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

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Vector Algebra

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

Answer: (C)

Given

∣ a + b ∣ = 21

and

∣ a - b ∣ = 3

. Squaring and adding:

2 (∣ a ∣^{2} + ∣ b ∣^{2}) = 30 \Rightarrow ∣ a ∣^{2} + ∣ b ∣^{2} = 15

. Squaring and subtracting:

4 a \cdot b = 12 \Rightarrow a \cdot b = 3

. From

a \cdot (a - b) = 0

∣ a ∣^{2} = a \cdot b = 3

. Thus

∣ a ∣^{2} = 3

and

∣ b ∣^{2} = 12

. The area of

△ O P R = \frac{1}{2} ∣ a \times b ∣

. Since

∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2} = 3 \cdot 12 - 9 = 27

, we get

∣ a \times b ∣ = 33

. Therefore area

= \frac{3 3}{2}

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Single correct option