The foot of the perpendicular drawn from the point P(5, 2, 3) to the line \(\vec{r} = (\hat{\imath}-\hat{\jmath}+2\hat{k})+t(2\hat{\imath}+\hat{\jmath}-2\hat{k})\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (3, 0, 0)
A general point of the line is
\(Q(1+2t,\ -1+t,\ 2-2t)\)
The vector from Q to P is
\(\overrightarrow{QP} = (4-2t)\hat{\imath}+(3-t)\hat{\jmath}+(1+2t)\hat{k}\)
At the foot of the perpendicular, \(\overrightarrow{QP}\) is perpendicular to the direction 2, 1, −2:
\(2(4-2t)+1(3-t)-2(1+2t) = 0\)
\(9-9t = 0\)
\(t = 1\)
\(Q = (1+2,\ -1+1,\ 2-2) = (3, 0, 0)\)
Why the other options are incorrect
All three points are on the line (t = 0, 2 and −1), but the joining vector from them to P is not perpendicular to the line:
Option (1) (1, −1, 2) — Incorrect. This is the point t = 0; \(\overrightarrow{QP} = (4, 3, 1)\) and \(8+3-2 = 9 \ne 0\).
Option (2) (5, 1, −2) — Incorrect. This is the point t = 2; \(\overrightarrow{QP} = (0, 1, 5)\) and \(0+1-10 = -9 \ne 0\).
Option (3) (−1, −2, 4) — Incorrect. This is the point t = −1; \(\overrightarrow{QP} = (6, 4, -1)\) and \(12+4+2 = 18 \ne 0\).