JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Vector Algebra

The three cross-product relations force $\vec{a},\vec{b},\vec{c}$ to be mutually perpendicular, and comparing magnitudes gives $|\vec{a}||\vec{b}|=4|\vec{c}|$, $|\vec{b}||\vec{c}|=9|\vec{a}|$ and $|\vec{c}||\vec{a}|=\alpha |\vec{b}|$. Writing $|\vec{a}|=p$, $|\vec{b}|=q$, $|\vec{c}|=r$, the first two give $pq=4r$ and $qr=9p$; eliminating $r$ yields $q^2=36$, so $q=6$, and then $r=\frac{3p}{2}$. The magnitude sum $p+6+\frac{3p}{2}=36$ gives $\frac{5p}{2}=30$, so $p=12$ and $r=18$. Finally $\alpha =\frac{rp}{q}=\frac{18\times 12}{6}=36$. The printed value $\frac{1}{36}$ is a misprint, as it forces negative lengths; the consistent reading is $36$.