Question

The distance of the point \( (2, -1, 0) \) from the plane \( 2x + y + 2z + 8 = 0 \) is

• A. \( \frac{17}{3} \) units
• B. \( \frac{13}{3} \) units
• C. \( \frac{7}{3} \) units
• D. \( \frac{11}{3} \) units

Hint:

For finding the distance from a point to a plane, use the distance formula and substitute the coordinates of the point and the coefficients of the plane equation.

Answer 1

The Correct Option is D

Step 1: Use the distance formula from a point to a plane.
The distance \( d \) of a point \( (x_1, y_1, z_1) \) from a plane \( Ax + By + Cz + D = 0 \) is given by the formula: \[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
Step 2: Substitute the values.
For the given plane \( 2x + y + 2z + 8 = 0 \) and the point \( (2, -1, 0) \), we have \( A = 2 \), \( B = 1 \), \( C = 2 \), and \( D = 8 \). The point is \( (x_1, y_1, z_1) = (2, -1, 0) \). Substitute these values into the distance formula: \[ d = \frac{|2(2) + 1(-1) + 2(0) + 8|}{\sqrt{2^2 + 1^2 + 2^2}} = \frac{|4 - 1 + 8|}{\sqrt{4 + 1 + 4}} = \frac{11}{3} \]
Step 3: Conclusion.
The distance of the point from the plane is \( \frac{11}{3} \) units.