The area of the triangle formed by A (1, 0, 0), B(0, 1, 0), C(1, 1, 1) is

Options

A. `1/2`
B. `sqrt(3)/4`
C. `sqrt(3)/2`
D. `1/4`

Answer

To find the area of the triangle formed by the points A(1, 0, 0), B(0, 1, 0), and C(1, 1, 1), we can use the formula for the area of a triangle given by vertices in 3D space. The area \( \Delta \) of the triangle can be calculated using the formula:

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

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

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

1. **Calculate \( \vec{AB} \)**:
\[ \vec{AB} = B - A = (0, 1, 0) - (1, 0, 0) = (-1, 1, 0) \]

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

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

To find the cross product \( \vec{AB} \times \vec{AC} \), we can use the determinant of a matrix formed by the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the components of the vectors \( \vec{AB} \) and \( \vec{AC} \):

\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} \]

Calculating this determinant:

\[ = \hat{i} \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & 0 \\ 0 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 1 \\ 0 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} = (1)(1) - (0)(1) = 1 \)
2. \( \begin{vmatrix} -1 & 0 \\ 0 & 1 \end{vmatrix} = (-1)(1) - (0)(0) = -1 \)
3. \( \begin{vmatrix} -1 & 1 \\ 0 & 1 \end{vmatrix} = (-1)(1) - (1)(0) = -1 \)

Thus, we have:

\[ \vec{AB} \times \vec{AC} = \hat{i}(1) - \hat{j}(-1) + \hat{k}(-1) = \hat{i} + \hat{j} - \hat{k} \]

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

Now we find the magnitude:

\[ \| \vec{AB} \times \vec{AC} \| = \sqrt{(1)^2 + (1)^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

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

Now we can find the area of the triangle:

\[ \Delta = \frac{1}{2} \| \vec{AB} \times \vec{AC} \| = \frac{1}{2} \sqrt{3} \]

Thus, the area of the triangle formed by the points A, B, and C is:

\[ \Delta = \frac{\sqrt{3}}{2} \]

### Final Answer:
The area of the triangle formed by A(1, 0, 0), B(0, 1, 0), and C(1, 1, 1) is \( \frac{\sqrt{3}}{2} \).

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