Lines in space: direction vectors, three forms of equation and the angle between lines
Answers are for parents and teachers.
1
Write the line through A(−2, 3, 1) and B(1, −1, 5) in vector, parametric and Cartesian form.
d = AB = (3, −4, 4). Vector: r = (−2, 3, 1) + t(3, −4, 4). Parametric: x = −2 + 3t, y = 3 − 4t, z = 1 + 4t. Cartesian: (x + 2)/3 = (y − 3)/(−4) = (z − 1)/4.
2
For the line (x − 1)/2 = (y + 2)/3 = (3 − z)/6 find a point, the direction cosines, and whether (5, 4, −9) lies on it.
(3 − z)/6 = (z − 3)/(−6), so the point is (1, −2, 3) and d = (2, 3, −6) with ‖d‖ = 7, cosines (2/7, 3/7, −6/7). For (5, 4, −9): (5 − 1)/2 = 2, (4 + 2)/3 = 2 and (−9 − 3)/(−6) = 2, all equal, so yes.
3
Find the angle between the lines with directions (3, 4, 12) and (1, −2, 2).
The dot product is 3 − 8 + 24 = 19, the norms are 13 and 3, so cos θ = 19/39 and θ ≈ 60.8°.
4
Do (0, 1, 2) + s(1, −1, 2) and (−1, 0, 3) + t(3, −1, 3) meet? If so, where?
x: s = −1 + 3t; y: 1 − s = −t. Substituting: 1 + 1 − 3t = −t, so t = 1 and s = 2. Third: 2 + 4 = 6 and 3 + 3 = 6, agree. They meet at (2, −1, 6).
5
Show that (1, 0, 0) + s(1, 2, 2) and (0, 1, 3) + t(2, −1, 0) are perpendicular and skew.
(1, 2, 2) · (2, −1, 0) = 2 − 2 + 0 = 0, so they are perpendicular. x: 1 + s = 2t; y: 2s = 1 − t. From these s = 1/5 and t = 3/5. Third: 2s = 2/5 but 3 + 0t = 3, which do not agree, so there is no common point: skew.