Vector Algebra
Position Vectors and Distances
Grade 12
Question:
<p>An equilateral triangle \( \Delta OAB \) 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: The locus of point P as parameter t varies is a straight line in the plane. Find the range of |AP| by expressing AP in terms of t and minimizing the distance from A to this line.
Step 1: Express AP in terms of the given condition. We have OP = (2-t)OA + tOB Therefore, AP = OP - OA = (2-t)OA + tOB - OA = (1-t)OA + tOB Step 2: Find |AP|^2 using the dot product. |AP|^2 = [(1-t)OA + tOB]·[(1-t)OA + tOB] = (1-t)^2|OA|^2 + 2t(1-t)(OA·OB) + t^2|OB|^2 Since the triangle is equilateral with side 1: |OA| = |OB| = 1 and OA·OB = cos(60°) = 1/2 |AP|^2 = (1-t)^2 + 2t(1-t)·(1/2) + t^2 = (1-t)^2 + t(1-t) + t^2 = 1 - 2t + t^2 + t - t^2 + t^2 = t^2 - t + 1 Step 3: Find the minimum value. To minimize |AP|^2, take the derivative: d(|AP|^2)/dt = 2t - 1 = 0 This gives t = 1/2 Minimum value: |AP|^2 = (1/2)^2 - (1/2) + 1 = 1/4 - 1/2 + 1 = 3/4 Therefore, |AP|_min = √(3/4) = √3/2 As t → ±∞, |AP| → ∞, so |AP| ∈ [√3/2, ∞) ∴ Answer: C (the specific value or range given as option C)
Correct Answer: C