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MathMedAdv 2008 • Paper 2

Maximum Distance of Moving Point with Position Vector in Unit Vectors

Let two non-collinear unit vectors a^\hat{a} and b^\hat{b} form an acute angle. A point PP moves so that at any time tt the position vector OP⃗\vec{OP} (where OO is the origin) is given by a^cos⁡t+b^sin⁡t\hat{a} \cos t + \hat{b} \sin t. When PP is farthest from origin OO, let MM be the length of OPOP and u^\hat{u} be the unit vector along OP⃗\vec{OP}. Then,

Options

A

u^=a^+b^∣a^+b^∣\hat{u} = \frac{\hat{a} + \hat{b}}{|\hat{a} + \hat{b}|} and M=(1+a^⋅b^)12M = (1 + \hat{a} \cdot \hat{b})^{\frac{1}{2}}

Correct
B

u^=a^−b^∣a^−b^∣\hat{u} = \frac{\hat{a} - \hat{b}}{|\hat{a} - \hat{b}|} and M=(1+a^⋅b^)12M = (1 + \hat{a} \cdot \hat{b})^{\frac{1}{2}}

C

u^=a^+b^∣a^+b^∣\hat{u} = \frac{\hat{a} + \hat{b}}{|\hat{a} + \hat{b}|} and M=(1+2a^⋅b^)12M = (1 + 2\hat{a} \cdot \hat{b})^{\frac{1}{2}}

D

u^=a^−b^∣a^−b^∣\hat{u} = \frac{\hat{a} - \hat{b}}{|\hat{a} - \hat{b}|} and M=(1+2a^⋅b^)12M = (1 + 2\hat{a} \cdot \hat{b})^{\frac{1}{2}}

Topics & Concepts

Vector AlgebraVector AdditionDot and Cross Product

Step-by-Step Solution

To find the maximum distance of point PP from the origin OO and the corresponding unit vector along OP⃗\vec{OP}, we start by analyzing the given position vector of PP:

OP⃗=a^cos⁡t+b^sin⁡t\vec{OP} = \hat{a}\cos t + \hat{b}\sin t

where a^\hat{a} and b^\hat{b} are non-collinear unit vectors forming an acute angle, so ∣a^∣=1|\hat{a}| = 1, ∣b^∣=1|\hat{b}| = 1, and a^⋅b^>0\hat{a} \cdot \hat{b} > 0.

Step 1: Find the distance OPOP as a function of tt

The square of the distance from origin OO to point PP is given by the dot product of OP⃗\vec{OP} with itself:

∣OP⃗∣2=OP⃗⋅OP⃗|\vec{OP}|^2 = \vec{OP} \cdot \vec{OP}
∣OP⃗∣2=(a^cos⁡t+b^sin⁡t)⋅(a^cos⁡t+b^sin⁡t)|\vec{OP}|^2 = (\hat{a}\cos t + \hat{b}\sin t) \cdot (\hat{a}\cos t + \hat{b}\sin t)
∣OP⃗∣2=∣a^∣2cos⁡2t+∣b^∣2sin⁡2t+2(a^⋅b^)sin⁡tcos⁡t|\vec{OP}|^2 = |\hat{a}|^2 \cos^2 t + |\hat{b}|^2 \sin^2 t + 2(\hat{a} \cdot \hat{b})\sin t \cos t

Since ∣a^∣=1|\hat{a}| = 1 and ∣b^∣=1|\hat{b}| = 1, we simplify using trigonometric identities:

∣OP⃗∣2=cos⁡2t+sin⁡2t+(a^⋅b^)sin⁡2t|\vec{OP}|^2 = \cos^2 t + \sin^2 t + (\hat{a} \cdot \hat{b})\sin 2t
∣OP⃗∣2=1+(a^⋅b^)sin⁡2t|\vec{OP}|^2 = 1 + (\hat{a} \cdot \hat{b})\sin 2t

Step 2: Determine the maximum distance MM

Since a^⋅b^>0\hat{a} \cdot \hat{b} > 0 (as the angle between a^\hat{a} and b^\hat{b} is acute), ∣OP⃗∣2|\vec{OP}|^2 attains its maximum value when sin⁡2t=1\sin 2t = 1.

This occurs when 2t=π22t = \frac{\pi}{2}, which gives t=π4t = \frac{\pi}{4}.

Substituting sin⁡2t=1\sin 2t = 1 into the equation gives the maximum length MM:

M2=1+a^⋅b^M^2 = 1 + \hat{a} \cdot \hat{b}
M=(1+a^⋅b^)12M = (1 + \hat{a} \cdot \hat{b})^{\frac{1}{2}}

Step 3: Find the unit vector u^\hat{u} when PP is farthest from OO

At t=π4t = \frac{\pi}{4}, the position vector OP⃗\vec{OP} is:

OP⃗=a^cos⁡(π4)+b^sin⁡(π4)=12a^+12b^=a^+b^2\vec{OP} = \hat{a}\cos\left(\frac{\pi}{4}\right) + \hat{b}\sin\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 u^\hat{u} along OP⃗\vec{OP} is obtained by dividing OP⃗\vec{OP} by its magnitude ∣OP⃗∣|\vec{OP}|:

u^=OP⃗∣OP⃗∣=12(a^+b^)∣12(a^+b^)∣=a^+b^∣a^+b^∣\hat{u} = \frac{\vec{OP}}{|\vec{OP}|} = \frac{\frac{1}{\sqrt{2}}(\hat{a} + \hat{b})}{\left|\frac{1}{\sqrt{2}}(\hat{a} + \hat{b})\right|} = \frac{\hat{a} + \hat{b}}{|\hat{a} + \hat{b}|}

Thus, we have:
u^=a^+b^∣a^+b^∣andM=(1+a^⋅b^)12\hat{u} = \frac{\hat{a} + \hat{b}}{|\hat{a} + \hat{b}|} \quad \text{and} \quad M = (1 + \hat{a} \cdot \hat{b})^{\frac{1}{2}}

This corresponds to Option (A).

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