The perpendicular distance of the point P(5, 1, 3) from the line \(\displaystyle\frac{x-1}{2}=\frac{y}{-1}=\frac{z-2}{2}\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3
The general point of the line is
\(Q = (2t+1,\ -t,\ 2t+2)\)
\(\overrightarrow{QP} = (4-2t,\ 1+t,\ 1-2t)\)
Q is the foot of the perpendicular when \(\overrightarrow{QP}\) is perpendicular to the direction 2, −1, 2:
\(2(4-2t)-(1+t)+2(1-2t) = 0\)
\(9-9t = 0\)
\(t = 1\)
So the foot is Q(3, −1, 4) and \(\overrightarrow{QP} = (2, 2, -1)\):
\(PQ = \sqrt{4+4+1} = 3\)
Why the other options are incorrect
Option (1) \(3\sqrt{3}\) — Incorrect. From A(1, 0, 2) on the line, \(AP^2 = 18\) and the part of AP along the line is 3; the distance is \(\sqrt{18-9} = 3\), while \(\sqrt{18+9} = 3\sqrt{3}\) adds the 9 instead of taking it away.
Option (2) \(3\sqrt{2}\) — Incorrect. This is the distance from P to the point (1, 0, 2) of the line, which is not the foot of the perpendicular.
Option (3) 1 — Incorrect. This is the value of t at the foot, not the distance.