MHT CET · Maths · Three Dimensional Geometry
If the lines \(\frac{x-1}{5}=\frac{y+1}{3}=\frac{3-z}{\lambda}\) and \(\frac{x+1}{4}=\frac{1-3 y}{15}=z+1\) are perpendicular to each other, then \(\lambda=\)
- A 2
- B 3
- C 5
- D 4
Answer & Solution
Correct Answer
(C) 5
Step-by-step Solution
Detailed explanation
Given lines are \(\frac{x-1}{5}=\frac{y+1}{3}=\frac{z-3}{-\lambda}\) and \(\frac{x+1}{4}=\frac{y-\frac{1}{3}}{-5}=\frac{z+1}{1}\)
d.r. of the given lines are \(5,3,-\lambda\) and \(4,-5,1\)
These lines are \(\perp\) er
\(\therefore 5(4)+3(-5)+(-\lambda)(1)=0 \Rightarrow 20-15-\lambda=\) \(0 \Rightarrow \lambda=5\)
d.r. of the given lines are \(5,3,-\lambda\) and \(4,-5,1\)
These lines are \(\perp\) er
\(\therefore 5(4)+3(-5)+(-\lambda)(1)=0 \Rightarrow 20-15-\lambda=\) \(0 \Rightarrow \lambda=5\)
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