JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the distance between planes , \(4x - 2y-4z + 1 = 0\) and \(4x -2y-4z+ d = 0\) is \(7,\) then \(d\) is
- A \(41\) or \(-42\)
- B \(42\) or \(-43\)
- C \(-41\) or \(43\)
- D \(-42\) or \(44\)
Answer & Solution
Correct Answer
(C) \(-41\) or \(43\)
Step-by-step Solution
Detailed explanation
Given planes are \(4 x-2 y-4 z+1=0\) and \(4 x-2 y-4 z+d=0\) They are parallel. Distance between them is \(\pm 7\) \(=\frac{d-1}{\sqrt{16+4+16}}\) \(\Rightarrow \frac{d-1}{6}=\pm 7\) \(\Rightarrow d=42+1\) or \(-42+1\) i.e. \(d=-41\) or 43.
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