Vectors & 3D Geometry
Angle bisector of vertex — identifying BC and B
MJAT_TS6_P2
Grade 12
Question:
Let $A(\frac{1}{2},1,3)$, $C(3,\frac{17}{4},\frac{13}{2})$. Angle bisector of $\angle ABC$: $\frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{2}$. Which is/are correct?
A) BC: $\dfrac{x-1}{4}=\dfrac{2y-5}{7}=\dfrac{z-9}{2}$
B) BC: $\dfrac{x-1}{2}=\dfrac{y+2}{7}=\dfrac{z+1}{3}$
C) $B=(\alpha,\beta,\gamma)$ with $\alpha+\beta+\gamma=15$
D) $B=(\alpha,\beta,\gamma)$ with $\alpha+\beta+\gamma=22$
Step-by-Step Solution
Key Concept: Find point $D$ where bisector meets $AC$ (divides $AC$ in ratio $BA:BC$). Then $B$ is determined by the condition that $BD$ is along the bisector direction and $|BA|/|BC|=|AD|/|DC|$.
Answer: A, D.
Correct Answer: AD