Consider a triangulat pyramid ABCD the position vector of whose angular points are `A(3, 0, 1), B(-1, 4, 1), C(5, 2, 3) and D(0, -5, 4)`. Let G be the point of intersection of the medians of the `triangle(BCD)`.
Q. Area of the `triangle(ABC)` (in sq. units) is

Options

A. `24`
B. `8sqrt(6)`
C. `4sqrt(6)`
D. None of these

Answer

To find the area of triangle ABC given the position vectors of points A, B, C, and D, we can follow these steps:

### Step 1: Identify the position vectors
The position vectors of the points are given as:
- A(3, 0, 1)
- B(-1, 4, 1)
- C(5, 2, 3)

### Step 2: Define the vectors for the sides of the triangle
We need to find the vectors for two sides of triangle ABC. We can choose vectors \( \vec{AB} \) and \( \vec{AC} \).

\[ \vec{AB} = \vec{B} - \vec{A} = (-1, 4, 1) - (3, 0, 1) = (-1 - 3, 4 - 0, 1 - 1) = (-4, 4, 0) \]

\[ \vec{AC} = \vec{C} - \vec{A} = (5, 2, 3) - (3, 0, 1) = (5 - 3, 2 - 0, 3 - 1) = (2, 2, 2) \]

### Step 3: Calculate the cross product of the two vectors
The area of triangle ABC can be found using the formula:
\[ \text{Area} = \frac{1}{2} \|\vec{AB} \times \vec{AC}\| \]

First, we calculate the cross product \( \vec{AB} \times \vec{AC} \).

\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & 4 & 0 \\ 2 & 2 & 2 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} 4 & 0 \\ 2 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} -4 & 0 \\ 2 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} -4 & 4 \\ 2 & 2 \end{vmatrix} \]

Calculating the minors:

\[ = \hat{i} (4 \cdot 2 - 0 \cdot 2) - \hat{j} (-4 \cdot 2 - 0 \cdot 2) + \hat{k} (-4 \cdot 2 - 4 \cdot 2) \]

\[ = \hat{i} (8) - \hat{j} (-8) + \hat{k} (-8 - 8) \]

\[ = 8\hat{i} + 8\hat{j} - 16\hat{k} \]

So,

\[ \vec{AB} \times \vec{AC} = (8, 8, -16) \]

### Step 4: Calculate the magnitude of the cross product
Now we find the magnitude of the cross product:

\[ \|\vec{AB} \times \vec{AC}\| = \sqrt{8^2 + 8^2 + (-16)^2} = \sqrt{64 + 64 + 256} = \sqrt{384} = 8\sqrt{6} \]

### Step 5: Calculate the area of triangle ABC
Now we can find the area:

\[ \text{Area} = \frac{1}{2} \|\vec{AB} \times \vec{AC}\| = \frac{1}{2} (8\sqrt{6}) = 4\sqrt{6} \]

### Final Answer
The area of triangle ABC is \( 4\sqrt{6} \) square units.

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