JEE Advanced 2025 Paper 2, Mathematics Section 3 Q4: Scalar and Vector Triple Product

JEE Advanced2025Paper 2Mathematics Section 3

Consider the vectors

$x = \hat{i} + 2 \hat{j} + 3 \hat{k}, y = 2 \hat{i} + 3 \hat{j} + \hat{k}, z = 3 \hat{i} + \hat{j} + 2 \hat{k} .$

For two distinct positive real numbers $α$ and $β$ , define

$X = α x + β y - z, Y = α y + β z - x, Z = α z + β x - y .$

If the vectors $X, Y$ , and $Z$ lie in a plane, the value of $α + β - 3$ is ________.

Your answer

Solution

Step 1: Coplanarity condition.

$[X Y Z] = 0$ can be written as the product of a scalar coefficient determinant and the box product $[x y z]$ . Since $[x y z] \neq = 0$ , we need

$α β - 1 - 1 α β β - 1 α = 0.$

Step 2: Expand.

$α^{3} + β^{3} + (- 1)^{3} - 3 (α) (β) (- 1) = 0 \Rightarrow α^{3} + β^{3} - 1 + 3 α β = 0$ .

Step 3: Use the factorisation identity.

$a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)$ with $a = α$ , $b = β$ , $c = - 1$ .

Since $α, β > 0$ and $α \neq = β$ , the quadratic factor is positive, so $α + β - 1 = 0$ .

Step 4: Compute.

$α + β - 3 = 1 - 3 = - 2$ .

Concept behind this question

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