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CUET · MATHS · PYQ PAPER 2023

The distance between the planes \(2x + 3y + 4z = 4\) and \(4x + 6y + 8z = 12\) is:

  1. A 2 units
  2. B 4 units
  3. C 8 units
  4. D \(\frac{2}{\sqrt{29}}\) units
Verified Solution

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 \(D = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}\)…