Question:
Let \(P(\alpha, \beta, \lambda)\) be the image of the point \(\mathrm{Q}(1, 2, 0)\) in the line \(\frac{x - 5}{3} = \frac{y - 12}{1} = \frac{z - 10}{2}\), then \((\mathrm{PQ})^2\) is equal to
Let \(P(\alpha, \beta, \lambda)\) be the image of the point \(\mathrm{Q}(1, 2, 0)\) in the line \(\frac{x - 5}{3} = \frac{y - 12}{1} = \frac{z - 10}{2}\), then \((\mathrm{PQ})^2\) is equal to
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Find the foot of the perpendicular F from Q to the line; the image P satisfies PQ = 2 QF.
Updated On: Nov 12, 2026
- 90
- 180
- 360
- 270
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The Correct Option is C
Solution and Explanation
A general point on the line is \(F(5+3t,\ 12+t,\ 10+2t)\); the direction vector is \((3,1,2)\).
\(\overrightarrow{QF}=(4+3t,\ 10+t,\ 10+2t)\) must be perpendicular to the line: \(3(4+3t)+(10+t)+2(10+2t)=0\Rightarrow 42+14t=0\Rightarrow t=-3\).
So \(F=(-4,\ 9,\ 4)\) and \(\overrightarrow{QF}=(-5,\ 7,\ 4)\), \(QF^2=25+49+16=90\).
F is the midpoint of PQ, so \(PQ=2\,QF\) and \(PQ^2=4\cdot90=360\).
Answer: (3) 360
\(\overrightarrow{QF}=(4+3t,\ 10+t,\ 10+2t)\) must be perpendicular to the line: \(3(4+3t)+(10+t)+2(10+2t)=0\Rightarrow 42+14t=0\Rightarrow t=-3\).
So \(F=(-4,\ 9,\ 4)\) and \(\overrightarrow{QF}=(-5,\ 7,\ 4)\), \(QF^2=25+49+16=90\).
F is the midpoint of PQ, so \(PQ=2\,QF\) and \(PQ^2=4\cdot90=360\).
Answer: (3) 360
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