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

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 \( \dfrac{2}{\sqrt{29}} \).