Distance From C to Angle Bisector of OA, OB is 9/√2, Find a

JEE Main 2025 — 28 January, Evening Shift. Previous Year Question.

Problem

Let $A$, $B$, $C$ be three points in the $xy$-plane, whose position vectors with respect to the origin $O$ are given by $\vec{OA} = \sqrt{3}\hat{i}+\hat{j}$, $\vec{OB} = \hat{i}+\sqrt{3}\hat{j}$, and $\vec{OC} = a\hat{i}+(1-a)\hat{j}$. If the distance of point $C$ from the line bisecting the angle between $\vec{OA}$ and $\vec{OB}$ is $\dfrac{9}{\sqrt{2}}$, then the sum of all possible values of $a$ is:

Key insight. A problem dressed up in vector notation is often a plain coordinate-geometry problem in disguise. Once $A$, $B$, and $C$ are read simply as points $(\sqrt3, 1)$, $(1, \sqrt3)$, and $(a, 1-a)$, the “angle bisector of two vectors from the origin” becomes a line whose equation can be spotted directly from symmetry.

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Approach

Whenever a vector problem is set in the $xy$-plane with everything given in terms of $\hat i$ and $\hat j$, it’s worth asking whether it’s actually easier as coordinate geometry — and here it is. $A=(\sqrt3,1)$ and $B=(1,\sqrt3)$ are reflections of each other across the line $y=x$, which means that line is exactly the angle bisector of $\vec{OA}$ and $\vec{OB}$. From there, the problem reduces to the standard formula for the perpendicular distance from a point to a line.

Solution

Step 1 — Identify the angle bisector by symmetry

The point $A=(\sqrt3, 1)$ makes an angle of $30°$ with the $x$-axis, since $\tan 30° = \dfrac{1}{\sqrt3}$. Point $B=(1,\sqrt3)$ is its mirror image, making an angle of $60°$. The bisector of the angle between $\vec{OA}$ and $\vec{OB}$ therefore sits exactly halfway, at $45°$ — which is the line:

$$y = x \quad \text{i.e.} \quad x – y = 0$$

Step 2 — Apply the point-to-line distance formula

Point $C = (a, 1-a)$. The distance from $C$ to the line $x – y = 0$ is:

$$D = \frac{|a – (1-a)|}{\sqrt{1^2+(-1)^2}} = \frac{|2a-1|}{\sqrt{2}}$$

Step 3 — Set the distance equal to the given value

$$\frac{|2a-1|}{\sqrt{2}} = \frac{9}{\sqrt{2}} \implies |2a-1| = 9$$

Step 4 — Solve both cases of the absolute value

$$2a – 1 = 9 \implies a = 5 \qquad \text{or} \qquad 2a-1 = -9 \implies a = -4$$

Step 5 — Sum the possible values

$$5 + (-4) = 1$$

Answer

Common mistakes

• Trying to work entirely in vector notation — computing the angle bisector formula $\dfrac{\vec{OA}}{|\vec{OA}|}+\dfrac{\vec{OB}}{|\vec{OB}|}$ directly works too, but takes longer than simply noticing the $30°$–$60°$ symmetry between the two given points.
• Dropping the absolute value too early. The distance formula involves $|2a-1|$, and skipping straight to $2a-1=9$ without considering the negative case loses one of the two valid values of $a$.

Practise next

• If $A=(2,0)$ and $B=(0,2)$, find the equation of the bisector of the angle between $\vec{OA}$ and $\vec{OB}$ by the same symmetry argument, then check it against the standard bisector formula.