A-Level Scalar and vector products: worked solution

Question

Points A, B, C have position vectors a, b, c; a is a unit vector and b⊥(b−a). Suppose further that |b|=1/2 and C lies on AB with AC:CB=2:1. Evaluate |a·c| and give its geometric meaning. [4]

Step 1: Turn the perpendicularity condition into a scalar product equation. b is perpendicular to (b - a), and perpendicular vectors have zero scalar product: b·(b - a) = 0 The scalar product distributes over subtraction, so expand: b·b - b·a = 0 Now use b·b = |b|² and b·a = a·b (the scalar product is commutative): |b|² = a·b Given |b| = 1/2, a·b = (1/2)² = 1/4. Step 2: Write c in terms of a and b using the ratio theorem. C lies on AB with AC : CB = 2 : 1, so C is two thirds of the way from A to B: c = a + (2/3)(b - a) = a - (2/3)a + (2/3)b = (1/3)a + (2/3)b. Step 3: Evaluate a·c. a·c = a·[(1/3)a + (2/3)b] = (1/3)(a·a) + (2/3)(a·b) = (1/3)|a|² + (2/3)(a·b) a is a unit vector, so |a| = 1 and |a|² = 1, and from Step 1 a·b = 1/4: = (1/3)(1) + (2/3)(1/4) = 1/3 + 1/6 = 2/6 + 1/6 = 1/2 Hence |a·c| = |1/2| = 1/2. Step 4: Give the geometric meaning. By definition a·c = |a||c|cos(theta), where theta is the angle between a and c. Because a is a UNIT vector, |a| = 1 and so |a·c| = |c||cos(theta)|, which is precisely the length of the projection of c onto a, i.e. the distance from O to the foot of the perpendicular dropped from C onto the line OA.