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MathMedAdv 2025 • Paper 2

Coplanar Vectors Linear Combination

Consider the vectors

x⃗=i^+2j^+3k^,y⃗=2i^+3j^+k^, andz⃗=3i^+j^+2k^.\vec{x} = \hat{i} + 2\hat{j} + 3\hat{k}, \\ \quad \vec{y} = 2\hat{i} + 3\hat{j} + \hat{k}, \text{ and} \\ \quad \vec{z} = 3\hat{i} + \hat{j} + 2\hat{k}.

For two distinct positive real numbers α\alpha and β\beta, define

X⃗=αx⃗+βy⃗−z⃗,Y⃗=αy⃗+βz⃗−x⃗, andZ⃗=αz⃗+βx⃗−y⃗.\vec{X} = \alpha \vec{x} + \beta \vec{y} - \vec{z}, \\ \quad \vec{Y} = \alpha \vec{y} + \beta \vec{z} - \vec{x}, \text{ and} \\ \quad \vec{Z} = \alpha \vec{z} + \beta \vec{x} - \vec{y}.

If the vectors X⃗,Y⃗\vec{X}, \vec{Y}, and Z⃗\vec{Z} lie in a plane, then the value of α+β−3\alpha + \beta - 3 is __________.

Official Numerical Answer-2

Topics & Concepts

Vector AlgebraScalar Triple ProductVector Addition

Step-by-Step Solution

To find the value of α+β−3\alpha + \beta - 3, we analyze the condition for the vectors X⃗,Y⃗,\vec{X}, \vec{Y}, and Z⃗\vec{Z} to be coplanar.

Step 1: Check the linear independence of x⃗,y⃗,\vec{x}, \vec{y}, and z⃗\vec{z}

The scalar triple product of the vectors x⃗,y⃗,\vec{x}, \vec{y}, and z⃗\vec{z} is given by the determinant of their components: [x⃗,y⃗,z⃗]=∣123231312∣[\vec{x}, \vec{y}, \vec{z}] = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{vmatrix}

Evaluating this determinant: [x⃗,y⃗,z⃗]=1(6−1)−2(4−3)+3(2−9)=5−2−21=−18[\vec{x}, \vec{y}, \vec{z}] = 1(6 - 1) - 2(4 - 3) + 3(2 - 9) = 5 - 2 - 21 = -18

Since [x⃗,y⃗,z⃗]≠0[\vec{x}, \vec{y}, \vec{z}] \neq 0, the vectors x⃗,y⃗,\vec{x}, \vec{y}, and z⃗\vec{z} are non-coplanar and form a basis in three-dimensional space.

Step 2: Set up the coplanarity condition for X⃗,Y⃗,\vec{X}, \vec{Y}, and Z⃗\vec{Z}

The vectors X⃗,Y⃗,\vec{X}, \vec{Y}, and Z⃗\vec{Z} are expressed as linear combinations of x⃗,y⃗,\vec{x}, \vec{y}, and z⃗\vec{z}: X⃗=αx⃗+βy⃗−z⃗\vec{X} = \alpha \vec{x} + \beta \vec{y} - \vec{z} Y⃗=−x⃗+αy⃗+βz⃗\vec{Y} = -\vec{x} + \alpha \vec{y} + \beta \vec{z} Z⃗=βx⃗−y⃗+αz⃗\vec{Z} = \beta \vec{x} - \vec{y} + \alpha \vec{z}

The scalar triple product [X⃗,Y⃗,Z⃗][\vec{X}, \vec{Y}, \vec{Z}] is related to [x⃗,y⃗,z⃗][\vec{x}, \vec{y}, \vec{z}] by: [X⃗,Y⃗,Z⃗]=∣αβ−1−1αββ−1α∣[x⃗,y⃗,z⃗][\vec{X}, \vec{Y}, \vec{Z}] = \begin{vmatrix} \alpha & \beta & -1 \\ -1 & \alpha & \beta \\ \beta & -1 & \alpha \end{vmatrix} [\vec{x}, \vec{y}, \vec{z}]

For X⃗,Y⃗,\vec{X}, \vec{Y}, and Z⃗\vec{Z} to lie in a plane (i.e., to be coplanar), their scalar triple product must be zero: [X⃗,Y⃗,Z⃗]=0[\vec{X}, \vec{Y}, \vec{Z}] = 0

Since [x⃗,y⃗,z⃗]=−18≠0[\vec{x}, \vec{y}, \vec{z}] = -18 \neq 0, we must have: ∣αβ−1−1αββ−1α∣=0\begin{vmatrix} \alpha & \beta & -1 \\ -1 & \alpha & \beta \\ \beta & -1 & \alpha \end{vmatrix} = 0

Step 3: Evaluate the determinant

Expanding the determinant along the first row: α(α2+β)−β(−α−β2)−1(1−αβ)=0\alpha(\alpha^2 + \beta) - \beta(-\alpha - \beta^2) - 1(1 - \alpha\beta) = 0 α3+αβ+αβ+β3−1+αβ=0\alpha^3 + \alpha\beta + \alpha\beta + \beta^3 - 1 + \alpha\beta = 0 α3+β3−1+3αβ=0\alpha^3 + \beta^3 - 1 + 3\alpha\beta = 0

Using the algebraic identity a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) with a=α,b=β,a = \alpha, b = \beta, and c=−1c = -1: α3+β3+(−1)3−3αβ(−1)=(α+β−1)(α2+β2+1−αβ+α+β)=0\alpha^3 + \beta^3 + (-1)^3 - 3\alpha\beta(-1) = (\alpha + \beta - 1)\left(\alpha^2 + \beta^2 + 1 - \alpha\beta + \alpha + \beta\right) = 0

Rewriting the second factor: α2+β2+1−αβ+α+β=12[(α−β)2+(α+1)2+(β+1)2]\alpha^2 + \beta^2 + 1 - \alpha\beta + \alpha + \beta = \frac{1}{2} \left[ (\alpha - \beta)^2 + (\alpha + 1)^2 + (\beta + 1)^2 \right]

Since α\alpha and β\beta are distinct positive real numbers (α,β>0\alpha, \beta > 0 and α≠β\alpha \neq \beta), the terms (α−β)2(\alpha - \beta)^2, (α+1)2(\alpha + 1)^2, and (β+1)2(\beta + 1)^2 are strictly positive. Therefore: α2+β2+1−αβ+α+β>0\alpha^2 + \beta^2 + 1 - \alpha\beta + \alpha + \beta > 0

Thus, the only solution to the equation is: α+β−1=0  ⟹  α+β=1\alpha + \beta - 1 = 0 \implies \alpha + \beta = 1

Step 4: Calculate the required expression

We need to find the value of α+β−3\alpha + \beta - 3: α+β−3=1−3=−2\alpha + \beta - 3 = 1 - 3 = -2

Coplanar Vectors Linear Combination | Mathematics PYQ Solution - JEE Challenger