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 length $M = |\overline{OP}|$ is given by: $M^2 = |\hat{a} \cos t + \

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

Vedclass · Answer

(A) The position vector is $\overline{OP} = \hat{a} \cos t + \hat{b} \sin t$. The length $M = |\overline{OP}|$ is given by: $M^2 = |\hat{a} \cos t + \hat{b} \sin t|^2 = |\hat{a}|^2 \cos^2 t + |\hat{b}|^2 \sin^2 t + 2(\hat{a} \cdot \hat{b}) \sin t \cos t$. Since $\hat{a}$ and $\hat{b}$ are unit vectors,$|\hat{a}| = |\hat{b}| = 1$. $M^2 = \cos^2 t + \sin^2 t + (\hat{a} \cdot \hat{b}) \sin 2t = 1 + (\hat{a} \cdot \hat{b}) \sin 2t$. For $M$ to be maximum,$\sin 2t$ must be $1$,so $2t = \frac{\pi}{2}$,which means $t = \frac{\pi}{4}$. Then $M = \sqrt{1 + \hat{a} \cdot \hat{b}}$. The unit vector $\hat{u}$ along $\overline{OP}$ is $\frac{\overline{OP}}{|\overline{OP}|}$. At $t = \frac{\pi}{4}$,$\overline{OP} = \hat{a} \cos(\frac{\pi}{4}) + \hat{b} \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}(\hat{a} + \hat{b})$. Thus,$\hat{u} = \frac{\frac{1}{\sqrt{2}}(\hat{a} + \hat{b})}{|\frac{1}{\sqrt{2}}(\hat{a} + \hat{b})|} = \frac{\hat{a} + \hat{b}}{|\hat{a} + \hat{b}|}$.

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