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

We can first define the coordinates of the tetrahedron vertices A(0,0,0), B(1,1,0), C(1,0,1), and D(0,1,1), and the cube's center point, O(1/2,1/2,1/2). Next, we compute the vectors OA and OB. By calculating the dot product of OA and OB and then finding the magnitudes of these vectors, we can use the dot product formula \(\cos\theta=\frac{OA \cdot OB}{|OA||OB|}\) to find the angle between the vectors. Plugging in the calculated values, we get \(\theta \approx 109.5^{\circ}\), which verifies that the angle made by connecting two vertices of the tetrahedron to a point at the center of the cube is the characteristic angle for tetrahedral molecules.

Step-by-step solution

Define the coordinates of the cube and tetrahedron vertices

We can define a cube with vertices at (0,0,0), (1,0,0), (0,1,0), (1,1,0), (0,0,1), (1,0,1), (0,1,1), (1,1,1). Let's choose four alternating corners as the vertices of the tetrahedron, for example: A(0,0,0), B(1,1,0), C(1,0,1) and D(0,1,1).

Calculate the coordinates of the center

To find the center of the cube, we can take the mean value of the coordinates of all the vertices. Calculate the center coordinates: \(O\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)\)

Compute the vectors from the center to the tetrahedron vertices

Now, we need to find the vectors from the center of the cube to two of the vertices of the tetrahedron. Let's take vertices A and B as an example: OA = A - O = (-1/2, -1/2, -1/2) OB = B - O = (1/2, 1/2, -1/2)

Calculate the dot product of OA and OB

Calculate the dot product of the vectors OA and OB: \(OA \cdot OB = -\frac{1}{2}\times\frac{1}{2} + (-\frac{1}{2})\times\frac{1}{2} + (-\frac{1}{2})(-\frac{1}{2}) = -\frac{1}{4}-\frac{1}{4}+\frac{1}{4} =-\frac{1}{4}\)

Compute the magnitudes of the vectors

Calculate the magnitudes of OA and OB: \(|OA| =\sqrt{(-\frac{1}{2})^2+(-\frac{1}{2})^2+(-\frac{1}{2})^2}=\sqrt{\frac{1}{4}+\frac{1}{4}+\frac{1}{4}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}\) \(|OB| =\sqrt{(\frac{1}{2})^2+(\frac{1}{2})^2+(-\frac{1}{2})^2}=\sqrt{\frac{1}{4}+\frac{1}{4}+\frac{1}{4}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}\)

Use dot product formula to find the angle

The dot product formula is: \(\cos\theta=\frac{OA \cdot OB}{|OA||OB|}\) Now, plug in the values from Steps 4 and 5: \(\cos\theta=\frac{-\frac{1}{4}}{\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}}\) \(\cos\theta=\frac{-\frac{1}{4}}{\frac{3}{4}}\) \(\cos\theta=-\frac{1}{3}\) Now, find the angle: \(\theta =\cos^{-1}(-\frac{1}{3}) = 109.471251^\circ\) Since the angle \(\theta\approx 109.5^\circ\), this verifies that the angle made by connecting two vertices of the tetrahedron to a point at the center of the cube is the characteristic angle for tetrahedral molecules.