JEE Main 2026 Jan 28 Shift 1, Mathematics Q19: Scalar (or Dot) Product of Two Vectors

Expand: $|\vec{a}-\vec{b}{|}^2+|\vec{b}-\vec{c}{|}^2+|\vec{c}-\vec{a}{|}^2=2(|\vec{a}{|}^2+|\vec{b}{|}^2+|\vec{c}{|}^2)-2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})=6-2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})$.

Setting this equal to 9: $\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}=-\frac{3}{2}$.

Note $|\vec{a}+\vec{b}+\vec{c}{|}^2=3+2(-3/2)=0$, so $\vec{a}+\vec{b}+\vec{c}=\vec{0}$. Hence $\vec{b}+\vec{c}=-\vec{a}$.

$|2\vec{a}+k(\vec{b}+\vec{c})|=|2\vec{a}-k\vec{a}|=|2-k||\vec{a}|=|2-k|$.

Set $|2-k|=3\Rightarrow k=5$ or $k=-1$. Positive value: $k=5$.

Option (4).