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MathematicsNumerical+4Question 12 of 48

JEE Advanced 2025 · Paper 2, Mathematics question 12

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,\qquad \vec y = 2\hat i + 3\hat j + \hat k,\qquad\text{and}\qquad \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,\qquad \vec Y = \alpha\vec y + \beta\vec z - \vec x,\qquad\text{and}\qquad \vec Z = \alpha\vec z + \beta\vec x - \vec y.

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

Answer and worked solution

Official answer: -2

Coefficient determinant. x⃗,y⃗,z⃗\vec x,\vec y,\vec z are independent (their determinant is −18≠0-18\ne0). So X⃗,Y⃗,Z⃗\vec X,\vec Y,\vec Z are coplanar exactly when the coefficient determinant vanishes:
∣αβ−1−1αββ−1α∣=α3+β3−1+3αβ=0.\begin{vmatrix}\alpha&\beta&-1\\-1&\alpha&\beta\\\beta&-1&\alpha\end{vmatrix}=\alpha^3+\beta^3-1+3\alpha\beta=0.

Factorise. This is a3+b3+c3−3abca^3+b^3+c^3-3abc with c=−1c=-1:
(α+β−1)⋅12[(α−β)2+(α+1)2+(β+1)2]=0.(\alpha+\beta-1)\cdot\tfrac12\left[(\alpha-\beta)^2+(\alpha+1)^2+(\beta+1)^2\right]=0.
The bracket is positive for distinct positive α,β\alpha,\beta.

So α+β=1\alpha+\beta=1, and α+β−3=−2\alpha+\beta-3=-2.

Topics:Scalar and vector triple productsDeterminants and properties
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