Solid Geometry – Problem 9: Prove that plane

In tetrahedron $D-ABC$, $AB\perp AC$. It is given that $AD\perp AB$ and $AD\perp AC$.

(1) Prove that $AD\perp$ plane $ABC$.

(2) Find the angle between line $AD$ and plane $ABC$.

Build a 3D rectangular coordinate system with $A$ as the origin, $\overrightarrow{AB}$ as the $x$-axis, $\overrightarrow{AC}$ as the $y$-axis, and $\overrightarrow{AD}$ as the $z$-axis. Take $$A(0,0,0),\ B(b,0,0),\ C(0,c,0),\ D(0,0,d),$$ where $b,c,d>0$. Then $$\overrightarrow{AB}=(b,0,0),\quad \overrightarrow{AC}=(0,c,0),\quad \overrightarrow{AD}=(0,0,d).$$ Compute: $$\overrightarrow{AD}\cdot\overrightarrow{AB}=0,\qquad \overrightarrow{AD}\cdot\overrightarrow{AC}=0.$$ Since $AB\cap AC=A$ and $AB\not\parallel AC$ (indeed $AB\perp AC$), line $AD$ is perpendicular to two intersecting lines in plane $ABC$. Therefore $$AD\perp\text{plane }ABC.$$ Hence the angle between line $AD$ and plane $ABC$ is $$90^\circ.$$

$AD\perp$ plane $ABC$, so the angle between line $AD$ and plane $ABC$ is $90^\circ$.