3D Line Vector Equations Practice

Write the vector equation of the line passing through point P(2, -1, 4) in the direction of vector d = (3, 5, -2).

r = (2, -1, 4) + t(3, 5, -2)

Find a vector equation for the line passing through points A(1, 0, 3) and B(4, 2, -1).

Direction vector d = B - A = (3, 2, -4). The line is r = (1, 0, 3) + t(3, 2, -4) or r = (4, 2, -1) + t(3, 2, -4).

Determine whether the point Q(7, 3, 5) lies on the line r = (1, 1, 1) + t(3, 1, 2).

Set up coordinates: 7 = 1 + 3t gives t = 2; 3 = 1 + t gives t = 2; 5 = 1 + 2t gives t = 2. Since t = 2 in all three coordinates, Q lies on the line.

Determine whether the point S(4, -2, 1) lies on the line r = (2, 0, 3) + t(1, -1, -2).

From x: 4 = 2 + t gives t = 2. From y: -2 = 0 - t gives t = 2. From z: 1 = 3 - 2t gives -2 = -2t, so t = 1. Since t values differ (2 vs 1), S does not lie on the line.

State whether the lines r1 = (1, 2, 3) + s(2, -4, 6) and r2 = (0, 5, -1) + t(-1, 2, -3) are parallel, intersecting, or skew.

Their direction vectors are (2, -4, 6) and (-1, 2, -3). Notice that (2, -4, 6) = -2(-1, 2, -3). Because the direction vectors are scalar multiples, the lines are parallel.

Find the point of intersection between line 1: r1 = (1, 0, 1) + s(1, 1, 0) and line 2: r2 = (0, 2, 3) + t(2, -1, -2).

Equating z-components: 1 = 3 - 2t gives t = 1. Equating y-components: s = 2 - t = 1. Check x: 1 + s = 2 and 0 + 2t = 2, which match. The point of intersection is (2, 1, 1).