TS EAMCET · Maths · Three Dimensional Geometry
If the plane \(-4 x-2 y+2 z+\alpha=0\) is at a distance of two units from the plane \(2 x+y-z+1=0\), then the product of all the possible values of \(\alpha\) is
- A -23
- B 42
- C -92
- D 72
Answer & Solution
Correct Answer
(C) -92
Step-by-step Solution
Detailed explanation
Rewrite plane equations for common normal: \(-4x-2y+2z+\alpha=0\), \(-4x-2y+2z-2=0\) Distance formula: \(d = \frac{|D_1-D_2|}{\sqrt{A^2+B^2+C^2}}\) Substitute values: \(2 = \frac{|\alpha - (-2)|}{\sqrt{(-4)^2 + (-2)^2 + (2)^2}}\) Simplify:…
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