MathIsimple

Solid Geometry – Problem 34: Find the distance from the origin to the plane

Question

Problem 34 diagram

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.

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