What is the area of triangle whose vertices are `(0,0,0),(1,2,3) and (-3,-2,1)`?

Options

A. `3sqrt5` sq units
B. `6sqrt5` sq units
C. 6 sq units
D. 12 sq units

Answer

To find the area of the triangle with vertices at points \( A(0,0,0) \), \( B(1,2,3) \), and \( C(-3,-2,1) \), we can use the formula for the area of a triangle in 3D space, which is given by:

\[ \text{Area} = \frac{1}{2} \| \vec{AB} \times \vec{AC} \| \]

where \( \vec{AB} \) and \( \vec{AC} \) are the vectors from point A to points B and C respectively, and \( \times \) denotes the cross product.

### Step 1: Calculate the vectors \( \vec{AB} \) and \( \vec{AC} \)

1. **Find vector \( \vec{AB} \)**:
\[ \vec{AB} = B - A = (1 - 0, 2 - 0, 3 - 0) = (1, 2, 3) \]

2. **Find vector \( \vec{AC} \)**:
\[ \vec{AC} = C - A = (-3 - 0, -2 - 0, 1 - 0) = (-3, -2, 1) \]

### Step 2: Calculate the cross product \( \vec{AB} \times \vec{AC} \)

The cross product of two vectors \( \vec{AB} = (x_1, y_1, z_1) \) and \( \vec{AC} = (x_2, y_2, z_2) \) can be calculated using the determinant of a matrix:

\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -3 & -2 & 1 \end{vmatrix} \]

Calculating this determinant:

\[ \vec{AB} \times \vec{AC} = \hat{i} \begin{vmatrix} 2 & 3 \\ -2 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 3 \\ -3 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ -3 & -2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} 2 & 3 \\ -2 & 1 \end{vmatrix} = (2)(1) - (3)(-2) = 2 + 6 = 8 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 1 & 3 \\ -3 & 1 \end{vmatrix} = (1)(1) - (3)(-3) = 1 + 9 = 10 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 1 & 2 \\ -3 & -2 \end{vmatrix} = (1)(-2) - (2)(-3) = -2 + 6 = 4 \]

Putting it all together, we have:

\[ \vec{AB} \times \vec{AC} = 8\hat{i} - 10\hat{j} + 4\hat{k} = (8, -10, 4) \]

### Step 3: Calculate the magnitude of the cross product

The magnitude of the vector \( \vec{AB} \times \vec{AC} \) is given by:

\[ \| \vec{AB} \times \vec{AC} \| = \sqrt{8^2 + (-10)^2 + 4^2} = \sqrt{64 + 100 + 16} = \sqrt{180} = 6\sqrt{5} \]

### Step 4: Calculate the area of the triangle

Now, we can find the area of the triangle:

\[ \text{Area} = \frac{1}{2} \| \vec{AB} \times \vec{AC} \| = \frac{1}{2} (6\sqrt{5}) = 3\sqrt{5} \]

Thus, the area of the triangle is:

\[ \text{Area} = 3\sqrt{5} \text{ square units} \]