Consider the cube ABCDEFGH with edge length 1. The space is equipped with the orthonormal frame $(\mathrm{A}; \overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AD}}, \overrightarrow{\mathrm{AE}})$. Point I is the midpoint of segment $[\mathrm{EF}]$, K is the center of square ADHE, and O is the midpoint of segment [AG].
The goal of the exercise is to calculate in two different ways the distance from point B to the plane (AIG).
Part 1. First method
- Give, without justification, the coordinates of points $\mathrm{A}$, $\mathrm{B}$, and G.
We admit that points I and K have coordinates $\mathrm{I}\left(\frac{1}{2}; 0; 1\right)$ and $\mathrm{K}\left(0; \frac{1}{2}; \frac{1}{2}\right)$.
\setcounter{enumi}{1} - Prove that the line (BK) is orthogonal to the plane (AIG).
- Verify that a Cartesian equation of the plane (AIG) is: $2x - y - z = 0$.
- Give a parametric representation of the line (BK).
- Deduce that the orthogonal projection L of point B onto the plane (AIG) has coordinates $\mathrm{L}\left(\frac{1}{3}; \frac{1}{3}; \frac{1}{3}\right)$.
- Determine the distance from point B to the plane (AIG).
Part 2. Second method
Recall that the volume $V$ of a pyramid is given by the formula $V = \frac{1}{3} \times b \times h$, where $b$ is the area of a base and $h$ is the height associated with this base.
- a. Justify that in the tetrahedron $\mathrm{ABIG}$, $[\mathrm{GF}]$ is the height relative to the base AIB. b. Deduce the volume of the tetrahedron ABIG.
- We admit that $\mathrm{AI} = \mathrm{IG} = \frac{\sqrt{5}}{2}$ and that $\mathrm{AG} = \sqrt{3}$. Prove that the area of the isosceles triangle AIG is equal to $\frac{\sqrt{6}}{4}$ square units.
- Deduce the distance from point B to the plane (AIG).