Points on the line \( \displaystyle\frac{x+2}{3}=\displaystyle\frac{y+1}{2}=\displaystyle\frac{z-3}{2} \) at a distance of 5 units from the point (1, 3, 3), are
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (−2, −1, 3) and (4, 3, 7)
A general point on the line is \( (3t-2,\ 2t-1,\ 2t+3) \). Its distance from (1, 3, 3) is 5:
\( \displaystyle (3t-3)^2+(2t-4)^2+(2t)^2=25 \)
\( \displaystyle 17t^2-34t+25=25 \)
\( \displaystyle 17t(t-2)=0 \)
\( \displaystyle t=0\ \text{or}\ t=2 \)
t = 0 gives (−2, −1, 3) and t = 2 gives (4, 3, 7).
Shortcut: The points must lie on the line, so put the options into the line’s equation. Only the points of Option (1) fit: (−2, −1, 3) gives 0 = 0 = 0, and (4, 3, 7) gives 2 = 2 = 2.