JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the line \(\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}\) intersect the lines \(\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}\) and \(\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}\) at the points \(A\) and \(B\) respectively. Then the distance of the mid-point of the line segment \(A B\) from the plane \(2 x-2 y+z=14\) is
- A \(4\)
- B \(\frac{10}{3}\)
- C \(3\)
- D \(\frac{11}{3}\)
Answer & Solution
Correct Answer
(A) \(4\)
Step-by-step Solution
Detailed explanation
\(\frac{x}{1}=\frac{y-6}{-2}=\frac{z+8}{5}=\lambda\) \(\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}=\mu\) \(\frac{x+3}{4}=\frac{y-3}{-3}=\frac{z-6}{1}=\gamma\) \(\text { Intersection of (1) \& (2) "A" }\) \((\lambda,-2 \lambda+6,5 \lambda-8) \&(4 \mu+5,3 \mu+7, \mu-2)\)…
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