Vector Algebra
Circumcentre from orthocentre–centroid–circumcentre collinearity
nta_pyq_2025_apr
Grade 12
Question:
Let the position vectors of three vertices of a triangle be $4\vec{p}+\vec{q}-3\vec{r}$, $-5\vec{p}+\vec{q}+2\vec{r}$ and $2\vec{p}-\vec{q}+2\vec{r}$. If the position vectors of the orthocentre and the circumcentre of the triangle are $\dfrac{\vec{p}+\vec{q}+\vec{r}}{4}$ and $\alpha\vec{p}+\beta\vec{q}+\gamma\vec{r}$ respectively, then $\alpha+2\beta+5\gamma$ is equal to:
Step-by-Step Solution
Key Concept: Use the relation $\overrightarrow{OG}=\tfrac{1}{3}(\overrightarrow{OH}+2\overrightarrow{OC})$ (centroid $G$ divides $HC$ in ratio $2:1$) where $G=\tfrac{\text{sum of vertices}}{3}$, to solve for the circumcentre.
Centroid $G=\dfrac{(4-5+2)\vec{p}+(1+1-1)\vec{q}+(-3+2+2)\vec{r}}{3}=\dfrac{\vec{p}+\vec{q}+\vec{r}}{3}$.
Euler line: $2(\alpha\vec{p}+\beta\vec{q}+\gamma\vec{r})+\dfrac{\vec{p}+\vec{q}+\vec{r}}{4}=3\cdot\dfrac{\vec{p}+\vec{q}+\vec{r}}{3}=\vec{p}+\vec{q}+\vec{r}$.
$8(\alpha\vec{p}+\beta\vec{q}+\gamma\vec{r})=3(\vec{p}+\vec{q}+\vec{r}) \Rightarrow \alpha=\beta=\gamma=\dfrac{3}{8}$.
$\alpha+2\beta+5\gamma=\dfrac{3}{8}+\dfrac{6}{8}+\dfrac{15}{8}=\dfrac{24}{8}=3$.
Correct Answer: 1