JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Vector Algebra

Place $O$ at the origin with $|\vec{u}|=|\vec{v}|=r$. From the figure $\overrightarrow{OP}=\vec{u}$ lies along the positive $y$-axis, $\overrightarrow{OQ}$ along the positive $x$-axis, and $\overrightarrow{OR}=\vec{v}$ bisects the right angle at $45^{\circ }$. Thus $\vec{u}=(0,r)$, $\vec{v}=\left(\frac{r}{\sqrt{2}},\frac{r}{\sqrt{2}}\right)$ and $\overrightarrow{OQ}=(r,0)$. Writing $\overrightarrow{OQ}=\alpha \vec{u}+\beta \vec{v}=\left(\frac{\beta r}{\sqrt{2}},\alpha r+\frac{\beta r}{\sqrt{2}}\right)=(r,0)$, the first component gives $\beta =\sqrt{2}$ and the second gives $\alpha +\frac{\beta }{\sqrt{2}}=0$, so $\alpha =-1$. Hence the roots are $\alpha =-1$ and ${\beta }^2=2$, and the quadratic with these roots is $(x+1)(x-2)=x^2-x-2=0$.