A-Level Vector geometry in three dimensions: worked solution

Question

With l: r = (2, 1, 0) + lambda(3, -2, 1), rooftop p: 2x + y - z = 7 and F = (8/3, 4/3, -1/3), the drone then glides along a path s on the rooftop perpendicular to l. Find a cartesian equation of s. A second rooftop parallel to p is a perpendicular distance d away; give two possible cartesian equations for it in terms of d.

PART 1: the glide path s. The path s must satisfy two conditions at once, and each one pins its direction down against a known vector: - s lies ON the rooftop plane p, so its direction is perpendicular to the normal of p, n = (2, 1, -1) - s is PERPENDICULAR to the line l, so its direction is perpendicular to l's direction (3, -2, 1) A vector perpendicular to two given vectors is their cross product, so that is what we compute. Step 1: check that F is a valid point of s by confirming it lies on the rooftop p: 2x + y - z = 7. 2(8/3) + (4/3) - (-1/3) = 16/3 + 4/3 + 1/3 = 21/3 = 7 It does, so F can be used as the point on s. Step 2: compute the direction of s as (3, -2, 1) x (2, 1, -1), using (a1, a2, a3) x (b1, b2, b3) = (a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1) i-component: (-2)(-1) - (1)(1) = 2 - 1 = 1 j-component: (1)(2) - (3)(-1) = 2 + 3 = 5 k-component: (3)(1) - (-2)(2) = 3 + 4 = 7 So the direction of s is (1, 5, 7). Step 3: write the cartesian equation from the point F(8/3, 4/3, -1/3) and direction (1, 5, 7), using the standard form (x - x0)/a = (y - y0)/b = (z - z0)/c. (x - 8/3)/1 = (y - 4/3)/5 = (z + 1/3)/7 Step 4: clear the thirds. Write x - 8/3 = (3x - 8)/3, y - 4/3 = (3y - 4)/3 and z + 1/3 = (3z + 1)/3, so the denominators become 3, 3 × 5 = 15 and 3 × 7 = 21. s: (3x - 8)/3 = (3y - 4)/15 = (3z + 1)/21 PART 2: the second rooftop. Step 5: a plane parallel to p has the SAME normal, so only the constant term changes: 2x + y - z = c for some constant c Step 6: the perpendicular distance between the parallel planes 2x + y - z = 7 and 2x + y - z = c is the difference of the constants divided by the length of the normal. |n| = √(2² + 1² + (-1)²) = √(4 + 1 + 1) = √6 distance = |c - 7|/√6 Step 7: set this equal to d and solve. The modulus gives two answers, because the second rooftop can be on either side of p. |c - 7|/√6 = d |c - 7| = √6 d c - 7 = √6 d or c - 7 = -√6 d c = 7 + √6 d or c = 7 - √6 d So the two possible cartesian equations are 2x + y - z = 7 + √6 d 2x + y - z = 7 - √6 d