JEE Advanced 2026 — Paper 2 — MATH-I solution

JEE Advanced2026Paper 2MATH-I

Let $a, b$ be two vectors, and let P, Q and R be the points with position vectors $a, b$ and $a + b$ , respectively, with respect to the origin O. If $∣ a + b ∣ = 21$ , $∣ a - b ∣ = 3$ , and $a$ and $(a - b)$ are perpendicular to each other, then the area of the triangle OPR is

A
$3$
B
$\frac{3}{2}$
C
$\frac{3 3}{2}$
D
$\frac{3}{2}$

Solution

From the two magnitudes, squaring and adding gives $2 (∣ a ∣^{2} + ∣ b ∣^{2}) = 21 + 9 = 30$ , so $∣ a ∣^{2} + ∣ b ∣^{2} = 15$ .

Squaring and subtracting yields $4 a \cdot b = 21 - 9 = 12$ , hence $a \cdot b = 3$ .

The perpendicularity condition $a \cdot (a - b) = 0$ gives $∣ a ∣^{2} = a \cdot b = 3$ , and then $∣ b ∣^{2} = 12$ .

Triangle OPR has sides $O P = a$ and $O R = a + b$ , so its area equals $\frac{1}{2} ∣ a \times (a + b) ∣ = \frac{1}{2} ∣ a \times b ∣$ .

Using $∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2} = 3 \cdot 12 - 9 = 27$ , we get $∣ a \times b ∣ = 33$ .

$∴$ Area of $△ O P R = \frac{3 3}{2}$ .

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