MathIsimple

Solid Geometry Advanced – Problem 10: Find the distance from the origin to the plane

Question

In space, let B(2,0,0)B(2,0,0), C(0,2,0)C(0,2,0), D(0,0,2)D(0,0,2). Find the distance from the origin O(0,0,0)O(0,0,0) to the plane BCDBCD.

Step-by-step solution

(1) The plane through the intercept points (2,0,0)(2,0,0), (0,2,0)(0,2,0), (0,0,2)(0,0,2) has equation x2+y2+z2=1x+y+z=2.\frac{x}{2}+\frac{y}{2}+\frac{z}{2}=1\quad\Longleftrightarrow\quad x+y+z=2. (2) The distance from O(0,0,0)O(0,0,0) to plane x+y+z2=0x+y+z-2=0 is d=0+0+0212+12+12=23.d=\frac{|0+0+0-2|}{\sqrt{1^2+1^2+1^2}}=\frac{2}{\sqrt3}.

Final answer

The distance from OO to plane BCDBCD is 23\dfrac{2}{\sqrt3}.

Marking scheme

Step 1 — Setup

Checkpoint: write the plane equation x+y+z=2x+y+z=2 from intercept form (2 pts)

Step 2 — Key Calculation

Checkpoint: use the point-to-plane distance formula correctly (3 pts)

Step 3 — Final Answer

Checkpoint: simplify to 23\frac{2}{\sqrt3} (2 pts)

Zero credit if: writes an incorrect plane equation.

Deductions: -1 pt for not simplifying the radical in the denominator.

Ask AI ✨