Lines and planes meeting, angles between planes, and distances
Answers are for parents and teachers.
1
Find where the line r = (2, −1, 0) + t(1, 3, 2) meets the plane x − y + 2z = 9.
The point is (2 + t, −1 + 3t, 2t). (2 + t) − (−1 + 3t) + 4t = 3 + 2t = 9, so t = 3 and the point is (5, 8, 6). Check: 5 − 8 + 12 = 9.
2
Find the angle between the planes x + 2y − 2z = 5 and 2x − 3y + 6z = 1.
The dot product is 2 − 6 − 12 = −16, the norms are 3 and 7, so cos θ = 16/21 and θ ≈ 40.4°.
3
Find the line of intersection of 3x + y − 2z = 4 and x − y + z = 0.
d = (3, 1, −2) × (1, −1, 1) = (1 × 1 − (−2)(−1), −(3 × 1 − (−2) × 1), 3 × (−1) − 1 × 1) = (−1, −5, −4), or (1, 5, 4). For z = 0: 3x + y = 4 and x = y give x = y = 1. The line is r = (1, 1, 0) + t(1, 5, 4). Check the direction: 3 + 5 − 8 = 0 and 1 − 5 + 4 = 0.
4
Find the distance from (1, 2, 0) to the plane 4x − 4y + 7z = 14.
|4 − 8 + 0 − 14|/√(16 + 16 + 49) = 18/9 = 2.
5
The plane 6x + 4y + 3z = 12 meets the axes at A, B and C. Write it in intercept form, then find the area of triangle ABC, the distance from the origin to the plane and the volume of the tetrahedron OABC.
x/2 + y/3 + z/4 = 1, so A(2, 0, 0), B(0, 3, 0), C(0, 0, 4). AB × AC = (−2, 3, 0) × (−2, 0, 4) = (12, 8, 6), of size √244 = 2√61, so the area is √61 ≈ 7.81. The distance from O is 12/√61. The volume is 2 × 3 × 4/6 = 4; check (1/3) × √61 × 12/√61 = 4.