The sine of angle formed by the lateral face ADC and plane of the base ABC of the terahedron ABCD, where `A=(3,-2,1), B=(3, 1,5), C=(4,0,3) and D=(1,0,0)`, is :

Options

A. `(2)/(sqrt(29))`
B. `(5)/(sqrt(29))`
C. `(3sqrt(3))/(sqrt(29))`
D. `(-2)/(sqrt(29))`

Answer

To find the sine of the angle formed by the lateral face ADC and the base ABC of the tetrahedron ABCD, we will follow these steps:

### Step 1: Identify the Points
Given points are:
- \( A = (3, -2, 1) \)
- \( B = (3, 1, 5) \)
- \( C = (4, 0, 3) \)
- \( D = (1, 0, 0) \)

### Step 2: Calculate Vectors
We need to calculate the vectors \( \overrightarrow{AD} \) and \( \overrightarrow{AC} \) for the lateral face ADC, and \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) for the base ABC.

1. **Calculate \( \overrightarrow{AD} \)**:
\[ \overrightarrow{AD} = D - A = (1 - 3, 0 + 2, 0 - 1) = (-2, 2, -1) \]

2. **Calculate \( \overrightarrow{AC} \)**:
\[ \overrightarrow{AC} = C - A = (4 - 3, 0 + 2, 3 - 1) = (1, 2, 2) \]

3. **Calculate \( \overrightarrow{AB} \)**:
\[ \overrightarrow{AB} = B - A = (3 - 3, 1 + 2, 5 - 1) = (0, 3, 4) \]

### Step 3: Find Normal Vectors
Next, we will find the normal vectors \( \mathbf{N_1} \) and \( \mathbf{N_2} \).

1. **Normal vector \( \mathbf{N_1} \) for face ADC**:
\[ \mathbf{N_1} = \overrightarrow{AD} \times \overrightarrow{AC} \]
\[ \mathbf{N_1} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 2 & -1 \\ 1 & 2 & 2 \end{vmatrix} \]
\[ = \mathbf{i}(2 \cdot 2 - (-1) \cdot 2) - \mathbf{j}(-2 \cdot 2 - (-1) \cdot 1) + \mathbf{k}(-2 \cdot 2 - 2 \cdot 1) \]
\[ = \mathbf{i}(4 + 2) - \mathbf{j}(-4 + 1) + \mathbf{k}(-4 - 2) \]
\[ = 6\mathbf{i} + 3\mathbf{j} - 6\mathbf{k} \]
\[ \mathbf{N_1} = (6, 3, -6) \]

2. **Normal vector \( \mathbf{N_2} \) for base ABC**:
\[ \mathbf{N_2} = \overrightarrow{AB} \times \overrightarrow{AC} \]
\[ \mathbf{N_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 3 & 4 \\ 1 & 2 & 2 \end{vmatrix} \]
\[ = \mathbf{i}(3 \cdot 2 - 4 \cdot 2) - \mathbf{j}(0 \cdot 2 - 4 \cdot 1) + \mathbf{k}(0 \cdot 2 - 3 \cdot 1) \]
\[ = \mathbf{i}(6 - 8) - \mathbf{j}(0 - 4) + \mathbf{k}(0 - 3) \]
\[ = -2\mathbf{i} + 4\mathbf{j} - 3\mathbf{k} \]
\[ \mathbf{N_2} = (-2, 4, -3) \]

### Step 4: Calculate Cross Product
Now we calculate \( \mathbf{N_1} \times \mathbf{N_2} \):
\[ \mathbf{N_1} \times \mathbf{N_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 6 & 3 & -6 \\ -2 & 4 & -3 \end{vmatrix} \]
\[ = \mathbf{i}(3 \cdot (-3) - (-6) \cdot 4) - \mathbf{j}(6 \cdot (-3) - (-6) \cdot (-2)) + \mathbf{k}(6 \cdot 4 - 3 \cdot (-2)) \]
\[ = \mathbf{i}(-9 + 24) - \mathbf{j}(-18 - 12) + \mathbf{k}(24 + 6) \]
\[ = 15\mathbf{i} + 30\mathbf{j} + 30\mathbf{k} \]

### Step 5: Calculate Magnitudes
1. **Magnitude of \( \mathbf{N_1} \)**:
\[ |\mathbf{N_1}| = \sqrt{6^2 + 3^2 + (-6)^2} = \sqrt{36 + 9 + 36} = \sqrt{81} = 9 \]

2. **Magnitude of \( \mathbf{N_2} \)**:
\[ |\mathbf{N_2}| = \sqrt{(-2)^2 + 4^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29} \]

3. **Magnitude of \( \mathbf{N_1} \times \mathbf{N_2} \)**:
\[ |\mathbf{N_1} \times \mathbf{N_2}| = \sqrt{15^2 + 30^2 + 30^2} = \sqrt{225 + 900 + 900} = \sqrt{2025} = 45 \]

### Step 6: Calculate Sine of the Angle
Using the formula:
\[ \sin(\theta) = \frac{|\mathbf{N_1} \times \mathbf{N_2}|}{|\mathbf{N_1}| \cdot |\mathbf{N_2}|} \]
\[ = \frac{45}{9 \cdot \sqrt{29}} = \frac{5}{\sqrt{29}} \]

### Final Answer
Thus, the sine of the angle formed by the lateral face ADC and the base ABC of the tetrahedron ABCD is:
\[ \sin(\theta) = \frac{5}{\sqrt{29}} \]