Question

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

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

Hint:

Logic Tip: The sum of two vectors scaled identically (like multiplying both by $1/\sqrt{2}$) will point in the exact same direction as the unscaled sum. Hence, the unit vector of $k\vec{A}$ is simply the unit vector of $\vec{A}$ (for any $k>0$).

Answer 1

The Correct Option is A

Concept:
The distance of a point from the origin is the magnitude of its position vector. We must find the maximum value of this magnitude by optimizing the trigonometric components, and then substitute the critical value of $t$ back to find the unit vector.
Step 1: Find an expression for the magnitude M.
The position vector is $\overline{OP} = \hat{a}\sin t + \hat{b}\cos t$. Since $\hat{a}$ and $\hat{b}$ are unit vectors, $|\hat{a}| = 1$ and $|\hat{b}| = 1$. $$M^2 = |\overline{OP}|^2 = (\hat{a}\sin t + \hat{b}\cos t) \cdot (\hat{a}\sin t + \hat{b}\cos t)$$ $$M^2 = |\hat{a}|^2\sin^2 t + |\hat{b}|^2\cos^2 t + 2(\hat{a}\cdot\hat{b})\sin t \cos t$$ $$M^2 = \sin^2 t + \cos^2 t + (\hat{a}\cdot\hat{b})(2\sin t \cos t)$$ Using the identity $\sin^2 t + \cos^2 t = 1$ and $2\sin t \cos t = \sin 2t$: $$M^2 = 1 + (\hat{a}\cdot\hat{b})\sin 2t$$ $$M = \sqrt{1 + (\hat{a}\cdot\hat{b})\sin 2t}$$
Step 2: Maximize M and find the optimal time t.
Since $\hat{a}$ and $\hat{b}$ form an acute angle, $\hat{a}\cdot\hat{b}>0$. Therefore, to maximize $M$, we must maximize $\sin 2t$. The maximum value of $\sin 2t$ is 1, which occurs when: $$2t = \frac{\pi}{2} \implies t = \frac{\pi}{4}$$ Substituting $\sin 2t = 1$ into our expression for M: $$M_{max} = \sqrt{1 + \hat{a}\cdot\hat{b}(1)} = (1 + \hat{a}\cdot\hat{b})^{\frac{1}{2$$
Step 3: Find the unit vector $\hat{u}$ at this maximum distance.
Substitute $t = \frac{\pi}{4}$ back into the position vector $\overline{OP}$: $$\overline{OP} = \hat{a}\sin\left(\frac{\pi}{4}\right) + \hat{b}\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2\hat{a} + \frac{1}{\sqrt{2\hat{b} = \frac{\hat{a} + \hat{b{\sqrt{2$$ The unit vector $\hat{u}$ is given by dividing the position vector by its magnitude at this point: $$\hat{u} = \frac{\overline{OP{|\overline{OP}|} = \frac{\frac{1}{\sqrt{2(\hat{a} + \hat{b})}{|\frac{1}{\sqrt{2(\hat{a} + \hat{b})|} = \frac{\hat{a} + \hat{b{|\hat{a} + \hat{b}|}$$