Vectors
PYP_JEE_ADV_2025_P2
Grade None
Question:
Consider the vectors
$$\vec{x} = \hat{i} + 2\hat{j} + 3\hat{k}, \quad \vec{y} = 2\hat{i} + 3\hat{j} + \hat{k}, \quad \text{and} \quad \vec{z} = 3\hat{i} + \hat{j} + 2\hat{k}.$$\n\nFor two distinct positive real numbers $\alpha$ and $\beta$, define\n\n$$\vec{X} = \alpha\vec{x} + \beta\vec{y} - \vec{z}, \quad \vec{Y} = \alpha\vec{y} + \beta\vec{z} - \vec{x}, \quad \text{and} \quad \vec{Z} = \alpha\vec{z} + \beta\vec{x} - \vec{y}.$$\n\nIf the vectors $\vec{X}$, $\vec{Y}$, and $\vec{Z}$ lie in a plane, then the value of $\alpha + \beta - 3$ is
Step-by-Step Solution
Key Concept: Circulant matrices and using the scalar triple product relation $[ ec{X} \ ec{Y} \ ec{Z}] = \det(A) [ ec{x} \ ec{y} \ ec{z}]$ to check coplanarity of linear combinations of vectors.
The vectors $\vec{X}, \vec{Y}, \vec{Z}$ lie in a plane if their scalar triple product is zero:
$$[\vec{X} \ \vec{Y} \ \vec{Z}] = 0$$
Expressing $\vec{X}, \vec{Y}, \vec{Z}$ as a matrix transformation on $\vec{x}, \vec{y}, \vec{z}$:
$$[\vec{X} \ \vec{Y} \ \vec{Z}] = \det\begin{pmatrix} \alpha & \beta & -1 \\ -1 & \alpha & \beta \\ \beta & -1 & \alpha \end{pmatrix} [\vec{x} \ \vec{y} \ \vec{z}]$$
Calculate the determinant of the original basis vectors:
$$\det\begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{pmatrix} = 1(6 - 1) - 2(4 - 3) + 3(2 - 9) = 5 - 2 - 21 = -18 \neq 0$$
Since $[\vec{x} \ \vec{y} \ \vec{z}] \neq 0$, the coplanarity determinant must be zero:
$$\det\begin{pmatrix} \alpha & \beta & -1 \\ -1 & \alpha & \beta \\ \beta & -1 & \alpha \end{pmatrix} = 0 \implies \alpha^3 + \beta^3 - 1 + 3\alpha\beta = 0$$
Factoring the equation:
$$(\alpha + \beta - 1)(\alpha^2 + \beta^2 + 1 - \alpha\beta + \alpha + \beta) = 0$$
Since $\alpha$ and $\beta$ are positive real numbers, the second quadratic term is strictly positive.
Therefore:
$$\alpha + \beta - 1 = 0 \implies \alpha + \beta = 1$$
Hence:
$$\alpha + \beta - 3 = 1 - 3 = -2$$
Correct Answer: