AP EAMCET · Maths · Three Dimensional Geometry
If the distance between the planes \(2 x+y+z+1=0\) and \(2 x+y+z+\alpha=0\) is 3 units, then product of all possible values of \(\alpha\) is
- A \(-43\)
- B \(43\)
- C \(53\)
- D \(-53\)
Answer & Solution
Correct Answer
(D) \(-53\)
Step-by-step Solution
Detailed explanation
Given two planes are \(2 x+y+z+1=0\) and \(2 x+y+z+\alpha=0\) Since, these are parallel planes So, distance between them…
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