$ABCDA'B'C'D'$ is a cube with side $4$, $M$ is the center of the face $A'B'C'D'$ and $H$ is a point on $AC$ such that $AH : HC = 1 : 3.$
a. Find the distance from $B$ to the plane $HMC$ using solid geometry.
I got that distance to be $2\sqrt{2}$
b. Find the distance from $B$ to the line containing $AM$ using solid geometry.
Let the intersection of $AC$ and $BD$ be $E$. Then I found that $AE=EC=BE=ED=2\sqrt{2}$, $AM=2\sqrt{6}$, $BM=2\sqrt{6}$.
Not sure where to go from here. Any suggestions?
$$MA=MB=\sqrt{4^2+(2\sqrt2)^2}=2\sqrt{6}.$$ Let $K$ be a midpoint of $AB$.
Thus, $$MK=\sqrt{24-2^2}=2\sqrt{5}$$
Id est, $x\cdot2\sqrt6=4\cdot2\sqrt5,$, which gives $x=\sqrt{\frac{40}{3}}$.