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Let two non-collinear vectors $\hat{a}$ and $\hat{b}$ form an acute angle. A point $\mathrm{P}$ moves, so that at any time $t$ the position vector $\overline{\mathrm{OP}}$, where $\mathrm{O}$ is origin, is given by $\hat{a} \sin t+\hat{b} \cos t$, when $P$ is farthest from origin $\mathrm{O}$, let $\mathrm{M}$ be the length of $\overline{\mathrm{OP}}$ and $\hat{\mathrm{u}}$ be the unit vector along $\overline{\mathrm{OP}}$, then
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Verified Answer
The correct answer is:
$\hat{\mathrm{u}}=\frac{\hat{\mathrm{a}}+\hat{\mathrm{b}}}{|\hat{\mathrm{a}}+\hat{\mathrm{b}}|}$ and $\mathrm{M}=(1+\hat{\mathrm{a}} \cdot \hat{\mathrm{b}})^{\frac{1}{2}}$
$\begin{aligned}
& \mathrm{M}=|\overrightarrow{\mathrm{OP}}| \\
& M=\sqrt{(\hat{a} \sin t+\hat{b} \cos t)^2} \\
& =\sqrt{(\hat{a} \sin t)^2+(\hat{b} \cos t)^2+2(\hat{a} \sin t) \cdot(\hat{b} \cos t)} \\
& =\sqrt{\sin ^2 t+\cos ^2 t+\hat{a} \cdot \hat{b}(2 \sin t \cos t)} \\
& =\sqrt{1+\hat{\mathrm{a}} \cdot \hat{\mathrm{b}}(\sin 2 \mathrm{t})} \\
& \text { Maximum value of } \sin 2 t=1 \\
& \therefore \quad 2 \mathrm{t}=\sin ^{-1}(1) \\
& \therefore \quad \mathrm{t}=\frac{\pi}{4} \\
& \therefore \quad \mathrm{M}=\sqrt{1+\hat{\mathrm{a}} \cdot \hat{\mathrm{b}}(1)} \\
& =(1+\hat{a} \cdot \hat{b})^{\frac{1}{2}} \\
& \text { Now, } \hat{\mathrm{u}}=\frac{\overline{\mathrm{OP}}}{|\overline{\mathrm{OP}}|} \\
& =\frac{\hat{a} \sin t+\hat{b} \cos t}{|\hat{a} \sin t+\hat{b} \cos t|} \\
& =\frac{\hat{a}\left(\frac{1}{\sqrt{2}}\right)+\hat{b}\left(\frac{1}{\sqrt{2}}\right)}{\left|\hat{a}\left(\frac{1}{\sqrt{2}}\right)+\hat{b}\left(\frac{1}{\sqrt{2}}\right)\right|}\\
&
\end{aligned}$
Unit vector of OP is
$\hat{\mathrm{u}}=\frac{\hat{\mathrm{a}}+\hat{\mathrm{b}}}{|(\hat{\mathrm{a}}+\hat{\mathrm{b}})|}$
& \mathrm{M}=|\overrightarrow{\mathrm{OP}}| \\
& M=\sqrt{(\hat{a} \sin t+\hat{b} \cos t)^2} \\
& =\sqrt{(\hat{a} \sin t)^2+(\hat{b} \cos t)^2+2(\hat{a} \sin t) \cdot(\hat{b} \cos t)} \\
& =\sqrt{\sin ^2 t+\cos ^2 t+\hat{a} \cdot \hat{b}(2 \sin t \cos t)} \\
& =\sqrt{1+\hat{\mathrm{a}} \cdot \hat{\mathrm{b}}(\sin 2 \mathrm{t})} \\
& \text { Maximum value of } \sin 2 t=1 \\
& \therefore \quad 2 \mathrm{t}=\sin ^{-1}(1) \\
& \therefore \quad \mathrm{t}=\frac{\pi}{4} \\
& \therefore \quad \mathrm{M}=\sqrt{1+\hat{\mathrm{a}} \cdot \hat{\mathrm{b}}(1)} \\
& =(1+\hat{a} \cdot \hat{b})^{\frac{1}{2}} \\
& \text { Now, } \hat{\mathrm{u}}=\frac{\overline{\mathrm{OP}}}{|\overline{\mathrm{OP}}|} \\
& =\frac{\hat{a} \sin t+\hat{b} \cos t}{|\hat{a} \sin t+\hat{b} \cos t|} \\
& =\frac{\hat{a}\left(\frac{1}{\sqrt{2}}\right)+\hat{b}\left(\frac{1}{\sqrt{2}}\right)}{\left|\hat{a}\left(\frac{1}{\sqrt{2}}\right)+\hat{b}\left(\frac{1}{\sqrt{2}}\right)\right|}\\
&
\end{aligned}$
Unit vector of OP is
$\hat{\mathrm{u}}=\frac{\hat{\mathrm{a}}+\hat{\mathrm{b}}}{|(\hat{\mathrm{a}}+\hat{\mathrm{b}})|}$
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