Question

If \( A(0, 4, 0) \), \( B(0, 0, 3) \), and \( C(0, 4, 3) \) are the vertices of \( \Delta ABC \), then its incenter is:

• A. \( (2, 0, 3) \)
• B. \( (3, 0, 2) \)
• C. \( (0, 3, 2) \)
• D. \( (0, 2, 3) \)

Hint:

To find the incenter in 3D, use the formula for the weighted average of the coordinates of the vertices, using the opposite side lengths as weights.

Answer 1

The Correct Option is C

Step 1: Formula for the incenter.
The incenter of a triangle in three-dimensional space can be found by taking the weighted average of the coordinates of the vertices, where the weights are the lengths of the sides opposite the respective vertices. Step 2: Find the lengths of the sides.
First, calculate the lengths of the sides of the triangle: \[ AB = \sqrt{(0 - 0)^2 + (4 - 0)^2 + (0 - 3)^2} = 5, \quad BC = \sqrt{(0 - 0)^2 + (0 - 4)^2 + (3 - 3)^2} = 4, \quad CA = \sqrt{(0 - 0)^2 + (4 - 0)^2 + (3 - 0)^2} = 5 \] Step 3: Find the coordinates of the incenter.
The coordinates of the incenter are given by: \[ I = \left( \frac{aA_x + bB_x + cC_x}{a + b + c}, \frac{aA_y + bB_y + cC_y}{a + b + c}, \frac{aA_z + bB_z + cC_z}{a + b + c} \right) \] where \( a = BC = 4 \), \( b = CA = 5 \), and \( c = AB = 5 \). Substituting the values, we find: \[ I = \left( \frac{4 \times 0 + 5 \times 0 + 5 \times 0}{4 + 5 + 5}, \frac{4 \times 4 + 5 \times 0 + 5 \times 4}{4 + 5 + 5}, \frac{4 \times 0 + 5 \times 3 + 5 \times 3}{4 + 5 + 5} \right) \] \[ I = \left( 0, 3, 2 \right) \] Step 4: Conclusion.
Thus, the incenter of the triangle is \( \boxed{(0, 3, 2)} \).