Question

The distance between the two planes $2x + 3y + 4z = 4$ and $4x + 6y + 8z = 12$ is:

• A. $2$ units
• B. $8$ units
• C. $\frac{2}{\sqrt{29}}$ units
• D. $4$ units

Answer 1

The Correct Option is C

The two planes are parallel since their normal vectors are proportional. The distance between the planes is given by: \[ \text{Distance} = \frac{|d_2 - d_1|}{\sqrt{A^2 + B^2 + C^2}}. \] Here, $A = 2, B = 3, C = 4, d_1 = 4, d_2 = 12/2 = 6$: \[ \text{Distance} = \frac{|6 - 4|}{\sqrt{2^2 + 3^2 + 4^2}} = \frac{2}{\sqrt{29}}. \]