In tetrahedron , suppose plane and . Prove that plane is perpendicular to plane .

Line is perpendicular to plane , so plane .

Solid Geometry Advanced Solution – Problem 3: Prove that plane is perpendicular to plane

Question

In tetrahedron $S-ABC$ , suppose $SA\perp$ plane $ABC$ and $AB\perp AC$ .

Prove that plane $SAB$ is perpendicular to plane $SAC$ .

Step-by-step solution

(1) From $SA\perp$ plane $ABC$ , we know $SA\perp AB,\qquad SA\perp AC.$ (2) The two lines $SA$ and $AC$ intersect and both lie in plane $SAC$ .

(3) Since $AB\perp SA$ and $AB\perp AC$ , line $AB$ is perpendicular to two intersecting lines in plane $SAC$ . Hence $AB\perp \text{plane }SAC.$ (4) Because $AB\subset$ plane $SAB$ , a plane that contains a line perpendicular to another plane must be perpendicular to that plane. Therefore $\text{plane }SAB\perp \text{plane }SAC.$

Final answer

Line $AB$ is perpendicular to plane $SAC$ , so plane $SAB\perp SAC$ .

Marking scheme

Step 1 — Setup

Checkpoint: correctly use $SA\perp$ plane $ABC$ to deduce $SA\perp AB$ and $SA\perp AC$ (2 pts)

Step 2 — Key Calculation

Checkpoint: show $AB\perp$ plane $SAC$ by “perpendicular to two intersecting lines” (3 pts)

Step 3 — Final Answer

Checkpoint: conclude plane $SAB\perp SAC$ from the existence of a line in plane $SAB$ perpendicular to plane $SAC$ (2 pts)

Zero credit if: mixes up line–plane and plane–plane perpendicular criteria.

Deductions: -1 pt for not explicitly stating which two intersecting lines lie in plane $SAC$ .

Solid Geometry Advanced – Problem 3: Prove that plane is perpendicular to plane