3. In the cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}, M$ is the midpoint of the edge [DD'], DD'= a. Find:

a) The distance from B to the plane (ACB').

b) The distance from M to the plane (ACB').

Prof. Popescu Mihai
## Subject 3.

In the cube $\mathrm{ABCDA}^{\prime} \mathrm{B}^{\prime} \mathrm{C}^{\prime} \mathrm{D}^{\prime}, \mathrm{M}$ is the midpoint of the edge [DD'], $\mathrm{DD}^{\prime}=\mathrm{a}$. The following are required:

a) The distance from B to the plane (ACB').

b) The distance from M to the plane (ACB').

a) Let $\{O\}=A C \cap B D$, according to the theorem of the three perpendiculars we have $B^{\prime} O \perp A C$

Let $B N \perp B^{\prime} O, N \in\left(B^{\prime} O\right)$, with the second reciprocal of the theorem of the three perpendiculars it is shown that $B N \perp\left(A B^{\prime} C\right)$

Therefore, $d\left[B,\left(A B^{\prime} C\right)\right]=B N$

$B^{\prime} O=\frac{a \sqrt{6}}{2}, B O=\frac{a \sqrt{2}}{2}, B N=\frac{a \sqrt{3}}{3}$

b) $M A=M C \Rightarrow \triangle M A C$ is isosceles $\Rightarrow M O \perp A C, \quad M O=\frac{a \sqrt{3}}{2}$

We calculate $M B^{\prime}=\frac{3 a}{2}$ and applying the Pythagorean theorem in $\triangle M O B^{\prime} \Rightarrow \triangle M O B^{\prime}$ is a right triangle with $m\left(M O B^{\prime}\right)=90^{\circ} \Rightarrow M O \perp O B^{\prime}$

We have $M O \perp\left(A B^{\prime} C\right)$ (justification) from which $d\left[M,\left(A B^{\prime} C\right)\right]=M O$