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 orthocenter and the circumcenter of the triangle are $\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{4}$ and $\alpha \overrightarrow{\mathrm{p}}+\beta \overrightarrow{\mathrm{q}}+\gamma \overrightarrow{\mathrm{r}}$ respectively, then $\alpha+2 \beta+5 \gamma$ is equal to:

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Vector Algebra

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

\frac{p + q + r}{4}

and

α p + β q + γ r

respectively, then

α + 2 β + 5 γ

is equal to:

Answer: (A)

The centroid

G

of the triangle with vertices

4 p + q - 3 r, - 5 p + q + 2 r, 2 p - q + 2 r

is their average:

G = \frac{( 4 - 5 + 2 ) p + ( 1 + 1 - 1 ) q + ( - 3 + 2 + 2 ) r}{3} = \frac{p + q + r}{3} .

For any triangle, the orthocenter

H

, centroid

G

, and circumcenter

C

are collinear on the Euler line with

G

dividing

H C

in the ratio

2 : 1

(from

H

), which gives the relation

H + 2 C = 3 G .

Here

H = \frac{p + q + r}{4}

and

C = α p + β q + γ r

. Substituting:

\frac{p + q + r}{4} + 2 (α p + β q + γ r) = 3 \cdot \frac{p + q + r}{3} = p + q + r .

2 (α p + β q + γ r) = (p + q + r) - \frac{p + q + r}{4} = \frac{3}{4} (p + q + r),

giving

α = β = γ = \frac{3}{8}

.

Therefore

α + 2 β + 5 γ = \frac{3}{8} (1 + 2 + 5) = \frac{24}{8} = 3.

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Single correct option