JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the foot of the perpendicular from the point \((1,2,4)\) on the line \(\frac{x+2}{4}=\frac{y-1}{2}=\frac{z+1}{3}\) be \(P.\) Then the distance of \(P\) from the plane \(3 x+4 y+12 z+23=0\)
- A \(5\)
- B \(\frac{50}{13}\)
- C \(4\)
- D \(\frac{63}{13}\)
Answer & Solution
Correct Answer
(A) \(5\)
Step-by-step Solution
Detailed explanation
\(\frac{ x +2}{4}=\frac{ y -1}{2}=\frac{ z +1}{3}=\lambda\) \(( x , y , z )=(4 \lambda-2,2 \lambda+1,3 \lambda-1)\) \(\overline{ AP }=(4 \lambda-3) \hat{ i }+(2 \lambda-1) \hat{ j }+(3 \lambda-5) \hat{ k }\) \(\overrightarrow{ b }=4 \hat{ i }+2 \hat{ j }+3 \hat{ k }\)…
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