Question

The vertices of a tetrahedron correspond to four alternating corners of a cube. By using analytical geometry, demonstrate that the angle made by connecting two of the vertices to a point at the center of the cube is \(109.5^{\circ}\), the characteristic angle for tetrahedral molecules.

Short answer

The vertices of the tetrahedron correspond to alternating corners of a cube with coordinates (0,0,0), (1,1,0), (1,0,1), and (0,1,1). The center of the cube has coordinates (0.5, 0.5, 0.5). We can find vectors A and B connecting vertices to the center, which are (0.5, 0.5, -0.5) and (0.5, -0.5, 0.5). The dot product of these vectors is -0.25, and their magnitudes are both \(\frac{\sqrt{3}}{2}\). Using the dot product and magnitudes, we find the cosine of the angle between the vectors to be -\(\frac{1}{3}\), and taking the arccosine of this value gives us the angle of approximately \(109.5^{\circ}\). This result confirms that the angle formed by connecting two vertices of the tetrahedron to the center of the cube is consistent with the characteristic angle for tetrahedral molecules.

Step-by-step solution

Find the coordinates of the alternating corners of the cube

Assume we have a cube with side length 1. Let the origin of our coordinate system be at one of the corners of the cube. The coordinates of the alternating corners will be (0,0,0), (1,1,0), (1,0,1), and (0,1,1). These points will be the vertices of the tetrahedron.

Find the coordinates of the center of the cube

The center of the cube is the midpoint of its opposite vertices. We can select two opposite corners and find the midpoint as follows: Center = \(((x_1 + x_2)/2, (y_1 + y_2)/2, (z_1 + z_2)/2)\) For this cube, the opposite corners are (0,0,0) and (1,1,1). So, the center of the cube has coordinates (0.5, 0.5, 0.5).

Find the vectors connecting the vertices with the center of the cube

We can find these vectors by subtracting the coordinates of the center from the coordinates of the vertices. Vector A = Vertex (1,1,0) - Center (0.5, 0.5, 0.5) = (0.5, 0.5, -0.5) Vector B = Vertex (1,0,1) - Center (0.5, 0.5, 0.5) = (0.5, -0.5, 0.5)

Calculate the dot product of the vectors

We can find the dot product of vectors A and B as follows: Dot Product (A, B) = [Ax * Bx] + [Ay * By] + [Az * Bz] Dot Product (A, B) = [(0.5 * 0.5) + (0.5 * -0.5) + (-0.5 * 0.5)] = -0.25

Calculate the magnitudes of the vectors

We can find the magnitudes of vectors A and B using the formula: Magnitude (vector) = \(\sqrt{x^2 + y^2 + z^2}\) Magnitude (A) = \(\sqrt{(0.5)^2 + (0.5)^2 + (-0.5)^2} = \frac{\sqrt{3}}{2}\) Magnitude (B) = \(\sqrt{(0.5)^2 + (-0.5)^2 + (0.5)^2} = \frac{\sqrt{3}}{2}\)

Calculate the cosine of the angle between the vectors

Now, we can find the cosine of the angle (θ) between vectors A and B using the dot product and the magnitudes: cos(θ) = (Dot Product) / (Magnitude (A) * Magnitude (B)) cos(θ) = (-0.25) / (\(\frac{\sqrt{3}}{2}\) * \(\frac{\sqrt{3}}{2}\)) = -\frac{1}{3}

Find the angle

Now we can find the angle between the two vectors by taking the arccosine of the cosine value we found. θ = arccos(-1/3) ≈ 109.5° The angle formed by connecting two of the vertices of the tetrahedron to a point at the center of the cube is approximately \(109.5^{\circ}\). This result is consistent with the characteristic angle for tetrahedral molecules.