JEE Advanced 2026 Paper 2Question 1
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 $|\vec{a}+\vec{b}| = \sqrt{21}$, $|\vec{a}-\vec{b}| = 3$, and $\vec{a}$ and $(\vec{a}-\vec{b})$ are perpendicular to each other, then the area of the triangle $OPR$ is
Vector Algebra
Solution
$|\vec a+\vec b|^2=21$, $|\vec a-\vec b|^2=9$ give $\vec a\cdot\vec b=3$, $|\vec a|^2+|\vec b|^2=15$; $\vec a\cdot(\vec a-\vec b)=0$ gives $|\vec a|^2=3$, $|\vec b|^2=12$. $|\vec a\times\vec b|^2=36-9=27$. Area $OPR=\frac12|\vec a\times(\vec a+\vec b)|=\frac12\cdot3\sqrt3=\frac{3\sqrt3}2$.