Let A(2,3,-1), B(4,1,0), C(-1,-1,11) be the v | vedclass

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(C) Given vertices of triangle $A(2,3,-1), B(4,1,0)$ and $C(-1,-1,11)$.First, calculate the lengths of sides $AB$ and $AC$:$AB = \sqrt{(4-2)^2 + (1-3)

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$(17,-38,49)$

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(C) Given vertices of triangle $A(2,3,-1), B(4,1,0)$ and $C(-1,-1,11)$.First, calculate the lengths of sides $AB$ and $AC$:$AB = \sqrt{(4-2)^2 + (1-3)^2 + (0+1)^2} = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4+4+1} = \sqrt{9} = 3$.$AC = \sqrt{(-1-2)^2 + (-1-3)^2 + (11+1)^2} = \sqrt{(-3)^2 + (-4)^2 + 12^2} = \sqrt{9+16+144} = \sqrt{169} = 13$.By the Angle Bisector Theorem, the bisector of $\angle BAC$ divides the opposite side $BC$ in the ratio of the sides containing the angle:$\frac{BD}{DC} = \frac{AB}{AC} = \frac{3}{13}$.Using the section formula, the coordinates of $D$ are:$D = \left( \frac{3(-1) + 13(4)}{3+13}, \frac{3(-1) + 13(1)}{3+13}, \frac{3(11) + 13(0)}{3+13} \right) = \left( \frac{-3+52}{16}, \frac{-3+13}{16}, \frac{33+0}{16} \right) = \left( \frac{49}{16}, \frac{10}{16}, \frac{33}{16} \right)$.The direction ratios of $AD$ are given by the vector $\vec{AD} = D - A$:$\vec{AD} = \left( \frac{49}{16} - 2, \frac{10}{16} - 3, \frac{33}{16} - (-1) \right) = \left( \frac{49-32}{16}, \frac{10-48}{16}, \frac{33+16}{16} \right) = \left( \frac{17}{16}, \frac{-38}{16}, \frac{49}{16} \right)$.Since direction ratios can be scaled by a constant, we multiply by $16$ to get $(17, -38, 49)$.Thus, option $C$ is correct.

Let $A(2,3,-1), B(4,1,0), C(-1,-1,11)$ be the vertices of a triangle $ABC$. Let $D$ be the point where the bisector of $\angle BAC$ meets the side $BC$. Then the direction ratios of $AD$ are

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