A-Level Vector geometry in three dimensions: worked solution

Question

A is (1, −2, 4) and π₁: 2x − y + 2z = 5, while π₂: x + 3y − az = 3 with a constant. Find, in terms of a, the vector equation of the line l where π₁ and π₂ meet.

The line of intersection lies in BOTH planes, so it is perpendicular to both normal vectors. A vector perpendicular to two given vectors is their cross product, so that gives the direction of l. Read the normals off the plane equations: n₁ = (2, −1, 2) from π₁: 2x − y + 2z = 5 n₂ = (1, 3, −a) from π₂: x + 3y − az = 3 Direction d = n₁ × n₂, taken component by component: i-component: (−1)(−a) − (2)(3) = a − 6 j-component: −[ (2)(−a) − (2)(1) ] = −(−2a − 2) = 2a + 2 k-component: (2)(3) − (−1)(1) = 6 + 1 = 7 d = (a − 6, 2a + 2, 7) The k-component is 7, which is never zero, so d ≠ 0 for every value of a: the planes are never parallel and the line always exists. Next find ONE point on the line, i.e. one point satisfying both plane equations. That is two equations in three unknowns, so one variable may be fixed freely - take z = 0: 2x − y = 5 x + 3y = 3 From the first equation, y = 2x − 5. Substitute into the second: x + 3(2x − 5) = 3 x + 6x − 15 = 3 7x = 18, so x = 18/7 Then y = 2(18/7) − 5 = 36/7 − 35/7 = 1/7 So (18/7, 1/7, 0) lies on both planes. (It satisfies π₂ for every a, since z = 0 kills the −az term: 18/7 + 3(1/7) = 21/7 = 3. ✓) A line through a known point with a known direction has vector equation r = (point) + λ(direction): l: r = (18/7, 1/7, 0) + λ(a − 6, 2a + 2, 7), λ ∈ ℝ