If the plane `x/2+y/3+z/4=1` cuts the coordinate axes in `A, B,C,` then the area of triangle `ABC` is

Options

A. `sqrt(19)` sq, units
B. `sqrt(41)` sq. units
C. `sqrt(61)` sq. units
D. None of these

Answer

To find the area of triangle ABC formed by the intersection of the plane \( \frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1 \) with the coordinate axes, we can follow these steps:

### Step 1: Find the intercepts of the plane with the coordinate axes.
The intercepts can be found by setting two of the variables to zero and solving for the third.

- **X-intercept (A)**: Set \( y = 0 \) and \( z = 0 \):
\[ \frac{x}{2} = 1 \implies x = 2 \implies A(2, 0, 0) \]

- **Y-intercept (B)**: Set \( x = 0 \) and \( z = 0 \):
\[ \frac{y}{3} = 1 \implies y = 3 \implies B(0, 3, 0) \]

- **Z-intercept (C)**: Set \( x = 0 \) and \( y = 0 \):
\[ \frac{z}{4} = 1 \implies z = 4 \implies C(0, 0, 4) \]

### Step 2: Use the formula for the area of a triangle formed by three points in 3D space.
The area \( A \) of triangle ABC can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \sqrt{a^2 b^2 + b^2 c^2 + c^2 a^2} \]
where \( a, b, c \) are the lengths of the sides opposite to vertices A, B, and C respectively.

### Step 3: Calculate the lengths of the sides.
Using the coordinates of points A, B, and C:
- Length \( a \) (opposite A):
\[ a = \sqrt{(0 - 0)^2 + (3 - 0)^2 + (0 - 4)^2} = \sqrt{0 + 9 + 16} = \sqrt{25} = 5 \]

- Length \( b \) (opposite B):
\[ b = \sqrt{(2 - 0)^2 + (0 - 3)^2 + (0 - 4)^2} = \sqrt{4 + 9 + 16} = \sqrt{29} \]

- Length \( c \) (opposite C):
\[ c = \sqrt{(2 - 0)^2 + (0 - 0)^2 + (0 - 4)^2} = \sqrt{4 + 0 + 16} = \sqrt{20} = 2\sqrt{5} \]

### Step 4: Substitute the lengths into the area formula.
Now substituting \( a = 5 \), \( b = \sqrt{29} \), and \( c = 2\sqrt{5} \) into the area formula:
\[ \text{Area} = \frac{1}{2} \sqrt{5^2 \cdot (\sqrt{29})^2 + (\sqrt{29})^2 \cdot (2\sqrt{5})^2 + (2\sqrt{5})^2 \cdot 5^2} \]
Calculating each term:
- \( 5^2 \cdot (\sqrt{29})^2 = 25 \cdot 29 = 725 \)
- \( (\sqrt{29})^2 \cdot (2\sqrt{5})^2 = 29 \cdot 20 = 580 \)
- \( (2\sqrt{5})^2 \cdot 5^2 = 20 \cdot 25 = 500 \)

Now, summing these:
\[ \text{Area} = \frac{1}{2} \sqrt{725 + 580 + 500} = \frac{1}{2} \sqrt{1805} \]

### Step 5: Final area calculation.
The final area of triangle ABC is:
\[ \text{Area} = \frac{1}{2} \sqrt{1805} \text{ square units} \]