Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The vertices of the triangle are $A(0,3,0)$, $B(0,0,4)$, and $C(0,3,4)$.

The lengths of the sides of the triangle are:
$a=BC=\sqrt{(0-0{)}^2+(3-0{)}^2+(4-4{)}^2}=3$
$b=AC=\sqrt{(0-0{)}^2+(3-3{)}^2+(4-0{)}^2}=4$
$c=AB=\sqrt{(0-0{)}^2+(0-3{)}^2+(4-0{)}^2}=5$

The coordinates of the incenter $I$ are given by the formula:
$I=\left(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c},\frac{a{z}_1+b{z}_2+c{z}_3}{a+b+c}\right)$

Substituting the values:
$x=\frac{3(0)+4(0)+5(0)}{3+4+5}=0$

$y=\frac{3(3)+4(0)+5(3)}{12}=\frac{9+0+15}{12}=2$

$z=\frac{3(0)+4(4)+5(4)}{12}=\frac{0+16+20}{12}=3$

The coordinates of the incenter are $(0,2,3)$.

In vector form, the position vector of the incenter is $2\hat{j}+3\hat{k}$.

Answer: $2\hat{j}+3\hat{k}$