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Perpendicular from point to line

Short answer

Let foot Q(2t,2+3t,3+4t). Dot (2,3,4)·PQ=0 gives t=1, Q(2,5,7). Length = √[(2-3)²+(5+1)²+(7-11)²]=√53. Perpendicular: (x-3)/-1 = (y+1)/6 = (z-11)/-4.

Find the length of perpendicular from the point (3,−1,11) to the line x/2 = (y−2)/3 = (z−3)/4. Also obtain the equation of the perpendicular.

Let P = (3, −1, 11) and the line L: x/2 = (y − 2)/3 = (z − 3)/4, which passes through A(0, 2, 3) and has direction ratios (2, 3, 4).

Foot of the perpendicular

A general point of L is Q(2t, 2 + 3t, 3 + 4t). Then

PQ = (2t − 3, 3t + 3, 4t − 8).

PQ is perpendicular to the direction (2, 3, 4):

2(2t − 3) + 3(3t + 3) + 4(4t − 8) = 0
4t − 6 + 9t + 9 + 16t − 32 = 0
29t − 29 = 0  →  t = 1

So the foot is Q(2, 5, 7).

Length of the perpendicular

PQ = √[(2 − 3)² + (5 + 1)² + (7 − 11)²]
   = √(1 + 36 + 16)
   = √53 ≈ 7.28

Equation of the perpendicular

It is the line through P(3, −1, 11) and Q(2, 5, 7). Direction ratios: (2 − 3, 5 + 1, 7 − 11) = (−1, 6, −4).

(x − 3)/(−1) = (y + 1)/6 = (z − 11)/(−4)

Check: dot product of (−1, 6, −4) with (2, 3, 4) is −2 + 18 − 16 = 0, so it is perpendicular to the line.

Answer: length = √53 units; equation: (x − 3)/(−1) = (y + 1)/6 = (z − 11)/(−4).

✓ Answer checked by Claude Opus 5.5 (high effort)

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