Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade 12
Question:
If $ABC$ be a triangle of sides $a, b, c$ with position vectors of $A, B, C$ as $\vec{a}, \vec{b}$ and $\vec{c}$ respectively, then the position vector of its incentre is:
$\frac{(\vec{a} + \vec{b} + \vec{c})}{3}$
$\frac{\vec{a} \times \vec{b} + \vec{b} + \vec{c} + \vec{c} \times \vec{a}}{a^2 + b^2 + c^2}$
$\frac{a\vec{a} + b\vec{b} + c\vec{c}}{a + b + c}$
None of these
Step-by-Step Solution
Key Concept: The incentre has position vector that is a weighted average of vertices with weights equal to opposite side lengths.
The incentre is the intersection of internal angle bisectors of angles $A$ and $B$: $\vec{r} = \vec{a} + t\left(\frac{\vec{c} - \vec{a}}{b} - \frac{\vec{b} - \vec{a}}{c}\right)$ and $\vec{r} = \vec{b} + \mu\left(\frac{\vec{a} - \vec{c}}{c} - \frac{\vec{c} - \vec{b}}{a}\right)$. Equating coefficients of $\vec{a}$, $\vec{b}$, and $\vec{c}$ yields the position vector of the incentre as $\vec{r} = \frac{a\vec{a} + b\vec{b} + c\vec{c}}{a + b + c}$.
Correct Answer: 3