Solid Geometry : Distance of a point to a plane and distance of a point to a line.

Question

$ABCDA'B'C'D'$ 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?

If you are allowed to use vector algebra then it would be much simpler. — jonsno, Jun 21 '17 at 19:33
I know, I got square root of 40/3 when I did it using vectors. — bob, Jun 21 '17 at 19:41

score 2 · Accepted Answer · answered Jun 21 '17 at 19:44

2

$$MA=MB=\sqrt{4^2+(2\sqrt2)^2}=2\sqrt{6}.$$ 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 $x=\sqrt{\frac{40}{3}}$.

answered Jun 21 '17 at 19:44

Michael Rozenberg

1 Answers1