2 \cos t \sin t \hat{a} \cdot \hat{b}\right)^{1 / 2} $

$ =(1+2 \cos t \sin t \hat{a} \cdot \hat{b})^{1 / 2} $

$ =(1+\sin 2 t \hat{a} \cdot \hat{b})^{1 / 2}$

$ \therefore|\overrightarrow{\mathrm{OP}}|_{\max }=(1+\hat{\mathrm{a}} \cdot \hat{\mathrm{b}})^{1 / 2} \text { when, } \mathrm{t}=\frac{\pi}{4} $

$ \hat{\mathrm{u}}=\frac{\hat{\mathrm{a}}+\hat{\mathrm{b}}}{\sqrt{2} \frac{|\hat{\mathrm{a}}+\hat{\mathrm{b}}|}{\sqrt{2}}} $

$ \Rightarrow \hat{\mathrm{u}}=\frac{\hat{\mathrm{a}}+\hat{\mathrm{b}}}{|\hat{\mathrm{a}}+\hat{\mathrm{b}}|}$

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