Question

A(1, 2, 1), B(2, 3, 2), C(3, 1, 3) and D(2, 1, 3) are the vertices of a tetrahedron. If \( \theta \) is the angle between the faces ABC and ABD then \( \cos \theta \) is:

• A. \( \frac{5}{\sqrt{14}} \)
• B. \( \frac{15}{8\sqrt{7}} \)
• C. \( \frac{3}{\sqrt{14}} \)
• D. \( \frac{5}{2\sqrt{7}} \)

Hint:

The cosine of the angle between two planes is calculated by using the normal vectors to the planes. The normal vectors are obtained by taking the cross product of the corresponding vectors on the planes.

Answer 1

The Correct Option is D

We are given the vertices \( A(1, 2, 1) \), \( B(2, 3, 2) \), \( C(3, 1, 3) \), and \( D(2, 1, 3) \) of a tetrahedron, and we are asked to find the cosine of the angle \( \theta \) between the faces \( ABC \) and \( ABD \).
Step 1: The angle \( \theta \) between two planes is given by the formula: \[ \cos \theta = \frac{n_1 \cdot n_2}{|n_1| |n_2|} \] where \( n_1 \) and \( n_2 \) are the normal vectors to the planes. For the face \( ABC \), the normal vector \( n_1 \) is the cross product of vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \), and for the face \( ABD \), the normal vector \( n_2 \) is the cross product of vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AD} \). 
Step 2: First, find the vectors: \[ \overrightarrow{AB} = B - A = (2 - 1, 3 - 2, 2 - 1) = (1, 1, 1) \] \[ \overrightarrow{AC} = C - A = (3 - 1, 1 - 2, 3 - 1) = (2, -1, 2) \] \[ \overrightarrow{AD} = D - A = (2 - 1, 1 - 2, 3 - 1) = (1, -1, 2) \] 
Step 3: Now, compute the cross products: For \( \overrightarrow{AB} \times \overrightarrow{AC} \): \[ \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 
1 & 1 & 1 
2 & -1 & 2 \end{vmatrix} = \hat{i}(1(2) - 1(-1)) - \hat{j}(1(2) - 1(2)) + \hat{k}(1(-1) - 1(2)) \] \[ = \hat{i}(2 + 1) - \hat{j}(2 - 2) + \hat{k}(-1 - 2) \] \[ = 3\hat{i} + 0\hat{j} - 3\hat{k} \] Thus, \( n_1 = (3, 0, -3) \). For \( \overrightarrow{AB} \times \overrightarrow{AD} \): \[ \overrightarrow{AB} \times \overrightarrow{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 
1 & 1 & 1 
1 & -1 & 2 \end{vmatrix} = \hat{i}(1(2) - 1(-1)) - \hat{j}(1(2) - 1(1)) + \hat{k}(1(-1) - 1(1)) \] \[ = \hat{i}(2 + 1) - \hat{j}(2 - 1) + \hat{k}(-1 - 1) \] \[ = 3\hat{i} - \hat{j} - 2\hat{k} \] Thus, \( n_2 = (3, -1, -2) \). 
Step 4: Now, calculate the dot product \( n_1 \cdot n_2 \): \[ n_1 \cdot n_2 = (3)(3) + (0)(-1) + (-3)(-2) = 9 + 0 + 6 = 15 \] Next, find the magnitudes of \( n_1 \) and \( n_2 \): \[ |n_1| = \sqrt{3^2 + 0^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \] \[ |n_2| = \sqrt{3^2 + (-1)^2 + (-2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \] Step 5: Now, calculate \( \cos \theta \): \[ \cos \theta = \frac{n_1 \cdot n_2}{|n_1| |n_2|} = \frac{15}{(3\sqrt{2})(\sqrt{14})} = \frac{15}{3\sqrt{28}} = \frac{5}{\sqrt{28}} = \frac{5}{2\sqrt{7}} \] Thus, the value of \( \cos \theta \) is \( \frac{5}{2\sqrt{7}} \).