Position Vectors 4p+q-3r, -5p+q+2r, 2p-q+2r: Find α+2β+5γ

JEE Main 2025 — 24 January, Evening Shift. Previous Year Question.

Problem

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 $\dfrac{\vec p+\vec q+\vec r}{4}$ and $\alpha\vec p+\beta\vec q+\gamma\vec r$ respectively, then $\alpha+2\beta+5\gamma$ is equal to:

Key insight. Finding the circumcenter of a triangle directly, from three vertices given in an unfamiliar vector basis, is a long computation. It’s far faster to find the centroid (a simple average of the three vertices) and then use the well-known Euler line fact that the centroid always divides the segment from orthocenter to circumcenter in a 2:1 ratio.

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Approach

Whenever a problem gives the orthocenter and asks for the circumcenter (or vice versa), the fastest route is almost always the Euler line: the centroid $G$, orthocenter $H$, and circumcenter $C$ of any triangle are collinear, with $G$ dividing $HC$ in the ratio $2:1$ (measuring from $H$ to $C$). Since the centroid is trivial to compute directly from the three given vertices, this turns an otherwise complicated geometric search into a single section-formula equation.

Solution

Step 1 — Compute the centroid

The centroid is the average of the three position vectors:

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

Adding the $\vec p$, $\vec q$, and $\vec r$ components separately: $4-5+2=1$ for $\vec p$, $1+1-1=1$ for $\vec q$, and $-3+2+2=1$ for $\vec r$:

$$G = \frac{\vec p+\vec q+\vec r}{3}$$

Step 2 — Apply the Euler line section formula

The centroid divides the segment from orthocenter $H$ to circumcenter $C$ in ratio $2:1$, so:

$$G = \frac{2\cdot C + 1\cdot H}{3}$$

Substituting $H = \dfrac{\vec p+\vec q+\vec r}{4}$ and $C = \alpha\vec p+\beta\vec q+\gamma\vec r$:

$$\frac{\vec p+\vec q+\vec r}{3} = \frac{2(\alpha\vec p+\beta\vec q+\gamma\vec r) + \dfrac{\vec p+\vec q+\vec r}{4}}{3}$$

Step 3 — Clear the denominator and simplify

Multiplying both sides by $3$:

$$\vec p+\vec q+\vec r = 2(\alpha\vec p+\beta\vec q+\gamma\vec r) + \frac{\vec p+\vec q+\vec r}{4}$$

Subtracting $\dfrac{\vec p+\vec q+\vec r}{4}$ from both sides:

$$\frac{3}{4}(\vec p+\vec q+\vec r) = 2(\alpha\vec p+\beta\vec q+\gamma\vec r)$$

Step 4 — Compare coefficients

Dividing both sides by $2$ and comparing the coefficient of $\vec p$ (and, by symmetry, $\vec q$ and $\vec r$):

$$\alpha = \beta = \gamma = \frac{3}{8}$$

Step 5 — Compute $\alpha+2\beta+5\gamma$

$$\alpha+2\beta+5\gamma = \frac{3}{8}+2\cdot\frac{3}{8}+5\cdot\frac{3}{8} = \frac{3+6+15}{8} = \frac{24}{8} = 3$$

Answer

Common mistakes

• Trying to compute the circumcenter directly from the three vertices by solving for a point equidistant from all three — this is a valid method but involves much heavier algebra than the Euler line shortcut.
• Getting the Euler line ratio backwards. The centroid divides the orthocenter-to-circumcenter segment $2:1$ (closer to the circumcenter), not $1:2$ — using the wrong ratio flips the final coefficients.

Practise next

• Given a triangle’s orthocenter and centroid directly, find its circumcenter using the same $2:1$ section-formula relationship.