Solid Geometry – Problem 7: Prove that plane

In the quadrilateral pyramid $P-ABCD$, $PD\perp$ plane $ABCD$, $AB\parallel CD$, $\angle ADC=90^\circ$, and $$AD=CD=PD=2AB=2.$$

(1) Prove that $AB\perp$ plane $PAD$.

(2) Find the cosine of the angle between planes $PAD$ and $PBC$.

(1) Since $PD\perp$ plane $ABCD$ and $AB\subset$ plane $ABCD$, we get $PD\perp AB$. Also $AB\parallel CD$, $\angle ADC=90^\circ$, so $AB\perp AD$. Because $AD\cap PD=D$, and $AD,PD\subset$ plane $PAD$, $$AB\perp\text{plane }PAD.$$

(2) Build a 3D rectangular coordinate system and take $$D(0,0,0),\ P(0,0,2),\ A(2,0,0),\ B(2,1,0),\ C(0,2,0).$$ From part (1), a normal vector of plane $PAD$ can be $$\vec n_1=\overrightarrow{AB}=(0,1,0).$$ Let $\vec n_2=(x,y,z)$ be a normal vector of plane $PBC$. From $$\vec n_2\cdot\overrightarrow{PB}=0,\qquad \vec n_2\cdot\overrightarrow{PC}=0$$ one can take $\vec n_2=(1,2,2)$. Hence for the angle $\theta$ between the two planes, $$\cos\theta=\frac{|\vec n_1\cdot\vec n_2|}{|\vec n_1||\vec n_2|}=\frac23.$$

(1) Because $AB$ is perpendicular to both $AD$ and $PD$, two intersecting lines in plane $PAD$, we conclude $$AB\perp\text{plane }PAD.$$

(2) Using normal vectors $\vec n_1=(0,1,0)$, $\vec n_2=(1,2,2)$, the cosine of the plane angle is $$\frac23.$$ So the required value is $\dfrac23$.