If a plane cuts off intercepts `OA = a, OB = b, OC = c` from the coordinate axes 9where `'O'` is the origin). then the area of the triangle `ABC` is equal to

Options

A. `(1)/(2)(ab+bc+ac)`
B. `(1)/(2)abc`
C. `(1)/(2)(a^(2)b^(2)+b^(2)c^(2)+c^(2)a^(2))^(1//2)`
D. `(1)/(2)(a+b+c)^(2)`

Explanation

To find the area of triangle ABC formed by the intercepts OA, OB, and OC from the coordinate axes, we can follow these steps:

### Step 1: Identify the coordinates of points A, B, and C.
- The intercepts given are OA = a, OB = b, and OC = c.
- Therefore, the coordinates of the points are:
- A (a, 0, 0)
- B (0, b, 0)
- C (0, 0, c)

### Step 2: Use the formula for the area of a triangle in 3D.
The area \( A \) of triangle ABC can be calculated using the formula:
\[ A = \frac{1}{2} \| \vec{AB} \times \vec{AC} \| \]
where \( \vec{AB} \) and \( \vec{AC} \) are the vectors from A to B and from A to C, respectively.

### Step 3: Calculate the vectors \( \vec{AB} \) and \( \vec{AC} \).
- The vector \( \vec{AB} \) is given by:
\[ \vec{AB} = B - A = (0, b, 0) - (a, 0, 0) = (-a, b, 0) \]
- The vector \( \vec{AC} \) is given by:
\[ \vec{AC} = C - A = (0, 0, c) - (a, 0, 0) = (-a, 0, c) \]

### Step 4: Compute the cross product \( \vec{AB} \times \vec{AC} \).
Using the determinant form for the cross product:
\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -a & b & 0 \\ -a & 0 & c \end{vmatrix} \]
Calculating this determinant:
\[ = \hat{i} \left(b \cdot c - 0 \cdot 0\right) - \hat{j} \left(-a \cdot c - 0 \cdot -a\right) + \hat{k} \left(-a \cdot 0 - (-a) \cdot b\right) \]
\[ = \hat{i} (bc) + \hat{j} (ac) + \hat{k} (ab) \]
Thus,
\[ \vec{AB} \times \vec{AC} = (bc, ac, ab) \]

### Step 5: Find the magnitude of the cross product.
The magnitude is given by:
\[ \| \vec{AB} \times \vec{AC} \| = \sqrt{(bc)^2 + (ac)^2 + (ab)^2} \]

### Step 6: Calculate the area of triangle ABC.
Substituting back into the area formula:
\[ A = \frac{1}{2} \| \vec{AB} \times \vec{AC} \| = \frac{1}{2} \sqrt{(bc)^2 + (ac)^2 + (ab)^2} \]

### Final Result:
The area of triangle ABC is:
\[ \text{Area} = \frac{1}{2} \sqrt{(bc)^2 + (ac)^2 + (ab)^2} \]