Consider a triangular pyramid ABCD the position vectors of whone agular points are `A(3,0,1),B(-1,4,1),C(5,3, 2) and D(0,-5,4)` Let G be the point of intersection of the medians of the triangle BCT. The length of the perpendicular from the vertex D on the opposite face

Options

A. (a) 24
B. (b) `8 sqrt6`
C. (c) `4sqrt6`
D. (d) none of these

Explanation

To solve the problem, we need to find the length of the perpendicular from vertex D to the opposite face of the triangular pyramid ABCD. The steps are as follows:

### Step 1: Identify the Points and Vectors
We have the position vectors of the points:
- A(3, 0, 1)
- B(-1, 4, 1)
- C(5, 3, 2)
- D(0, -5, 4)

### Step 2: Find the Vectors AB and AC
To find the vectors AB and AC, we use the formula for the vector between two points \( P(x_1, y_1, z_1) \) and \( Q(x_2, y_2, z_2) \):
\[ \overrightarrow{PQ} = (x_2 - x_1) \hat{i} + (y_2 - y_1) \hat{j} + (z_2 - z_1) \hat{k} \]

Calculating \( \overrightarrow{AB} \):
\[ \overrightarrow{AB} = B - A = (-1 - 3) \hat{i} + (4 - 0) \hat{j} + (1 - 1) \hat{k} = -4 \hat{i} + 4 \hat{j} + 0 \hat{k} = -4 \hat{i} + 4 \hat{j} \]

Calculating \( \overrightarrow{AC} \):
\[ \overrightarrow{AC} = C - A = (5 - 3) \hat{i} + (3 - 0) \hat{j} + (2 - 1) \hat{k} = 2 \hat{i} + 3 \hat{j} + 1 \hat{k} \]

### Step 3: Find the Normal Vector to the Plane ABC
To find the normal vector \( \overrightarrow{n} \) to the plane formed by points A, B, and C, we can take the cross product of vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \):
\[ \overrightarrow{n} = \overrightarrow{AB} \times \overrightarrow{AC} \]

Calculating the cross product:
\[ \overrightarrow{AB} = (-4, 4, 0), \quad \overrightarrow{AC} = (2, 3, 1) \]
Using the determinant formula:
\[ \overrightarrow{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & 4 & 0 \\ 2 & 3 & 1 \end{vmatrix} \]

Calculating the determinant:
\[ \overrightarrow{n} = \hat{i}(4 \cdot 1 - 0 \cdot 3) - \hat{j}(-4 \cdot 1 - 0 \cdot 2) + \hat{k}(-4 \cdot 3 - 4 \cdot 2) \]
\[ = 4\hat{i} + 4\hat{j} - 24\hat{k} \]
Thus, \( \overrightarrow{n} = (4, 4, -24) \).

### Step 4: Find the Length of the Perpendicular from D to the Plane ABC
The length of the perpendicular from point D to the plane can be calculated using the formula:
\[ \text{Distance} = \frac{| \overrightarrow{n} \cdot \overrightarrow{AD} |}{|\overrightarrow{n}|} \]
where \( \overrightarrow{AD} = D - A = (0 - 3, -5 - 0, 4 - 1) = (-3, -5, 3) \).

Calculating the dot product:
\[ \overrightarrow{n} \cdot \overrightarrow{AD} = (4, 4, -24) \cdot (-3, -5, 3) = 4(-3) + 4(-5) + (-24)(3) = -12 - 20 - 72 = -104 \]
Taking the absolute value:
\[ | \overrightarrow{n} \cdot \overrightarrow{AD} | = 104 \]

Calculating the magnitude of \( \overrightarrow{n} \):
\[ |\overrightarrow{n}| = \sqrt{4^2 + 4^2 + (-24)^2} = \sqrt{16 + 16 + 576} = \sqrt{608} = 4\sqrt{38} \]

### Step 5: Calculate the Distance
Now substituting back into the distance formula:
\[ \text{Distance} = \frac{104}{4\sqrt{38}} = \frac{26}{\sqrt{38}} = \frac{26\sqrt{38}}{38} = \frac{13\sqrt{38}}{19} \]

### Final Answer
The length of the perpendicular from vertex D to the opposite face ABC is \( \frac{13\sqrt{38}}{19} \).

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