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 plane of the base ABC of the tetrahedron ABCD, we will follow these steps:

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

### Step 2: Find Vectors for the Planes
We need to find two vectors for each of the planes.

**For Plane ABC:**
- Vector \( \overrightarrow{AB} = B - A = (3, 1, 5) - (3, -2, 1) = (0, 3, 4) \)
- Vector \( \overrightarrow{AC} = C - A = (4, 0, 3) - (3, -2, 1) = (1, 2, 2) \)

**For Plane ADC:**
- Vector \( \overrightarrow{AD} = D - A = (1, 0, 0) - (3, -2, 1) = (-2, 2, -1) \)
- Vector \( \overrightarrow{AC} \) is the same as before: \( (1, 2, 2) \)

### Step 3: Find Normal Vectors
The normal vector of a plane can be found using the cross product of two vectors lying in that plane.

**Normal Vector for Plane ABC (\( \overrightarrow{n_1} \)):**
\[ \overrightarrow{n_1} = \overrightarrow{AB} \times \overrightarrow{AC} \]
Calculating the cross product:
\[ \overrightarrow{n_1} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 3 & 4 \\ 1 & 2 & 2 \end{vmatrix} = \hat{i}(3 \cdot 2 - 4 \cdot 2) - \hat{j}(0 \cdot 2 - 4 \cdot 1) + \hat{k}(0 \cdot 2 - 3 \cdot 1) \]
\[ = \hat{i}(6 - 8) - \hat{j}(0 - 4) + \hat{k}(0 - 3) = -2\hat{i} + 4\hat{j} - 3\hat{k} \]

Thus, \( \overrightarrow{n_1} = (-2, 4, -3) \).

**Normal Vector for Plane ADC (\( \overrightarrow{n_2} \)):**
\[ \overrightarrow{n_2} = \overrightarrow{AD} \times \overrightarrow{AC} \]
Calculating the cross product:
\[ \overrightarrow{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 2 & -1 \\ 1 & 2 & 2 \end{vmatrix} = \hat{i}(2 \cdot 2 - (-1) \cdot 2) - \hat{j}(-2 \cdot 2 - (-1) \cdot 1) + \hat{k}(-2 \cdot 2 - 2 \cdot 1) \]
\[ = \hat{i}(4 + 2) - \hat{j}(-4 + 1) + \hat{k}(-4 - 2) = 6\hat{i} + 3\hat{j} - 6\hat{k} \]

Thus, \( \overrightarrow{n_2} = (6, 3, -6) \).

### Step 4: Calculate the Cross Product of Normal Vectors
Now we calculate \( \overrightarrow{n_1} \times \overrightarrow{n_2} \):
\[ \overrightarrow{n_1} \times \overrightarrow{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 4 & -3 \\ 6 & 3 & -6 \end{vmatrix} \]
Calculating this determinant:
\[ = \hat{i}(4 \cdot -6 - (-3) \cdot 3) - \hat{j}(-2 \cdot -6 - (-3) \cdot 6) + \hat{k}(-2 \cdot 3 - 4 \cdot 6) \]
\[ = \hat{i}(-24 + 9) - \hat{j}(12 - 18) + \hat{k}(-6 - 24) = -15\hat{i} + 6\hat{j} - 30\hat{k} \]

### Step 5: Find Magnitudes
Now we find the magnitudes of \( \overrightarrow{n_1} \), \( \overrightarrow{n_2} \), and \( \overrightarrow{n_1} \times \overrightarrow{n_2} \):
\[ |\overrightarrow{n_1}| = \sqrt{(-2)^2 + 4^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29} \]
\[ |\overrightarrow{n_2}| = \sqrt{6^2 + 3^2 + (-6)^2} = \sqrt{36 + 9 + 36} = \sqrt{81} = 9 \]
\[ |\overrightarrow{n_1} \times \overrightarrow{n_2}| = \sqrt{(-15)^2 + 6^2 + (-30)^2} = \sqrt{225 + 36 + 900} = \sqrt{1161} \]

### Step 6: Calculate Sine of the Angle
Using the formula for sine of the angle between two planes:
\[ \sin \theta = \frac{|\overrightarrow{n_1} \times \overrightarrow{n_2}|}{|\overrightarrow{n_1}| \cdot |\overrightarrow{n_2}|} \]
Substituting the values:
\[ \sin \theta = \frac{\sqrt{1161}}{\sqrt{29} \cdot 9} \]

### Final Result
Thus, the sine of the angle formed by the lateral face ADC and the plane of the base ABC is:
\[ \sin \theta = \frac{\sqrt{1161}}{9\sqrt{29}} \]