JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Vector Algebra

From $\vec{a}\times \vec{b}=\vec{c}$, the vector $\vec{c}$ is perpendicular to $\vec{a}$ and $\vec{b}$; from $\vec{b}\times \vec{c}=\vec{a}$, the vector $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{c}$. The three are therefore mutually perpendicular, so $|\vec{c}|=|\vec{a}||\vec{b}|$ and $|\vec{a}|=|\vec{b}||\vec{c}|$. Eliminating $|\vec{c}|$ gives $|\vec{b}|=1$, and with $|\vec{a}|=2$ this gives $|\vec{c}|=2$. The projection of $\vec{a}$ on $\vec{b}\times \vec{c}=\vec{a}$ is $|\vec{a}|=2$; the triple product $[\vec{a}\vec{b}\vec{c}]=\vec{a}\cdot \vec{a}=4$ and equals $[\vec{c}\vec{a}\vec{b}]$ by cyclic symmetry, so their sum is $8$; and $(\vec{b}+\vec{c})\times (\vec{b}-\vec{c})=-2(\vec{b}\times \vec{c})=-2\vec{a}$, whose cross product with $\vec{a}$ is the zero vector. Mutual perpendicularity kills the cross terms in the remaining statement, so

The statement that fails is ${\left|3\vec{a}+\vec{b}-2\vec{c}\right|}^2=51$.