Question

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

• A. $\frac{1-x}{2} = \frac{y-(-1)}{2} = \frac{z}{3}$
• B. $\frac{x+1}{2} = \frac{y-1}{2} = \frac{z}{3}$
• C. $\frac{x-1}{2} = \frac{y+1}{2} = \frac{z}{3}$
• D. $\frac{x-2}{1} = \frac{y+3}{2} = \frac{z}{3}$

Hint:

The direction of the internal angle bisector of two vectors $\vec{p}$ and $\vec{q}$ is always along the vector $\frac{\vec{p}}{|\vec{p}|} + \frac{\vec{q}}{|\vec{q}|}$ (the sum of their unit vectors). Just add them up and scale to remove fractions!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The internal bisector of $\angle A$ divides the opposite side $BC$ in the ratio of the sides $AB$ and $AC$.

Step 2: Formula Application:
Distance $AB = \sqrt{(3-1)^2 + (5+1)^2 + (3-0)^2} = \sqrt{4+36+9} = 7$. Distance $AC = \sqrt{(-11-1)^2 + (-5+1)^2 + (6-0)^2} = \sqrt{144+16+36} = 14$. Ratio $AB:AC = 7:14 = 1:2$.

Step 3: Explanation:
The bisector meets $BC$ at point $D$, which divides $BC$ in ratio $1:2$. $D = \left( \frac{1(-11) + 2(3)}{1+2}, \frac{1(-5) + 2(5)}{1+2}, \frac{1(6) + 2(3)}{1+2} \right) = \left( \frac{-5}{3}, \frac{5}{3}, 4 \right)$. The line $AD$ passes through $(1, -1, 0)$ and $(-5/3, 5/3, 4)$. Direction ratios: $(1 - (-5/3), -1 - 5/3, 0 - 4) = (8/3, -8/3, -4)$. Simplified DRs: $(2, -2, -3)$ or $(2, 2, 3)$ depending on vector direction.

Step 4: Final Answer:
The equation is $\frac{x-1}{2} = \frac{y+1}{2} = \frac{z}{3}$.