MHT CET · Maths · Three Dimensional Geometry
If the lines \(\frac{1-x}{2}=\frac{y-8}{\lambda}=\frac{z-5}{2}\) and \(\frac{x-11}{5}=\frac{y-3}{3}=\frac{z-1}{1}\) are perpendicular, then \(\lambda=\)
- A 4
- B \(-4\)
- C \(\frac{8}{3}\)
- D \(\frac{-8}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{8}{3}\)
Step-by-step Solution
Detailed explanation
We have \(\frac{1-x}{2}=\frac{y-8}{\lambda}=\frac{z-5}{2} \Rightarrow \frac{x-1}{-2}=\frac{y-8}{\lambda}=\frac{z-5}{2}\)
Direction ratio of given lines are \(-2, \lambda, 2\) and \(5,3,1\).
Given lines are \(\perp\) er. \(\therefore-2(5)+\lambda(3)+2(1)=0\)
\(-10+3 \lambda+2=0 \Rightarrow \lambda=\frac{8}{3}\)
Direction ratio of given lines are \(-2, \lambda, 2\) and \(5,3,1\).
Given lines are \(\perp\) er. \(\therefore-2(5)+\lambda(3)+2(1)=0\)
\(-10+3 \lambda+2=0 \Rightarrow \lambda=\frac{8}{3}\)
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