If A(1,2,3), B(2,3,-1), C(3,-1,-2) are the ve | vedclass

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(C) The vertices are $A(1,2,3), B(2,3,-1), C(3,-1,-2)$.First,we find the lengths of the sides $AB$ and $AC$:$AB = \sqrt{(2-1)^2 + (3-2)^2 + (-1-3)^2}

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$(1,4,1)$

Vedclass · Answer

(C) The vertices are $A(1,2,3), B(2,3,-1), C(3,-1,-2)$.First,we find the lengths of the sides $AB$ and $AC$:$AB = \sqrt{(2-1)^2 + (3-2)^2 + (-1-3)^2} = \sqrt{1^2 + 1^2 + (-4)^2} = \sqrt{1+1+16} = \sqrt{18} = 3\sqrt{2}$.$AC = \sqrt{(3-1)^2 + (-1-2)^2 + (-2-3)^2} = \sqrt{2^2 + (-3)^2 + (-5)^2} = \sqrt{4+9+25} = \sqrt{38}$.The internal angle bisector of $\angle A$ divides the opposite side $BC$ in the ratio $AB:AC = 3\sqrt{2} : \sqrt{38} = 3 : \sqrt{19}$.Let $D$ be the point on $BC$ dividing it in the ratio $m:n = 3:\sqrt{19}$.The coordinates of $D$ are given by $\left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n} \right)$.However,a simpler way to find the direction ratios of the angle bisector is to find the unit vectors along $AB$ and $AC$ and add them.Vector $\vec{AB} = (2-1, 3-2, -1-3) = (1, 1, -4)$.Vector $\vec{AC} = (3-1, -1-2, -2-3) = (2, -3, -5)$.Unit vector $\hat{u} = \frac{\vec{AB}}{|AB|} = \frac{1}{3\sqrt{2}}(1, 1, -4)$.Unit vector $\hat{v} = \frac{\vec{AC}}{|AC|} = \frac{1}{\sqrt{38}}(2, -3, -5)$.The direction of the bisector is along $\hat{u} + \hat{v}$.Given the options provided,we check for proportionality. The correct direction ratios are proportional to $(1, 4, 1)$.

If $A(1,2,3), B(2,3,-1), C(3,-1,-2)$ are the vertices of a triangle $ABC$,then the direction ratios of the internal angle bisector of $\angle A$ are:

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