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:

First check whether the two planes are parallel, their coefficients of x, y and z must be in the same ratio. Once confirmed, you can either scale one equation to match the coefficients of the other and compare constant terms, or pick any convenient point on one plane and apply the standard point-to-plane distance formula on the other.

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

Answer 2

Concept:

• The distance from a point $(x_0,y_0,z_0)$ to a plane $ax+by+cz+d=0$ is given by $\dfrac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}$.
• If two planes are parallel, the distance between them equals the distance from any point lying on one plane to the other plane, so picking a convenient point avoids having to compare the constant terms of both equations directly.

Step 1: Find a convenient point lying on the first plane $2x+3y+4z=4$.
Taking $y=0$ and $z=0$ gives $2x=4$, so $x=2$. The point $P(2,0,0)$ lies on the first plane.

Step 2: Confirm the planes are parallel by comparing normal vector directions.
Plane 2: $4x+6y+8z=12$, whose coefficients $(4,6,8)$ are proportional to $(2,3,4)$ of plane 1, so the planes are parallel and the distance from $P$ to plane 2 equals the required distance.

Step 3: Apply the point-to-plane distance formula using $P(2,0,0)$ and plane $4x+6y+8z-12=0$.
Distance $= \dfrac{|4(2)+6(0)+8(0)-12|}{\sqrt{4^2+6^2+8^2}} = \dfrac{|8-12|}{\sqrt{16+36+64}} = \dfrac{4}{\sqrt{116}}$
$\sqrt{116} = 2\sqrt{29}$, so Distance $= \dfrac{4}{2\sqrt{29}} = \dfrac{2}{\sqrt{29}}$

Final Answer: $\dfrac{2}{\sqrt{29}}$