JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Vector Algebra

$\vec{a}+5\vec{b}$ collinear with $\vec{c}$: $\vec{a}+5\vec{b}=\lambda \vec{c}$.
$\vec{b}+6\vec{c}$ collinear with $\vec{a}$: $\vec{b}+6\vec{c}=\mu \vec{a}$.

From the first, $\vec{c}=\frac{\vec{a}+5\vec{b}}{\lambda }$. Substituting into the second:


Since $\vec{a},\vec{b}$ are non-collinear (independent), both coefficients vanish: $\lambda =-30$ and $\mu =6/\lambda =-\frac{1}{5}$.

So $\vec{c}=\frac{\vec{a}+5\vec{b}}{-30}=-\frac{\vec{a}}{30}-\frac{\vec{b}}{6}$.

Using $\vec{a}+\alpha \vec{b}+\beta \vec{c}=0$:


So $\beta =30$ and $\alpha =\beta /6=5$.



Key confirmed: A.