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Distance of point from plane parallel to line

Short answer

Line through (1,-2,3) parallel to (2,3,-6): (1+2t, -2+3t, 3-6t). Intersection with x-y+z=5 gives t=1/7. Distance = |t|√(2²+3²+(-6)²) = (1/7)×7 = 1 unit.

Find the distance of the point (1,−2,3) from the plane x−y+z = 5 measured parallel to the line x/2 = y/3 = z/(−6).

Given. Point P(1, −2, 3), plane x − y + z = 5, and the direction of the line x/2 = y/3 = z/(−6), which is (2, 3, −6).

Method

Draw the line from P parallel to (2, 3, −6) and find where it meets the plane. The distance from P to that point is the required distance.

A point on the line through P: (1 + 2t, −2 + 3t, 3 − 6t).

Meeting the plane

(1 + 2t) − (−2 + 3t) + (3 − 6t) = 5
1 + 2t + 2 − 3t + 3 − 6t = 5
6 − 7t = 5
t = 1/7

The meeting point is Q(1 + 2/7, −2 + 3/7, 3 − 6/7) = (9/7, −11/7, 15/7).

Distance

PQ = |t| √(2² + 3² + (−6)²) = (1/7)(7) = 1

Direct check: PQ = (2/7, 3/7, −6/7), and √(4 + 9 + 36)/7 = 7/7 = 1.

Answer: the distance is 1 unit.

✓ Answer checked by Claude Opus 5.5 (high effort)

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