A-Level Vector geometry in three dimensions: worked solution

Question

Plane π2 has normal (1,5,1). Find plane π3, parallel to π2 and passing through B(2,−3,7), and hence show that π2 and π3 are a distance 20/(3√3) apart.

Step 1: write down the normal of π3. Parallel planes have the same normal vector, so π3 also has normal n = (1, 5, 1) and its equation has the form r·(1, 5, 1) = D for some constant D. Only D is unknown. Step 2: use the point B to find D. B(2, −3, 7) lies in π3, so its position vector satisfies the equation: D = (2, −3, 7)·(1, 5, 1) = 2(1) + (−3)(5) + 7(1) = 2 − 15 + 7 = −6. So π3: r·(1, 5, 1) = −6. Step 3: find the distance between the planes. For two parallel planes r·n = D1 and r·n = D2 written with the SAME normal vector n, the perpendicular distance between them is distance = |D1 − D2| / |n|. (The reason: take any point on one plane; its perpendicular distance to the other is |r·n − D|/|n|, and r·n is the constant of its own plane.) Here π2 is r·(1, 5, 1) = 14 and π3 is r·(1, 5, 1) = −6. |n| = √(1² + 5² + 1²) = √(1 + 25 + 1) = √27. Simplify the surd: √27 = √(9 × 3) = 3√3. distance = |14 − (−6)| / (3√3) = 20/(3√3), which is the required result.