Solid Geometry – Problem 8: Find the distance from point to plane

In the quadrilateral pyramid $P-ABCD$, base $ABCD$ is a square. Given $PA=AB=2$, and $PA\perp$ plane $ABCD$. Points $E,F$ are the midpoints of $PA,AB$, respectively, and lines $AC$, $DF$ intersect at point $O$.

(1) Find the distance from point $B$ to plane $DEF$.

(2) Find the sine of the angle between line $PC$ and plane $DEF$.

(1) Since $PA\perp$ plane $ABCD$, and $AB,AD\subset$ plane $ABCD$, we get $PA\perp AB$, $PA\perp AD$. Also base $ABCD$ is a square, so $AB\perp AD$. Hence we can build a 3D rectangular coordinate system with $A$ as origin and lines containing $AB,AD,AP$ as the $x,y,z$-axes. Take $$A(0,0,0),\ B(2,0,0),\ D(0,2,0),\ E(0,0,1),\ F(1,0,0).$$ Then $$\overrightarrow{DE}=(0,-2,1),\quad \overrightarrow{FE}=(-1,0,1),\quad \overrightarrow{BF}=(-1,0,0).$$ Let $\vec n=(x,y,z)$ be a normal vector of plane $DEF$. From $$\vec n\cdot\overrightarrow{DE}=0,\qquad \vec n\cdot\overrightarrow{FE}=0$$ one can take $\vec n=(2,1,2)$. Therefore the distance from $B$ to plane $DEF$ is $$d=\frac{|\overrightarrow{BF}\cdot\vec n|}{|\vec n|}=\frac{2}{3}.$$

(2) From $P(0,0,2),\ C(2,2,0)$, $$\overrightarrow{PC}=(2,2,-2).$$ Let $\theta$ be the angle between line $PC$ and plane $DEF$. Since $\vec n$ is a normal vector of the plane, $$\sin\theta=\frac{|\overrightarrow{PC}\cdot\vec n|}{|\overrightarrow{PC}|\,|\vec n|} =\frac{\sqrt3}{9}.$$

(1) After constructing coordinates and a normal vector of plane $DEF$, the point-to-plane distance is $$d(B,\,DEF)=\frac23.$$

(2) Using $\overrightarrow{PC}=(2,2,-2)$ and the same plane normal, the line-plane angle satisfies $$\sin\theta=\frac{\sqrt3}{9}.$$ So the required sine value is $\dfrac{\sqrt3}{9}$.