Coplanar lines and point of intersection

Lines.

• L₁: (x + 4)/3 = (y + 6)/5 = (z − 1)/(−2).
• L₂: 3x − 2y + z + 5 = 0 and 2x + 3y + 4z − 4 = 0 (a line given as two planes).

Approach

Two lines are coplanar if they intersect (or are parallel). We test whether L₁ meets L₂ by putting a general point of L₁ into both planes of L₂.

A general point of L₁: (3t − 4, 5t − 6, 1 − 2t).

Plane 1: 3(3t − 4) − 2(5t − 6) + (1 − 2t) + 5 = 0
         9t − 12 − 10t + 12 + 1 − 2t + 5 = 0
         −3t + 6 = 0   →  t = 2

Plane 2: 2(3t − 4) + 3(5t − 6) + 4(1 − 2t) − 4 = 0
         6t − 8 + 15t − 18 + 4 − 8t − 4 = 0
         13t − 26 = 0  →  t = 2

Both planes give the same value t = 2, so the point of L₁ with t = 2 lies on both planes, i.e. on L₂. The lines meet, so they are coplanar. (They are not parallel: L₂ has direction (3, −2, 1) × (2, 3, 4) = (−11, −10, 13), which is not proportional to (3, 5, −2).)

Point of intersection

With t = 2: x = 3(2) − 4 = 2, y = 5(2) − 6 = 4, z = 1 − 4 = −3.

Check in L₂: 3(2) − 2(4) + (−3) + 5 = 0 ✓ and 2(2) + 3(4) + 4(−3) − 4 = 0 ✓.

Answer: the lines are coplanar and intersect at (2, 4, −3).