Vector Algebra
Position vectors and distance
Grade 12
Question:
<p>An equilateral triangle \( \triangle OAB \) has side length 1, \( P \) is a point on the plane of the triangle. If \( \overrightarrow{OP} = (2-t)\overrightarrow{OA} + t\overrightarrow{OB},\ t \in R \), then the possible value of \( |\overrightarrow{AP}| \) can be:</p>
<p>(a) \( \dfrac{1}{2} \)</p>
<p>(b) \( \dfrac{1}{\sqrt{2}} \)</p>
<p>(c) \( \dfrac{\sqrt{3}}{2} \)</p>
<p>(d) \( 2 \)</p>
Step-by-Step Solution
Key Concept: Point P traces a line in the plane as parameter t varies; finding the range of |AP| requires identifying this locus and computing the distance from fixed point A to this line.
Step 1: Express AP in terms of parameter t. $\overrightarrow{AP} = \overrightarrow{OP} - \overrightarrow{OA} = (2-t)\overrightarrow{OA} + t\overrightarrow{OB} - \overrightarrow{OA}$ $\overrightarrow{AP} = (1-t)\overrightarrow{OA} + t\overrightarrow{OB}$ Step 2: Compute |AP|^2 using the dot product. $|\overrightarrow{AP}|^2 = (1-t)^2|\overrightarrow{OA}|^2 + t^2|\overrightarrow{OB}|^2 + 2t(1-t)\overrightarrow{OA} \cdot \overrightarrow{OB}$ Since the triangle is equilateral with side 1: $|\overrightarrow{OA}| = |\overrightarrow{OB}| = 1$ and $\overrightarrow{OA} \cdot \overrightarrow{OB} = 1 \cdot 1 \cdot \cos(60°) = \frac{1}{2}$ $|\overrightarrow{AP}|^2 = (1-t)^2 + t^2 + 2t(1-t) \cdot \frac{1}{2}$ $= (1-t)^2 + t^2 + t(1-t) = 1 - 2t + t^2 + t^2 + t - t^2$ $= t^2 - t + 1 = \left(t - \frac{1}{2}\right)^2 + \frac{3}{4}$ Step 3: Find the range of |AP|. The minimum value occurs at $t = \frac{1}{2}$: $|\overrightarrow{AP}|_{\min}^2 = \frac{3}{4}$, so $|\overrightarrow{AP}|_{\min} = \frac{\sqrt{3}}{2}$ As $t \to \pm\infty$, we have $|\overrightarrow{AP}| \to \infty$ Therefore: $|\overrightarrow{AP}| \in \left[\frac{\sqrt{3}}{2}, +\infty\right)$ ∴ The possible value of |AP| is any value ≥ $\frac{\sqrt{3}}{2}$ (Answer: C)
Correct Answer: C