ExamBro
ExamBro
JEE Mains · Maths · STD 12 - 11. three dimension geometry

यदि रेखा \(\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-1}{-1}\), समतल \(2 x +3 y - z +13=0\) को बिन्दु \(P\) पर काटती है तथा समतल \(3 x + y +4 z =16\) को बिन्दु \(Q\) पर काटती है, तो \(PQ\) बराबर हैं

  1. A \(2\sqrt {14} \)
  2. B \(14\)
  3. C \(2\sqrt {7} \)
  4. D \(\sqrt {14} \)
Verified Solution

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…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app