A-Level Scalar and vector products: worked solution

Question

Relative to origin O, points A and B have position vectors a and b. The magnitude of a is k units and b is a unit vector, and the angle between a and b is π/6 radians. Point M lies on segment AB with AM:AB=3:4. Find the exact area of triangle OAM in terms of k.

Step 1: find the area of triangle OAB. With O the origin, OA = a and OB = b, and the area of a triangle formed by two vectors from a common point is half the magnitude of their cross product: area of triangle OAB = (1/2)|a × b| = (1/2)|a||b| sin θ, where θ is the angle between a and b. Step 2: substitute the given data. |a| = k, |b| = 1 (b is a unit vector) and θ = π/6, and sin(π/6) = 1/2, so area of triangle OAB = (1/2)(k)(1)(1/2) = k/4. Step 3: relate triangle OAM to triangle OAB. M lies on the segment AB, so triangles OAM and OAB share the vertex O and have their bases AM and AB on the same straight line. The perpendicular distance from O to that line is the same for both, so their areas are in the same ratio as their bases: (area of triangle OAM)/(area of triangle OAB) = AM/AB = 3/4. Step 4: combine the two results. area of triangle OAM = (3/4) × (k/4) = 3k/16 units².