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 Question:
We must find the 3D line equation for the internal angle bisector of a triangle. The direction vector of an angle bisector is the sum of the unit vectors of the two forming sides.

Step 2: Detailed Explanation:
The bisector passes through vertex A, so the line must use point $(1, -1, 0)$.
Let's find the direction vectors of the sides emanating from A.
Vector $\vec{AB$:}
$\vec{AB} = (3-1)\hat{i} + (5 - (-1))\hat{j} + (3-0)\hat{k} = 2\hat{i} + 6\hat{j} + 3\hat{k}$
Magnitude $|\vec{AB}| = \sqrt{2^2 + 6^2 + 3^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7$
Unit vector $\hat{u}_{AB} = \frac{2}{7}\hat{i} + \frac{6}{7}\hat{j} + \frac{3}{7}\hat{k}$
Vector $\vec{AC$:}
$\vec{AC} = (-11-1)\hat{i} + (-5 - (-1))\hat{j} + (6-0)\hat{k} = -12\hat{i} - 4\hat{j} + 6\hat{k}$
Magnitude $|\vec{AC}| = \sqrt{(-12)^2 + (-4)^2 + 6^2} = \sqrt{144 + 16 + 36} = \sqrt{196} = 14$
Unit vector $\hat{u}_{AC} = \frac{-12}{14}\hat{i} - \frac{4}{14}\hat{j} + \frac{6}{14}\hat{k} = \frac{-6}{7}\hat{i} - \frac{2}{7}\hat{j} + \frac{3}{7}\hat{k}$
Direction of the Internal Bisector ($\vec{d}$):
The direction vector is proportional to the sum of the unit vectors:
$\vec{d} = \hat{u}_{AB} + \hat{u}_{AC}$
$\vec{d} = \left(\frac{2}{7} - \frac{6}{7}\right)\hat{i} + \left(\frac{6}{7} - \frac{2}{7}\right)\hat{j} + \left(\frac{3}{7} + \frac{3}{7}\right)\hat{k}$
$\vec{d} = \frac{-4}{7}\hat{i} + \frac{4}{7}\hat{j} + \frac{6}{7}\hat{k}$
We can scale this vector to remove fractions. Multiply by $7/2$:
$\vec{d} \propto -2\hat{i} + 2\hat{j} + 3\hat{k}$
Equation of the line:
Using point $A(1, -1, 0)$ and direction vector $\langle -2, 2, 3 \rangle$:
$\frac{x - 1}{-2} = \frac{y - (-1)}{2} = \frac{z - 0}{3}$
Rearranging to match the options. Multiply the first fraction's numerator and denominator by -1:
$\frac{1 - x}{2} = \frac{y - (-1)}{2} = \frac{z}{3}$

Step 3: Final Answer:
The equation matches exactly with option (a).