Solid Geometry – Problem 17: find the distance from vertex to plane

In cube $ABCD-A_1B_1C_1D_1$ with side length $2$, find the distance from vertex $A$ to plane $BCD_1$.

Set up coordinates: $$A(0,0,0),\ B(2,0,0),\ C(2,2,0),\ D(0,2,0),\ D_1(0,2,2).$$ In plane $BCD_1$, take $$\overrightarrow{BC}=(0,2,0),\qquad \overrightarrow{BD_1}=(-2,2,2).$$ A normal vector is $$\vec n=\overrightarrow{BC}\times\overrightarrow{BD_1}=(4,0,4)\sim(1,0,1).$$ Using point $B(2,0,0)$, the plane equation is $$(1,0,1)\cdot(x-2,\,y-0,\,z-0)=0\ \Rightarrow\ x+z=2.$$ Therefore the distance from $A(0,0,0)$ to this plane is $$d=\frac{|0+0-2|}{\sqrt{1^2+0^2+1^2}}=\frac{2}{\sqrt2}=\sqrt2.$$

The distance from $A$ to plane $BCD_1$ is $\sqrt2$.