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, β2) to the line r = (1, 0, 0) + t(2, β1, 2), the line through A meeting it at right angles, and the distance.
C = (1 + 2t, βt, 2t), AC = (2t + 1, βt β 3, 2t + 2). AC Β· d = 4t + 2 + t + 3 + 4t + 4 = 9t + 9 = 0, so t = β1, C = (β1, 1, β2) and AC = (β1, β2, 0). The line is r = (0, 3, β2) + s(β1, β2, 0) and the distance is β5.
2
Write the equation of the plane through (β3, 1, 2) with normal vector (4, β2, 1) in general form.
n Β· A = β12 β 2 + 2 = β12, so 4x β 2y + z = β12, that is 4x β 2y + z + 12 = 0.
3
Find the plane through D(1, β1, 2), E(3, 0, β1) and F(2, 2, 1).
DE = (2, 1, β3), DF = (1, 3, β1). DE Γ DF = (1 Γ (β1) β (β3) Γ 3, β(2 Γ (β1) β (β3) Γ 1), 2 Γ 3 β 1 Γ 1) = (8, β1, 5). n Β· D = 8 + 1 + 10 = 19, so 8x β y + 5z = 19. Check E: 24 β 0 β 5 = 19; F: 16 β 2 + 5 = 19.
4
Find the plane containing (1, 0, 2) + s(1, 1, 0) and (2, 3, 0) + t(1, β1, 2), which meet at (3, 2, 2).
n = (1, 1, 0) Γ (1, β1, 2) = (1 Γ 2 β 0 Γ (β1), β(1 Γ 2 β 0 Γ 1), 1 Γ (β1) β 1 Γ 1) = (2, β2, β2), or (1, β1, β1). n Β· P = 3 β 2 β 2 = β1, so x β y β z = β1, that is x β y β z + 1 = 0. Check (1, 0, 2): 1 β 0 β 2 = β1 and (2, 3, 0): 2 β 3 β 0 = β1.
5
Use the vector product to find the distance from P(2, 1, β1) to the line through A(0, 1, 1) with direction (2, 2, β1).
AP = (2, 0, β2). AP Γ d = (0 Γ (β1) β (β2) Γ 2, β(2 Γ (β1) β (β2) Γ 2), 2 Γ 2 β 0 Γ 2) = (4, β2, 4), of size 6. Divide by βdβ = 3: the distance is 2.