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

Step 1: Understanding the Question:
The problem asks for the shortest distance between two planes in three-dimensional space.
Two planes are parallel if their normal vectors are proportional, i.e., the coefficients of \(x, y,\) and \(z\) can be made identical.
Step 2: Key Formula or Approach:
For two parallel planes given by \(ax + by + cz + d_1 = 0\) and \(ax + by + cz + d_2 = 0\), the distance \(d\) is:
\[ d = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}} \]
Step 3: Detailed Explanation:
First, we express both equations in a comparable form.
Plane 1: \(2x + 3y + 4z - 4 = 0\)
Plane 2: \(4x + 6y + 8z - 12 = 0\)
Dividing the second equation by \(2\), we get:
\[ 2x + 3y + 4z - 6 = 0 \]
Now, both equations have the same coefficients \(a=2, b=3, c=4\).
Comparing with the standard form:
\(d_1 = -4\) and \(d_2 = -6\).
Substituting these values into the distance formula:
\[ d = \frac{|-4 - (-6)|}{\sqrt{2^2 + 3^2 + 4^2}} \]
\[ d = \frac{|-4 + 6|}{\sqrt{4 + 9 + 16}} \]
\[ d = \frac{2}{\sqrt{29}} \]
Step 4: Final Answer:
The distance between the two planes is \( \dfrac{2}{\sqrt{29}} \).