JEE Mathematics 2025 Question with Solution

Given: $\overrightarrow{OA} = \sqrt{3}\hat{i} + \hat{j}$, $\overrightarrow{OB} = \hat{i} + \sqrt{3}\hat{j}$, and $C = (a, 1-a)$.

Find: The sum of all possible real values of $a$.

The vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$ are symmetric about the line $y = x$, so the line bisecting the angle between them is

$$x - y = 0$$

The distance of a point $\left(x_1, y_1\right)$ from the line $x - y = 0$ is

$$\frac{|x_1 - y_1|}{\sqrt{2}}$$

For the point $C(a, 1-a)$,

$$\text{Distance} = \frac{|a - (1-a)|}{\sqrt{2}} = \frac{|2a - 1|}{\sqrt{2}}$$

Given that this distance is $\frac{9}{\sqrt{2}}$,

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

So,

$$|2a - 1| = 9$$

Hence,

$$2a - 1 = 9 \quad \text{or} \quad 2a - 1 = -9$$

Thus,

$$a = 5 \quad \text{or} \quad a = -4$$

Their sum is

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

Therefore, the correct option is A, and the required sum is $1$.

Given: The equation of the angle bisector is taken as $x - y = 0$.

Find: All possible values of $a$ and their sum.

From the solution working, the distance condition is applied to point $C(a,1-a)$ as

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

which gives

$$|2a-1| = 9$$

So,

$$2a-1 = 9 \quad \text{or} \quad 2a-1 = -9$$

Therefore,

$$a = 5 \quad \text{or} \quad a = -4$$

Hence, the sum is

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

Therefore, the required sum is $1$.

The solution also contains an inconsistent intermediate expression involving $|a(1-a)|$, but the later working and the geometric distance formula correctly use $|2a-1|$. Using the correct distance from $y=x$ leads to the final answer $1$.