Let two non-collinear vectors hat{a} and hat{ | vedclass

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(A) Given $\overline{OP} = \hat{a} \sin t + \hat{b} \cos t$. $M = |\overline{OP}| = \sqrt{(\hat{a} \sin t + \hat{b} \cos t)^2}$. Since $\hat{a}$ and $

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

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(A) Given $\overline{OP} = \hat{a} \sin t + \hat{b} \cos t$. $M = |\overline{OP}| = \sqrt{(\hat{a} \sin t + \hat{b} \cos t)^2}$. Since $\hat{a}$ and $\hat{b}$ are unit vectors,$|\hat{a}| = 1$ and $|\hat{b}| = 1$. $M^2 = \sin^2 t |\hat{a}|^2 + \cos^2 t |\hat{b}|^2 + 2 \sin t \cos t (\hat{a} \cdot \hat{b}) = \sin^2 t + \cos^2 t + \sin 2t (\hat{a} \cdot \hat{b}) = 1 + \sin 2t (\hat{a} \cdot \hat{b})$. For $P$ to be farthest from $O$,$M$ must be maximum. Since $\hat{a}$ and $\hat{b}$ form an acute angle,$\hat{a} \cdot \hat{b} > 0$. Thus,$M$ is maximum when $\sin 2t = 1$,which implies $2t = \frac{\pi}{2}$,so $t = \frac{\pi}{4}$. At $t = \frac{\pi}{4}$,$\sin t = \cos t = \frac{1}{\sqrt{2}}$. $M = \sqrt{1 + 1(\hat{a} \cdot \hat{b})} = (1 + \hat{a} \cdot \hat{b})^{\frac{1}{2}}$. $\hat{u} = \frac{\overline{OP}}{|\overline{OP}|} = \frac{\hat{a} \sin t + \hat{b} \cos t}{M} = \frac{\hat{a} \frac{1}{\sqrt{2}} + \hat{b} \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} |\hat{a} + \hat{b}|} = \frac{\hat{a} + \hat{b}}{|\hat{a} + \hat{b}|}$.

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