A-Level Scalar and vector products: worked solution

Question

Explain, for arbitrary vectors m and n, why m . (m x n) equals 0. [1] Relative to origin O, points A, B, C have position vectors a, b, c, and O, A, B, C are non-coplanar. M is the midpoint of AC and p is the plane OAB. (a) If R is chosen so MR is perpendicular to p, show R lies on the line l given by r = k(a + c) + lambda(a x b), lambda real, determining k. [2] (b) Given a and b are mutually perpendicular unit vectors satisfying a.c = -2 and b.c = 4, express, using a and b, the position vector at which l crosses p. [3]

Opening result [1] By the definition of the vector (cross) product, m x n is perpendicular to BOTH m and n. Two perpendicular vectors have scalar product zero, so m . (m x n) = 0 (In the special cases where m or n is the zero vector, or m and n are parallel, m x n is itself the zero vector and the scalar product is again 0.) (a) [2] Step 1: write down the position vector of M. M is the midpoint of AC, so OM = (1/2)(a + c) Step 2: find a normal to the plane p. The plane p is the plane OAB. It passes through the origin O and contains the directions a and b, so a vector perpendicular to it is n = a x b Step 3: interpret the condition on R. "MR is perpendicular to p" means MR is perpendicular to every direction lying in p, which is the same as saying MR is PARALLEL to the normal. So MR = λ(a x b) for some real λ Step 4: add the two vectors to reach R. OR = OM + MR = (1/2)(a + c) + λ(a x b) This is exactly the given form r = k(a + c) + λ(a x b), so R lies on l with k = 1/2 (b) [3] Now a and b are perpendicular unit vectors, so a.a = 1, b.b = 1, a.b = 0, and we are given a.c = -2, b.c = 4 Step 1: write the crossing point in two different ways. The plane p = OAB passes through O and is spanned by a and b, so every point of p has position vector r = αa + βb for some scalars α, β ... (1) And from part (a), every point of l has position vector r = (1/2)(a + c) + λ(a x b) ... (2) At the crossing point both descriptions hold. Step 2: take the scalar product of both forms with a. The key move is that a . (a x b) = 0 by the opening result, so the unknown λ disappears. From (2): r.a = (1/2)(a.a + c.a) + λ(a x b).a = (1/2)(1 + (-2)) + 0 = (1/2)(-1) = -1/2 From (1): r.a = α(a.a) + β(b.a) = α(1) + β(0) = α Equating: α = -1/2 Step 3: take the scalar product of both forms with b. Again b . (a x b) = 0, since a x b is perpendicular to b as well. From (2): r.b = (1/2)(a.b + c.b) + λ(a x b).b = (1/2)(0 + 4) + 0 = 2 From (1): r.b = α(a.b) + β(b.b) = α(0) + β(1) = β Equating: β = 2 Step 4: assemble the answer. r = -(1/2)a + 2b