Lines intersect: point and plane through them

Two lines intersect if a common point exists; then they are coplanar.

Line 1: (x−1)/2 = (y+1)/(−3) = (z+10)/8 = t, so (1 + 2t, −1 − 3t, −10 + 8t). Line 2: (x−4)/1 = (y+3)/(−4) = (z+1)/7 = s, so (4 + s, −3 − 4s, −1 + 7s).

Intersection

Equate the coordinates:

1 + 2t = 4 + s          → 2t − s = 3     ... (i)
−1 − 3t = −3 − 4s       → −3t + 4s = −2  ... (ii)
−10 + 8t = −1 + 7s      → 8t − 7s = 9    ... (iii)

From (i), s = 2t − 3. Put in (ii): −3t + 8t − 12 = −2, so t = 2 and s = 1.

Check (iii): 8(2) − 7(1) = 9 ✓. All three equations are satisfied, so the lines intersect.

Point of intersection: t = 2 gives (5, −7, 6). (s = 1 gives (5, −7, 6) as well.)

Plane through them

Normal = d₁ × d₂ with d₁ = (2, −3, 8), d₂ = (1, −4, 7):

n = ((−3)(7) − 8(−4), 8(1) − 2(7), 2(−4) − (−3)(1))
  = (−21 + 32, 8 − 14, −8 + 3) = (11, −6, −5)

Plane through (5, −7, 6):

11(x − 5) − 6(y + 7) − 5(z − 6) = 0 → 11x − 6y − 5z − 55 − 42 + 30 = 0.

Answer: the lines meet at (5, −7, 6) and lie in the plane 11x − 6y − 5z = 67.

Check: the point (1, −1, −10) of line 1: 11 + 6 + 50 = 67 ✓.