Find the foot of perpendicular from origin to the line \(\displaystyle\frac{x-1}{3}=\frac{y-2}{-2}=\frac{z+1}{3}\).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\left(\displaystyle\frac{17}{11}, \frac{18}{11}, \frac{-5}{11}\right)\)
Let the foot of the perpendicular be the point P of the line for the parameter value t:
\(P = (1+3t,\ 2-2t,\ -1+3t)\)
OP must be perpendicular to the direction (3, −2, 3) of the line:
\(3(1+3t) - 2(2-2t) + 3(-1+3t) = 0\)
\(22t - 4 = 0\)
\( \displaystyle t = \frac{2}{11} \)
Substituting:
\( \displaystyle P = \left(1+\frac{6}{11},\ 2-\frac{4}{11},\ -1+\frac{6}{11}\right) = \left(\frac{17}{11}, \frac{18}{11}, \frac{-5}{11}\right) \)