CUET · MATHS · PYQ PAPER 2023
The distance between the planes \(2x + 3y + 4z = 4\) and \(4x + 6y + 8z = 12\) is:
- A 2 units
- B 4 units
- C 8 units
- D \(\frac{2}{\sqrt{29}}\) units
Answer & Solution
Correct Answer
(D) \(\frac{2}{\sqrt{29}}\) units
Step-by-step Solution
Detailed explanation
The given planes are \(2x + 3y + 4z = 4\) and \(4x + 6y + 8z = 12\). Normalize the second equation: \(\frac{4x}{2} + \frac{6y}{2} + \frac{8z}{2} = \frac{12}{2} \Rightarrow 2x + 3y + 4z = 6\). Distance \(D = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}\)…
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