The perpendicular distance of point (1, 1, 3) on the line \( \displaystyle\frac{x}{2}=\displaystyle\frac{y+1}{3}=\displaystyle\frac{-z-3}{-5} \) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \sqrt{3} \) units
In standard form, \( \frac{-z-3}{-5}=\frac{z+3}{5} \), so the line passes through (0, −1, −3) with direction (2, 3, 5).
A general point of the line is
\( \displaystyle Q=(2t,\ 3t-1,\ 5t-3) \)
With P(1, 1, 3), the vector PQ is
\( \displaystyle \overrightarrow{PQ}=(2t-1,\ 3t-2,\ 5t-6) \)
For the foot of the perpendicular, PQ is perpendicular to the line:
\( \displaystyle 2(2t-1)+3(3t-2)+5(5t-6)=0 \)
\( \displaystyle 38t-38=0\ \Rightarrow\ t=1 \)
So the foot is Q(2, 2, 2), and
\( \displaystyle PQ=\sqrt{(2-1)^2+(2-1)^2+(2-3)^2} \)
\( \displaystyle =\sqrt{3} \)
The distance is \( \sqrt{3} \) units.