Perpendiculars to a line, distances from a line and the equation of a plane
Answers are for parents and teachers.
1
Find the foot of the perpendicular from A(0, 3, −2) to the line r = (1, 0, 0) + t(2, −1, 2), the line through A meeting it at right angles, and the distance.
C = (1 + 2t, −t, 2t), AC = (2t + 1, −t − 3, 2t + 2). AC · d = 4t + 2 + t + 3 + 4t + 4 = 9t + 9 = 0, so t = −1, C = (−1, 1, −2) and AC = (−1, −2, 0). The line is r = (0, 3, −2) + s(−1, −2, 0) and the distance is √5.
2
Write the equation of the plane through (−3, 1, 2) with normal vector (4, −2, 1) in general form.
n · A = −12 − 2 + 2 = −12, so 4x − 2y + z = −12, that is 4x − 2y + z + 12 = 0.
3
Find the plane through D(1, −1, 2), E(3, 0, −1) and F(2, 2, 1).
DE = (2, 1, −3), DF = (1, 3, −1). DE × DF = (1 × (−1) − (−3) × 3, −(2 × (−1) − (−3) × 1), 2 × 3 − 1 × 1) = (8, −1, 5). n · D = 8 + 1 + 10 = 19, so 8x − y + 5z = 19. Check E: 24 − 0 − 5 = 19; F: 16 − 2 + 5 = 19.
4
Find the plane containing (1, 0, 2) + s(1, 1, 0) and (2, 3, 0) + t(1, −1, 2), which meet at (3, 2, 2).
n = (1, 1, 0) × (1, −1, 2) = (1 × 2 − 0 × (−1), −(1 × 2 − 0 × 1), 1 × (−1) − 1 × 1) = (2, −2, −2), or (1, −1, −1). n · P = 3 − 2 − 2 = −1, so x − y − z = −1, that is x − y − z + 1 = 0. Check (1, 0, 2): 1 − 0 − 2 = −1 and (2, 3, 0): 2 − 3 − 0 = −1.
5
Use the vector product to find the distance from P(2, 1, −1) to the line through A(0, 1, 1) with direction (2, 2, −1).
AP = (2, 0, −2). AP × d = (0 × (−1) − (−2) × 2, −(2 × (−1) − (−2) × 2), 2 × 2 − 0 × 2) = (4, −2, 4), of size 6. Divide by ‖d‖ = 3: the distance is 2.