JEE Main 2026 Jan 22 Shift 1, Mathematics Q1: Scalar (or Dot) Product of Two Vectors

In parallelogram ABCD, $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}=3\hat{i}+6\hat{j}+(\lambda -5)\hat{k}$.

The projection of $\vec{v}$ on $\overrightarrow{AC}$ has length $\frac{\vec{v}\cdot \overrightarrow{AC}}{|\overrightarrow{AC}|}=1$, so $\vec{v}\cdot \overrightarrow{AC}=|\overrightarrow{AC}|$.

$\vec{v}\cdot \overrightarrow{AC}=3+6+(\lambda -5)=\lambda +4$ and $|\overrightarrow{AC}{|}^2=9+36+(\lambda -5{)}^2$.

Squaring: $(\lambda +4{)}^2=45+(\lambda -5{)}^2$, giving ${\lambda }^2+8\lambda +16={\lambda }^2-10\lambda +70$, so $18\lambda =54$ and $\lambda =3$.

Quadratic becomes $9x^2-18x+5=0$ with roots $\frac{1}{3},\frac{5}{3}$. Since $\alpha >\beta$: $\alpha =\frac{5}{3},\beta =\frac{1}{3}$.

$2\alpha -\beta =\frac{10}{3}-\frac{1}{3}=3$.