JEE Advanced 2025 Paper 2 · Q12 · Vector Algebra

Question

Consider the vectors $\vec{x} = \hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{y} = 2\hat{i} + 3\hat{j} + \hat{k}$, $\vec{z} = 3\hat{i} + \hat{j} + 2\hat{k}$. For two distinct positive real numbers $\alpha$ and $\beta$, define
$$\vec{X} = \alpha \vec{x} + \beta \vec{y} - \vec{z},$$
$$\vec{Y} = \alpha \vec{y} + \beta \vec{z} - \vec{x},$$
$$\vec{Z} = \alpha \vec{z} + \beta \vec{x} - \vec{y}.$$

If the vectors $\vec{X}, \vec{Y}, \vec{Z}$ lie in a plane, then the value of $\alpha + \beta - 3$ is __________.

SolveKar AI · Accepted Answer

$\vec{X}, \vec{Y}, \vec{Z}$ are coplanar iff the $3 	imes 3$ matrix of their coefficients (in basis $\vec{x}, \vec{y}, \vec{z}$) has determinant zero (since $[\vec{x}, \vec{y}, \vec{z}]$ scalar triple product is non-zero, vectors $\vec{x}, \vec{y}, \vec{z}$ are linearly independent). Coefficient matrix: $\begin{pmatrix} \alpha & \beta & -1 \ -1 & \alpha & \beta \ \beta & -1 & \alpha \end{pmatrix}$. Its determinant equals $\alpha^3 + \beta^3 - 1 + 3\alpha\beta = (\alpha + \beta - 1)(\alpha^2 + \beta^2 + 1 - \alpha\beta + \alpha + \beta)$. The second factor $= \frac{1}{2}((\alpha - \beta)^2 + (\alpha + 1)^2 + (\beta + 1)^2) > 0$ for positive $\alpha, \beta$. So $\alpha + \beta = 1$ $	o$ $\alpha + \beta - 3 = -2$.

JEE Advanced 2025 Paper 2 · Q12 · Vector Algebra

Solution