To solve the problem, we will follow these steps:
### Step 1: Define the Tetrahedron
Let’s define the vertices of the regular tetrahedron \(ABCD\) in a 3D coordinate system. We can place the points as follows:
- \(D = (0, 0, 0)\)
- \(A = (1, 1, 1)\)
- \(B = (1, -1, -1)\)
- \(C = (-1, 1, -1)\)
### Step 2: Find Midpoints \(P\) and \(Q\)
- The midpoint \(P\) of edge \(AC\) can be calculated as:
\[
P = \left(\frac{1 + (-1)}{2}, \frac{1 + 1}{2}, \frac{1 + (-1)}{2}\right) = (0, 1, 0)
\]
- The midpoint \(Q\) of edge \(AB\) can be calculated as:
\[
Q = \left(\frac{1 + 1}{2}, \frac{1 + (-1)}{2}, \frac{1 + (-1)}{2}\right) = (1, 0, 0)
\]
### Step 3: Find the Centroid \(G\) of Face \(BCD\)
The centroid \(G\) of triangle \(BCD\) can be calculated as:
\[
G = \left(\frac{B_x + C_x + D_x}{3}, \frac{B_y + C_y + D_y}{3}, \frac{B_z + C_z + D_z}{3}\right)
\]
Substituting the coordinates:
\[
G = \left(\frac{1 + (-1) + 0}{3}, \frac{-1 + 1 + 0}{3}, \frac{-1 + (-1) + 0}{3}\right) = \left(0, 0, -\frac{2}{3}\right)
\]
### Step 4: Find Vectors \(\vec{PG}\) and \(\vec{DQ}\)
- The vector \(\vec{PG}\) is given by:
\[
\vec{PG} = G - P = \left(0, 0, -\frac{2}{3}\right) - (0, 1, 0) = (0, -1, -\frac{2}{3})
\]
- The vector \(\vec{DQ}\) is given by:
\[
\vec{DQ} = Q - D = (1, 0, 0) - (0, 0, 0) = (1, 0, 0)
\]
### Step 5: Calculate the Angle \(\theta\) Between Vectors \(\vec{PG}\) and \(\vec{DQ}\)
To find the angle \(\theta\) between the vectors, we use the dot product formula:
\[
\vec{PG} \cdot \vec{DQ} = |\vec{PG}| |\vec{DQ}| \cos(\theta)
\]
Calculating the dot product:
\[
\vec{PG} \cdot \vec{DQ} = (0)(1) + (-1)(0) + \left(-\frac{2}{3}\right)(0) = 0
\]
Calculating the magnitudes:
\[
|\vec{PG}| = \sqrt{0^2 + (-1)^2 + \left(-\frac{2}{3}\right)^2} = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3}
\]
\[
|\vec{DQ}| = \sqrt{(1)^2 + (0)^2 + (0)^2} = 1
\]
Substituting into the dot product equation:
\[
0 = \frac{\sqrt{13}}{3} \cdot 1 \cdot \cos(\theta)
\]
Since the dot product is zero, it implies that:
\[
\cos(\theta) = 0 \Rightarrow \theta = \frac{\pi}{2} \text{ (90 degrees)}
\]
### Final Answer
The angle \(\theta\) between the vectors \(\vec{PG}\) and \(\vec{DQ}\) is \(90^\circ\).
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