WBJEE · Maths · Three Dimensional Geometry
The value of \(\lambda\) for which the straight line \(\frac{x-\lambda}{3}=\frac{y-1}{2+\lambda}=\frac{z-3}{-1}\) may lie on the plane \(x-2 y=0,\) is
- A 2
- B 0
- C \(-\frac{1}{2}\)
- D there is no such \(\lambda\)
Answer & Solution
Correct Answer
(D) there is no such \(\lambda\)
Step-by-step Solution
Detailed explanation
If a line \(\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}\) lies on a plane \(a_{2} x+b_{2} y+c_{2} z=d,\) then (i) \(a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0\) (ii) \(a_{2} x_{1}+b_{2} y_{1}+c_{2} z_{1}=d\) According to the question.…
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