Vectors — Area of Triangle
PYP_JEE_ADV_2026_P2
Grade None
Question:
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
$\sqrt{3}$
$\dfrac{\sqrt{3}}{2}$
$\dfrac{3\sqrt{3}}{2}$
$\dfrac{3}{2}$
Step-by-Step Solution
Key Concept: The area of the triangle with two sides $\vec{OP}=\vec{a}$ and $\vec{OR}=\vec{a}+\vec{b}$ equals $\frac{1}{2}|\vec{a}\times\vec{b}|$, since $\vec{a}\times(\vec{a}+\vec{b})=\vec{a}\times\vec{b}$.
**Step 1: Set up equations from given conditions**
$|\vec{a}+\vec{b}|^2=|\vec{a}|^2+2\vec{a}\cdot\vec{b}+|\vec{b}|^2=21$ and $|\vec{a}-\vec{b}|^2=|\vec{a}|^2-2\vec{a}\cdot\vec{b}+|\vec{b}|^2=9$. Adding: $|\vec{a}|^2+|\vec{b}|^2=15$. Subtracting: $4\vec{a}\cdot\vec{b}=12 \Rightarrow \vec{a}\cdot\vec{b}=3$.
**Step 2: Use perpendicularity condition**
$\vec{a}\perp(\vec{a}-\vec{b}) \Rightarrow \vec{a}\cdot(\vec{a}-\vec{b})=0 \Rightarrow |\vec{a}|^2=\vec{a}\cdot\vec{b}=3$. So $|\vec{b}|^2=12$.
**Step 3: Compute area of triangle OPR**
$O$ is origin, $P=\vec{a}$, $R=\vec{a}+\vec{b}$. Area $=\tfrac{1}{2}|\vec{OP}\times\vec{OR}|=\tfrac{1}{2}|\vec{a}\times(\vec{a}+\vec{b})|=\tfrac{1}{2}|\vec{a}\times\vec{b}|$. $|\vec{a}\times\vec{b}|^2=|\vec{a}|^2|\vec{b}|^2-(\vec{a}\cdot\vec{b})^2=3\times12-9=27$. Area $=\tfrac{\sqrt{27}}{2}=\dfrac{3\sqrt{3}}{2}$.
Correct Answer: C