Let the position vectors of three vertices of a triangle be 4 p + q -3 r ,-5 p + q +2 r and 2 p - q +2 r . If the position vectors of the orthocenter and the circumcenter of the triangle are p + q + r 4 and α p +β q +γ

Correct answer: A. 3 We know that aligned & O (orthocentre) p + q + r 4 & C( circum centre) α p +β q +γ r & C (centroid) = p + q + r 3 aligned by relation aligned & 2(α p +β q +γ r )+ p + q + r 4 =3( p + q + r 3 ) & 8(α p +β q +γ r )=3( p + q + r ) & 8 α=3,8 β=3,8 γ=3 & α=(3)/(8), β=(3)/(8), γ=(3)/(8) & α+2 β+3 γ & (3)/(8)+(6)/(8)+(15)/(8)=(24)/(8)=3 aligned

JEE Main 202524 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2025 · 24 January · Evening Shift · Mathematics · Vector Algebra

Let the position vectors of three vertices of a triangle be 4 p + q...

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 :

Options

Correct answer

A. 3

Step-by-step 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}

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