JEE Main 2024 (Online) 9th April Evening ShiftQuestion 1
Let $\vec{a}=2\hat{i}+\alpha\hat{j}+\hat{k}$, $\vec{b}=-\hat{i}+\hat{k}$, $\vec{c}=\beta\hat{j}-\hat{k}$, where $\alpha$ and $\beta$ are integers and $\alpha\beta=-6$. Let the values of the ordered pair $(\alpha,\beta)$, for which the area of the parallelogram of diagonals $\vec{a}+\vec{b}$ and $\vec{b}+\vec{c}$ is $\frac{\sqrt{21}}{2}$, be $(\alpha_1,\beta_1)$ and $(\alpha_2,\beta_2)$. Then $\alpha_1^{2}+\beta_1^{2}-\alpha_2\beta_2$ is equal to
Vector Algebra
Solution
$\vec a+\vec b=\hat i+\alpha\hat j+2\hat k$, $\vec b+\vec c=-\hat i+\beta\hat j$. Area $=\tfrac12|(\vec a+\vec b)\times(\vec b+\vec c)|$, and the cross product is $(-2\beta,-2,\alpha\beta+\alpha)$. $4\beta^2+4+\alpha^2(\beta+1)^2=21$. With $\alpha\beta=-6$: $\alpha^2\beta^2+2\alpha^2\beta+\alpha^2=36-12\alpha+\alpha^2$, so $4\beta^2+\alpha^2-12\alpha+40=21$. Integer pairs with $\alpha\beta=-6$ give $(\alpha,\beta)=(-3,2)$ and $(3,-2)$ (checked). $\alpha_1^2+\beta_1^2-\alpha_2\beta_2=9+4+6=19$ (same for either order). Answer: (B).