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

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(A) The position vector is $\overline{OP} = \hat{a} \cos t + \hat{b} \sin t$.The square of the length is $|\overline{OP}|^2 = (\hat{a} \cos t + \hat{b

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The position vector is $\overline{OP} = \hat{a} \cos t + \hat{b} \sin t$.The square of the length is $|\overline{OP}|^2 = (\hat{a} \cos t + \hat{b} \sin t) \cdot (\hat{a} \cos t + \hat{b} \sin t) = \cos^2 t + \sin^2 t + 2 \sin t \cos t (\hat{a} \cdot \hat{b}) = 1 + \sin(2t) (\hat{a} \cdot \hat{b})$.Since $\hat{a}$ and $\hat{b}$ form an acute angle,$\hat{a} \cdot \hat{b} > 0$. Thus,$|\overline{OP}|$ is maximum when $\sin(2t) = 1$,i.e.,$2t = \frac{\pi}{2} \Rightarrow t = \frac{\pi}{4}$.At $t = \frac{\pi}{4}$,$M = |\overline{OP}| = (1 + \hat{a} \cdot \hat{b})^{1/2}$.The vector $\overline{OP}$ at $t = \frac{\pi}{4}$ is $\frac{1}{\sqrt{2}}(\hat{a} + \hat{b})$.The unit vector $\hat{u}$ is $\frac{\overline{OP}}{|\overline{OP}|} = \frac{\frac{1}{\sqrt{2}}(\hat{a} + \hat{b})}{\frac{1}{\sqrt{2}}|\hat{a} + \hat{b}|} = \frac{\hat{a} + \hat{b}}{|\hat{a} + \hat{b}|}$.

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