Coplanar lines with point of intersection

Lines.

• L₁: (x − 1)/2 = (y − 2)/3 = (z − 3)/4.
• L₂: 4x − 3y + 1 = 0 and 5x − 3z + 2 = 0.

Method 1: test for a common point

A general point of L₁ is (1 + 2t, 2 + 3t, 3 + 4t). Put it into both planes of L₂:

4(1 + 2t) − 3(2 + 3t) + 1 = 4 + 8t − 6 − 9t + 1 = −1 − t = 0   →  t = −1
5(1 + 2t) − 3(3 + 4t) + 2 = 5 + 10t − 9 − 12t + 2 = −2 − 2t = 0 →  t = −1

Both give t = −1, so this point of L₁ lies on L₂. The lines intersect, so they are coplanar.

Method 2: determinant (as a check)

Write L₂ in symmetric form. From 4x − 3y + 1 = 0: y = (4x + 1)/3. From 5x − 3z + 2 = 0: z = (5x + 2)/3. So with x = 3m: (x, y, z) = (3m, 4m + 1/3, 5m + 2/3). Direction (3, 4, 5), point C(0, 1/3, 2/3).

With A(1, 2, 3): AC = (−1, −5/3, −7/3) ∝ (−3, −5, −7).

| −3  −5  −7 |
|  2   3   4 | = −3(15 − 16) + 5(10 − 12) − 7(8 − 9)
|  3   4   5 |
             = 3 − 10 + 7 = 0

Point of intersection

With t = −1: (1 − 2, 2 − 3, 3 − 4) = (−1, −1, −1).

Answer: the lines are coplanar and intersect at (−1, −1, −1).