Let the position vectors of three vertices of a triangle be $4 \vec{p}+\vec{q}-3 \vec{r},-5…

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 $2 \overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{q}}+2 \overrightarrow{\mathrm{r}}$. If the position vectors of the orthocenter and the circumcenter of the triangle are $\frac{\vec{p}+\vec{q}+\vec{r}}{4}$ and $\alpha \vec{p}+\beta \vec{q}+\gamma \vec{r}$ respectively, then $\alpha+2 \beta+5 \gamma$ is equal to :
  1. 3
  2. 4
  3. 1
  4. 6

Solution

We know that

$\begin{aligned}
& \mathrm{O} \text { (orthocentre) } \frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{4} \\ & \mathrm{C}(\text { circum centre) } \alpha \overrightarrow{\mathrm{p}}+\beta \overrightarrow{\mathrm{q}}+\gamma \overrightarrow{\mathrm{r}} \\ & \mathrm{C} \text { (centroid) }=\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{3}
\end{aligned}$
by relation
$\begin{aligned}
& \Rightarrow 2(\alpha \overrightarrow{\mathrm{p}}+\beta \overrightarrow{\mathrm{q}}+\gamma \overrightarrow{\mathrm{r}})+\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{4}=3\left(\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{3}\right) \\ & \Rightarrow 8(\alpha \overrightarrow{\mathrm{p}}+\beta \overrightarrow{\mathrm{q}}+\gamma \overrightarrow{\mathrm{r}})=3(\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}) \\ & \Rightarrow 8 \alpha=3,8 \beta=3,8 \gamma=3 \\ & \alpha=\frac{3}{8}, \beta=\frac{3}{8}, \gamma=\frac{3}{8} \\ & \therefore \alpha+2 \beta+3 \gamma \\ & \frac{3}{8}+\frac{6}{8}+\frac{15}{8}=\frac{24}{8}=3
\end{aligned}$ ,

Asked in: JEE Main 2025 (24 Jan Shift 2)

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