JEE Main 2025 Jan 28 Shift 2, Mathematics Q2: Applications of Vectors

JEE Main2025Jan 28, Shift 2Mathematics

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

A
$1$
B
$\frac{9}{2}$
C
$0$
D
$2$

Solution

Both $O A$ and $O B$ have equal magnitude $3 + 1 = 2$ , so the internal angle bisector is along the sum $O A + O B = (3 + 1) (\hat{i} + \hat{j})$ , which is the line $y = x$ , or equivalently $x - y = 0$ .

Distance from $C = (a, 1 - a)$ to this line:

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

$\Rightarrow ∣2 a - 1∣ = 9 \Rightarrow a = 5$ or $a = - 4$ .

$∴$ Sum $= 5 + (- 4) = 1$ .

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JEE Main 2025 Jan 28 Shift 2, Mathematics Q2: Applications of Vectors

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