If O(0,0,0), A(3,0,0), B(0,4,0) form a triang | vedclass

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(D) The vertices of the triangle are $O(0,0,0)$,$A(3,0,0)$,and $B(0,4,0)$.Let the side lengths be $a, b, c$ opposite to vertices $A, B, O$ respectivel

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$(1,1,0)$

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(D) The vertices of the triangle are $O(0,0,0)$,$A(3,0,0)$,and $B(0,4,0)$.Let the side lengths be $a, b, c$ opposite to vertices $A, B, O$ respectively.$a = |\overrightarrow{AB}| = \sqrt{(0-3)^2 + (4-0)^2 + (0-0)^2} = \sqrt{9+16} = 5$.$b = |\overrightarrow{OB}| = \sqrt{(0-0)^2 + (4-0)^2 + (0-0)^2} = 4$.$c = |\overrightarrow{OA}| = \sqrt{(3-0)^2 + (0-0)^2 + (0-0)^2} = 3$.The coordinates of the incentre $I$ are given by $\left( \frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c}, \frac{az_1 + bz_2 + cz_3}{a+b+c} \right)$.Substituting the values: $I = \left( \frac{5(0) + 4(3) + 3(0)}{5+4+3}, \frac{5(0) + 4(0) + 3(4)}{5+4+3}, \frac{5(0) + 4(0) + 3(0)}{5+4+3} \right)$.$I = \left( \frac{12}{12}, \frac{12}{12}, \frac{0}{12} \right) = (1, 1, 0)$.

If $O(0,0,0), A(3,0,0), B(0,4,0)$ form a triangle,then the incentre of triangle $OAB$ is

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