If the triangle ABC whose vertices are `A(-1, 1, 1), B(1, -1, 1) and C(1, 1, -1)` is projected on xy-plane, then the area of the projection triangles is…..

Answer

To find the area of the projection of triangle ABC onto the XY-plane, we can follow these steps:

### Step 1: Identify the vertices of the triangle
The vertices of triangle ABC are given as:
- A(-1, 1, 1)
- B(1, -1, 1)
- C(1, 1, -1)

### Step 2: Project the vertices onto the XY-plane
When projecting onto the XY-plane, we ignore the z-coordinates of the vertices. Thus, the projected vertices are:
- A'(-1, 1)
- B'(1, -1)
- C'(1, 1)

### Step 3: Use the formula for the area of a triangle given by its vertices
The area \( A \) of a triangle with vertices at points \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

### Step 4: Substitute the coordinates of the projected vertices into the formula
Using the projected vertices A'(-1, 1), B'(1, -1), and C'(1, 1):
- \( x_1 = -1, y_1 = 1 \)
- \( x_2 = 1, y_2 = -1 \)
- \( x_3 = 1, y_3 = 1 \)

Substituting these values into the area formula:
\[ A = \frac{1}{2} \left| -1((-1) - 1) + 1(1 - 1) + 1(1 - (-1)) \right| \]

### Step 5: Simplify the expression
Calculating each term:
- First term: \( -1 \cdot (-2) = 2 \)
- Second term: \( 1 \cdot 0 = 0 \)
- Third term: \( 1 \cdot 2 = 2 \)

Putting it all together:
\[ A = \frac{1}{2} \left| 2 + 0 + 2 \right| = \frac{1}{2} \left| 4 \right| = \frac{4}{2} = 2 \]

### Final Result
The area of the projection of triangle ABC onto the XY-plane is \( 2 \) square units.

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