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← JEE Advanced 2026 · Paper 2
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JEE Advanced 2026 · Paper 2, Mathematics question 1

Let a⃗,b⃗\vec a,\vec b be two vectors, and let P,QP,Q and RR be the points with position vectors a⃗\vec a, b⃗\vec b and a⃗+b⃗\vec a+\vec b, respectively, with respect to the origin OO. If ∣a⃗+b⃗∣=21|\vec a+\vec b|=\sqrt{21}, ∣a⃗−b⃗∣=3|\vec a-\vec b|=3, and a⃗\vec a and (a⃗−b⃗)(\vec a-\vec b) are perpendicular to each other, then the area of the triangle OPROPR is

Choose one option
Answer and worked solution

Official answer: C

Dot products. From ∣a⃗+b⃗∣2=21|\vec a+\vec b|^2=21 and ∣a⃗−b⃗∣2=9|\vec a-\vec b|^2=9:
∣a⃗∣2+∣b⃗∣2=15,a⃗⋅b⃗=3.|\vec a|^2+|\vec b|^2=15,\qquad \vec a\cdot\vec b=3.
a⃗⊥(a⃗−b⃗)\vec a\perp(\vec a-\vec b) gives ∣a⃗∣2=a⃗⋅b⃗=3|\vec a|^2=\vec a\cdot\vec b=3, so ∣b⃗∣2=12|\vec b|^2=12.

Area. Area(OPR)=12∣a⃗×(a⃗+b⃗)∣=12∣a⃗×b⃗∣\text{Area}(OPR)=\frac12|\vec a\times(\vec a+\vec b)|=\frac12|\vec a\times\vec b|, and
∣a⃗×b⃗∣2=∣a⃗∣2∣b⃗∣2−(a⃗⋅b⃗)2=36−9=27.|\vec a\times\vec b|^2=|\vec a|^2|\vec b|^2-(\vec a\cdot\vec b)^2=36-9=27.
So the area is 332\dfrac{3\sqrt3}2.

Topics:Cross productDot product
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