Question

Determine the distance of the point \( (1,2,3) \) from the plane \( 2x + 3y - z = 7 \).

• A. \( \frac{2}{\sqrt{14}} \)
• B. \( \frac{4}{\sqrt{14}} \)
• C. \( \frac{1}{\sqrt{14}} \)
• D. \( \frac{3}{\sqrt{14}} \)

Hint:

Always convert the plane equation to the form \(Ax + By + Cz + D = 0\) before applying the point–plane distance formula.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem involves finding the shortest perpendicular distance from a given point in 3D space to a specified plane.
The topic is 3D Geometry.
Step 2: Key Formula or Approach:
The distance \( d \) from point \( (x_1, y_1, z_1) \) to plane \( Ax + By + Cz + D = 0 \) is:
\[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
Step 3: Detailed Explanation:
First, rewrite the plane equation as \( 2x + 3y - z - 7 = 0 \).
Here, \( A=2, B=3, C=-1, D=-7 \).
The point is \( (1, 2, 3) \).
Substituting these into the distance formula:
\[ d = \frac{|2(1) + 3(2) + (-1)(3) - 7|}{\sqrt{2^2 + 3^2 + (-1)^2}} \]
\[ d = \frac{|2 + 6 - 3 - 7|}{\sqrt{4 + 9 + 1}} \]
\[ d = \frac{|-2|}{\sqrt{14}} = \frac{2}{\sqrt{14}} \]
Step 4: Final Answer:
According to the calculation, the distance is \( \frac{2}{\sqrt{14}} \).