Question

Find the distance between the planes \(2x + 3y + 4z = 4\) and \(4x + 6y + 8z = 12\).

• A. \( \dfrac{1}{\sqrt{29}} \)
• B. \( \dfrac{2}{\sqrt{29}} \)
• C. \( \dfrac{3}{\sqrt{29}} \)
• D. \( \dfrac{4}{\sqrt{29}} \)

Hint:

To find the distance between two planes, first confirm that the planes are parallel (their \(x, y, z\) coefficients are proportional). Then apply \[ d = \frac{|d_2-d_1|}{\sqrt{a^2+b^2+c^2}} \] directly.

Answer 1

The Correct Option is B

Concept: The distance between two parallel planes \[ ax + by + cz + d_1 = 0 \quad \text{and} \quad ax + by + cz + d_2 = 0 \] is given by \[ d = \frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}} \] This formula is applicable when the planes have the same normal vector, meaning the coefficients of \(x, y, z\) are proportional.

Step 1: Check whether the planes are parallel. Given planes: \[ 2x + 3y + 4z = 4 \] \[ 4x + 6y + 8z = 12 \] Divide the second equation by \(2\): \[ 2x + 3y + 4z = 6 \] Since both equations now have the same coefficients, the planes are parallel.

Step 2: Write both equations in standard form. \[ 2x + 3y + 4z - 4 = 0 \] \[ 2x + 3y + 4z - 6 = 0 \] Thus, \[ d_1 = -4, \quad d_2 = -6 \]

Step 3: Apply the distance formula. \[ d = \frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}} \] \[ d = \frac{|(-6) - (-4)|}{\sqrt{2^2 + 3^2 + 4^2}} \] \[ d = \frac{2}{\sqrt{4 + 9 + 16}} \] \[ d = \frac{2}{\sqrt{29}} \]