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

Concept: The distance of a point \( (x_1, y_1, z_1) \) from a plane \[ Ax + By + Cz + D = 0 \] is given by \[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

Step 1: Write the plane in standard form. Given plane: \[ 2x + 3y - z = 7 \] \[ 2x + 3y - z - 7 = 0 \] Thus, \[ A=2,\quad B=3,\quad C=-1,\quad D=-7 \] Point: \[ (1,2,3) \]

Step 2: Substitute the values in the distance formula. \[ d = \frac{|2(1) + 3(2) - 1(3) - 7|} {\sqrt{2^2 + 3^2 + (-1)^2}} \] \[ d = \frac{|2 + 6 - 3 - 7|} {\sqrt{4 + 9 + 1}} \] \[ d = \frac{|-2|}{\sqrt{14}} \]

Step 3: Simplify the result. \[ d = \frac{2}{\sqrt{14}} \]