The foot of perpendicular from the point P (1, 2, −3) to the line \(\displaystyle\frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (1, 1, −1)
Let each ratio of the line equal \(\lambda\). Then a general point of the line is
\(Q(2\lambda - 1,\ -2\lambda + 3,\ -\lambda)\)
The vector from P to Q is
\(\overrightarrow{PQ} = (2\lambda - 2,\ -2\lambda + 1,\ -\lambda + 3)\)
At the foot of the perpendicular, \(\overrightarrow{PQ}\) is perpendicular to the direction (2, −2, −1):
\(2(2\lambda - 2) - 2(-2\lambda + 1) - 1(-\lambda + 3) = 0\)
\(4\lambda - 4 + 4\lambda - 2 + \lambda - 3 = 0\)
\(9\lambda - 9 = 0,\quad \lambda = 1\)
So the foot is
\(Q(2 - 1,\ -2 + 3,\ -1) = (1, 1, -1)\)