Solid Geometry – Problem 21: Find the volume of the pyramid

In a right regular hexagonal pyramid $P-ABCDEF$, $ABCDEF$ is a regular hexagon with side length $1$. Let $O$ be the center of the base and $PO\perp$ plane $ABCDEF$ with $PO=\sqrt3$.

(1) Find the volume of the pyramid.

(2) Find the lateral surface area.

(3) Find the cosine of the dihedral angle along edge $AB$ between planes $PAB$ and $ABCDEF$.

Set up coordinates with the base in $z=0$ and center at the origin: $$O(0,0,0),\ P(0,0,\sqrt3),\ A(1,0,0),\ B\left(\frac12,\frac{\sqrt3}{2},0\right).$$ (1) The base area can be computed by splitting into 6 congruent triangles $\triangle OAB$. Compute $$\overrightarrow{OA}=(1,0,0),\quad \overrightarrow{OB}=\left(\frac12,\frac{\sqrt3}{2},0\right),$$ so $$|\overrightarrow{OA}\times\overrightarrow{OB}|=\frac{\sqrt3}{2}.$$ Thus $$[OAB]=\frac12\cdot\frac{\sqrt3}{2}=\frac{\sqrt3}{4},\quad [ABCDEF]=6\cdot\frac{\sqrt3}{4}=\frac{3\sqrt3}{2}.$$ The volume is $$V=\frac13\cdot[ABCDEF]\cdot PO=\frac13\cdot\frac{3\sqrt3}{2}\cdot\sqrt3=\frac32.$$

(2) Let $M$ be the midpoint of $AB$. Then $$M\left(\frac34,\frac{\sqrt3}{4},0\right),\qquad PM^2=\left(\frac34\right)^2+\left(\frac{\sqrt3}{4}\right)^2+(\sqrt3)^2=\frac{15}{4},$$ so $PM=\frac{\sqrt{15}}{2}$. In a right regular pyramid, $PM$ is the slant height of each lateral face, so the lateral area is $$S_{\text{lat}}=6\cdot\frac12\cdot AB\cdot PM=6\cdot\frac12\cdot1\cdot\frac{\sqrt{15}}{2}=\frac{3\sqrt{15}}{2}.$$

(3) Plane $ABCDEF$ is $z=0$, so a normal vector is $\vec n_1=(0,0,1)$. For plane $PAB$, use $$\overrightarrow{AB}=\left(-\frac12,\frac{\sqrt3}{2},0\right),\qquad \overrightarrow{AP}=(-1,0,\sqrt3).$$ A normal vector is $$\vec n_2=\overrightarrow{AB}\times\overrightarrow{AP}=\left(\frac32,\frac{\sqrt3}{2},\frac{\sqrt3}{2}\right).$$ Then $$\cos\theta=\frac{|\vec n_1\cdot\vec n_2|}{|\vec n_1||\vec n_2|}=\frac{\frac{\sqrt3}{2}}{\frac{\sqrt{15}}{2}}=\frac{1}{\sqrt5}.$$

(1) $V=\dfrac32$.

(2) $S_{\text{lat}}=\dfrac{3\sqrt{15}}{2}$.

(3) For the dihedral angle $\theta$, $\cos\theta=\dfrac{1}{\sqrt5}$.