Question

Find the distance from the point \[ (1,1,1) \] to the plane \[ x+y+z=3. \]

• A. \(1\)
• B. \(0\)
• C. \(\sqrt3\)
• D. \(3\)

Hint:

If a point satisfies the equation of a plane, then its perpendicular distance from the plane is zero.

Answer 1

The Correct Option is B

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

Step 1: Write the plane in standard form Given plane: \[ x+y+z=3 \] or \[ x+y+z-3=0 \] Comparing with \[ Ax+By+Cz+D=0 \] we get \[ A=1,\quad B=1,\quad C=1,\quad D=-3 \]

Step 2: Substitute the coordinates of the point Point: \[ (1,1,1) \] Substituting into the formula: \[ d= \frac{|1(1)+1(1)+1(1)-3|} {\sqrt{1^2+1^2+1^2}} \] \[ = \frac{|3-3|} {\sqrt3} \] \[ = \frac0{\sqrt3} \] \[ =0 \]

Step 3: Interpret the result Since the distance is zero, the point lies exactly on the plane. Checking: \[ 1+1+1=3 \] which satisfies the plane equation. Therefore, the point lies on the plane itself. Final Answer: \[ \boxed{0} \]