Question:
The equation of the perpendicular line from the point \((2,-3,1)\) to the line \(\frac{x+1}{2} = \frac{y-3}{3} = \frac{z+2}{-1}\) is
The equation of the perpendicular line from the point \((2,-3,1)\) to the line \(\frac{x+1}{2} = \frac{y-3}{3} = \frac{z+2}{-1}\) is
Show Hint
On the YOZ plane x = 0, so find the parameter where the x coordinate vanishes.
Updated On: Oct 1, 2026
- \(\frac{x-2}{24} = \frac{y+3}{13} = \frac{z-1}{9}\)
- \(\frac{x-2}{24} = \frac{y-3}{-13} = \frac{z-1}{9}\)
- \(\frac{x+2}{-24} = \frac{y+3}{13} = \frac{z+1}{-9}\)
- \(\frac{x-2}{-24} = \frac{y+3}{13} = \frac{z-1}{-9}\)
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The Correct Option is D
Solution and Explanation
Step 1: Equation of the line:
Points \((3, 5, -7)\) and \((-2, 1, 8)\). A general point is \((3 - 5t,\ 5 - 4t,\ -7 + 15t)\).
Step 2: YOZ plane:
On the YOZ plane \(x = 0\): \(3 - 5t = 0\), so \(t = \frac35\).
Step 3: Coordinates:
\(y = 5 - \frac{12}{5} = \frac{13}{5}\) and \(z = -7 + 9 = 2\). The point is \(\left(0, \frac{13}{5}, 2\right)\).
Final Answer:
The point is \(\left(0, \frac{13}{5}, 2\right)\), option (C). \[ \boxed{\left(0,\frac{13}{5},2\right)} \]
Points \((3, 5, -7)\) and \((-2, 1, 8)\). A general point is \((3 - 5t,\ 5 - 4t,\ -7 + 15t)\).
Step 2: YOZ plane:
On the YOZ plane \(x = 0\): \(3 - 5t = 0\), so \(t = \frac35\).
Step 3: Coordinates:
\(y = 5 - \frac{12}{5} = \frac{13}{5}\) and \(z = -7 + 9 = 2\). The point is \(\left(0, \frac{13}{5}, 2\right)\).
Final Answer:
The point is \(\left(0, \frac{13}{5}, 2\right)\), option (C). \[ \boxed{\left(0,\frac{13}{5},2\right)} \]
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