The points on the line \( \displaystyle\frac{x+2}{3}=\displaystyle\frac{y+1}{2}=\displaystyle\frac{z-3}{2} \) at a distance of \( \sqrt{17} \) from the point (-2, -1, 3) are:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (1, 1, 5) and (-5, -3, 1)
Putting each fraction equal to t gives the general point of the line:
\( \displaystyle Q(3t-2,\ 2t-1,\ 2t+3) \)
The given point P(−2, −1, 3) is on the line (t = 0). Its distance from Q is
\( \displaystyle PQ=\sqrt{(3t)^2+(2t)^2+(2t)^2} \)
\( \displaystyle =\sqrt{17t^2}=\sqrt{17}\,|t| \)
\( \displaystyle \sqrt{17}\,|t|=\sqrt{17}\ \Rightarrow\ t=\pm 1 \)
\( \displaystyle t=1:\ (1,\ 1,\ 5) \)
\( \displaystyle t=-1:\ (-5,\ -3,\ 1) \)
Shortcut: Check whether the option points lie on the line: (5, 3, 1) gives \( \frac{7}{3}\ne\frac{4}{2} \), (1, 1, −5) gives \( \frac{z-3}{2}=-4 \), not 1, and (1, −1, 5) gives \( \frac{y+1}{2}=0\ne 1 \). Only both points of Option (3) lie on the line.