Solid Geometry – Problem 4: Prove that plane

In tetrahedron $P-ABC$, $PA\perp$ plane $ABC$, $AC=2$, $AB=4$, $\angle ABC=\dfrac\pi6$.

(1) Prove that plane $PAC\perp PBC$.

(2) If the cosine of the dihedral angle $A-PB-C$ is $\dfrac{\sqrt6}{4}$, find $PA$.

(1) In $\triangle ABC$, by the sine law, $$\frac{AB}{\sin\angle ACB}=\frac{AC}{\sin\angle ABC}$$ that is, $$\frac{4}{\sin\angle ACB}=\frac{2}{\sin(\pi/6)}=4,$$ so $\sin\angle ACB=1$, hence $\angle ACB=\dfrac\pi2$, therefore $CA\perp CB$.

Also, $PA\perp$ plane $ABC$, so $PA\perp CB$. Since $CA,PA\subset$ plane $PAC$, and $CA\cap PA=A$, we have $CB\perp$ plane $PAC$. Since $CB\subset$ plane $PBC$, $$\text{plane }PAC\perp\text{plane }PBC.$$

(2) Build a 3D rectangular coordinate system with lines containing $CB,CA$ as the $x,y$-axes and $C$ as origin. Take $$C(0,0,0),\ A(0,2,0),\ B(2\sqrt3,0,0),\ P(0,2,m),$$ where $m=PA$. Let normal vectors of planes $APB$, $PBC$ be $\vec n_1=(1,\sqrt3,0)$, $\vec n_2=(0,m,-2)$, respectively. From the dihedral-cosine condition $$\frac{|\vec n_1\cdot\vec n_2|}{|\vec n_1||\vec n_2|}=\frac{\sqrt6}{4}$$ we get $$\frac{\sqrt3 m}{2\sqrt{m^2+4}}=\frac{\sqrt6}{4} \Rightarrow m^2=4.$$ Taking the positive value, $PA=2$.

(1) By proving $CB\perp CA$ and $CB\perp PA$, we conclude $CB\perp$ plane $PAC$, so $$\text{plane }PAC\perp\text{plane }PBC.$$

(2) Using face normal vectors around edge $PB$ and the given dihedral cosine, we solve $m^2=4$. Since $m=PA>0$, the final result is $$PA=2.$$