Question

Ex. 7. Using vector method, find the incenter of the triangle whose vertices are A(0, 3, 0), B(0, 0, 4) and C(0, 3, 4).

Answer 1

(0, 2, 3)

Answer 2

Solution:

1. Compute the side lengths:
    • Length BC (opposite A):
     $BC = \sqrt{(0-0)^2+(3-0)^2+(4-4)^2} = 3$
    • Length AC (opposite B):
     $AC = \sqrt{(0-0)^2+(3-3)^2+(4-0)^2} = 4$
    • Length AB (opposite C):
     $AB = \sqrt{(0-0)^2+(3-0)^2+(0-4)^2} = 5$

2. Use the formula for the incenter with vertices $A$, $B$, $C$ and side lengths $a$, $b$, $c$ (opposite to $A$, $B$, $C$ respectively):
     $I = \frac{aA + bB + cC}{a+b+c} = \frac{3(0,3,0) + 4(0,0,4) + 5(0,3,4)}{3+4+5}$

3. Evaluate the numerator:
     $3(0,3,0) = (0,9,0)$
     $4(0,0,4) = (0,0,16)$
     $5(0,3,4) = (0,15,20)$
     Sum: $(0,9+15,0+16+20) = (0,24,36)$

4. Divide by the perimeter $12$:
     $I = \left(0,\frac{24}{12},\frac{36}{12}\right) = (0, 2, 3)$

Answer:
The incenter of the triangle is $(0, 2, 3)$.