Vectors
Vectors
Allen Star Batch
Grade 12
Question:
Let $ABC$ be a triangle and $\vec{a},\vec{b},\vec{c}$ be the position vectors of the point $A, B, C$ respectively. External bisectors of $\angle B$ and $\angle C$ meet at $P$ with the sides of the triangle as $a, b, c$, the position vector of $P$ becomes :
$\frac{(-b)\vec{b}+(-c)\vec{c}}{(b+c)}$
$\frac{a\vec{a}+(-b)\vec{b}+(-c)\vec{c}}{(a-b-c)}$
$\left(\frac{\vec{a}+\vec{b}+\vec{c}}{3}\right)(abc)$
$\frac{a\vec{a}+b\vec{b}+c\vec{c}}{(\vec{a}+\vec{b}+\vec{c})}$
Step-by-Step Solution
Key Concept: The external angle bisectors at vertices B and C of a triangle intersect at the excenter P opposite to vertex A. Using the section formula with signed ratios, P divides the configuration such that the position vector is weighted by side lengths with opposite signs: $\vec{p} = \frac{a\vec{a} - b\vec{b} - c\vec{c}}{a - b - c}$, where a, b, c are side lengths opposite to vertices A, B, C respectively.
Vectors along $AB$, $BC$, and $CA$ are $\vec{b} - \vec{a}$, $\vec{c} - \vec{b}$, and $\vec{a} - \vec{c}$ respectively. The angle bisectors at $B$ and $C$ are $\vec{r} = \vec{b} + \mu\left(\frac{\vec{b} - \vec{a}}{c} + \frac{\vec{c} - \vec{b}}{a}\right)$ and $\vec{r} = \vec{c} + t\left(\frac{\vec{c} - \vec{b}}{a} + \frac{\vec{a} - \vec{c}}{b}\right)$, giving the incenter position.
Correct Answer: 2