JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the line \(\frac{{x - 2}}{3} = \frac{{y + 1}}{2} = \frac{{z - 1}}{1}\) intersects the plane \(2x + 3y -z + 13 = 0\) at a point \(P\) and the plane \(3x + y + 4z = 16\) at a point \(Q\), then \(PQ\) is equal to
- A \(2\sqrt {14} \)
- B \(14\)
- C \(2\sqrt {7} \)
- D \(\sqrt {14} \)
Answer & Solution
Correct Answer
(A) \(2\sqrt {14} \)
Step-by-step Solution
Detailed explanation
\(\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-1}{-1}=\lambda\) \(x=3 \lambda+2, y=2 \lambda-1, z=-\lambda+1\) Intersection with plane \(2 x+3 y-z+13=0\) \(2(3 \lambda+2)+3(2 \lambda-1)-(-\lambda+1)+13=0\) \(13 \lambda+13=0\) \(\lambda=-1\) \(\therefore P(-1,-3,2)\) Intersection with…
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