JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the line passing through the points, \(P(2,-1,2)\) and \(Q(5,3,4)\) meet the plane \(x-y+z=4\) at the point \(R\). Then the distance of the point \(R\) from the plane \(x+2 y+3 z+2=0\) measured parallel to the line \(\frac{x-7}{2}=\frac{y+3}{2}=\frac{z-2}{1}\) is equal to
- A \(\sqrt{31}\)
- B \(\sqrt{189}\)
- C \(\sqrt{61}\)
- D \(3\)
Answer & Solution
Correct Answer
(D) \(3\)
Step-by-step Solution
Detailed explanation
Line : \(\frac{x-5}{3}=\frac{y-3}{4}=\frac{z-4}{2}=\lambda\) \(R (3 \lambda+5,4 \lambda+3,2 \lambda+4)\) \(\therefore 3 \lambda+5-4 \lambda-3+2 \lambda+4=4\) \(\lambda+6=4 \therefore \lambda=-2\) \(\therefore R \equiv(-1,-5,0)\) Line:…
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