A triangle ABC is formed by vertices A(1, -1, | vedclass

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Question

(A) The internal angle bisector of $\angle A$ passes through $A(1, -1, 0)$ and divides the side $BC$ in the ratio of the lengths of the adjacent sides

Vedclass · Accepted Answer

$\frac{x-1}{2} = \frac{y+1}{2} = \frac{z}{3}$

Vedclass · Answer

(A) The internal angle bisector of $\angle A$ passes through $A(1, -1, 0)$ and divides the side $BC$ in the ratio of the lengths of the adjacent sides $AB$ and $AC$.First,calculate the lengths of sides $AB$ and $AC$:$AB = \sqrt{(3-1)^2 + (5-(-1))^2 + (3-0)^2} = \sqrt{2^2 + 6^2 + 3^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7$.$AC = \sqrt{(-11-1)^2 + (-5-(-1))^2 + (6-0)^2} = \sqrt{(-12)^2 + (-4)^2 + 6^2} = \sqrt{144 + 16 + 36} = \sqrt{196} = 14$.The ratio $AB:AC = 7:14 = 1:2$.The angle bisector divides $BC$ in the ratio $1:2$. Let $D$ be the point on $BC$ such that $BD:DC = 1:2$. Using the section formula:$D = \left( \frac{1(-11) + 2(3)}{1+2}, \frac{1(-5) + 2(5)}{1+2}, \frac{1(6) + 2(3)}{1+2} \right) = \left( \frac{-5}{3}, \frac{5}{3}, \frac{12}{3} \right) = \left( -\frac{5}{3}, \frac{5}{3}, 4 \right)$.The direction vector of the line $AD$ is $\vec{AD} = \left( -\frac{5}{3} - 1, \frac{5}{3} - (-1), 4 - 0 \right) = \left( -\frac{8}{3}, \frac{8}{3}, 4 \right)$.Multiplying by $\frac{3}{4}$ to simplify the direction ratios,we get $(-2, 2, 3)$.Thus,the equation of the line passing through $A(1, -1, 0)$ with direction ratios $(-2, 2, 3)$ is $\frac{x-1}{-2} = \frac{y+1}{2} = \frac{z}{3}$,which is equivalent to $\frac{x-1}{2} = \frac{y+1}{-2} = \frac{z}{-3}$.Comparing with the options,the correct direction vector is proportional to $(2, -2, -3)$ or $(-2, 2, 3)$. The provided options seem to contain a typo in the signs. Given the standard form,option $A$ is the closest representation.

$A$ triangle $ABC$ is formed by vertices $A(1, -1, 0)$,$B(3, 5, 3)$,and $C(-11, -5, 6)$. The equation of the internal angle bisector of $\angle A$ is:

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