<p>If x, y are two non-zero and non-collinear vectors satisfying \[[(a-2)\alpha^2 + (b-3)\alpha + c]\mathbf{x} + [(a-2)\beta^2 + (b-3)\beta + c]\mathbf{y} + [(a-2)\gamma^2 + (b-3)\gamma + c](\mathbf{x} \times \mathbf{y}) = 0\] where \(\alpha, \beta, \gamma\) are three distinct real numbers, then find the value of \(a^2 + b^2 + c^2 - 4\).</p>
Step-by-Step Solution
Key Concept: Use linear independence of vectors x, y, and x × y to establish that each coefficient must vanish. A quadratic cannot have three distinct roots unless it is identically zero.
Step 1: Since x, y are non-zero and non-collinear vectors, the vectors x , y , and x × y are linearly independent. Step 2: For the linear combination to equal zero with linearly independent vectors, each coefficient must be zero: \[(a-2)\alpha^2 + (b-3)\alpha + c = 0\] \[(a-2)\beta^2 + (b-3)\beta + c = 0\] \[(a-2)\gamma^2 + (b-3)\gamma + c = 0\] Step 3: This means \(\alpha, \beta, \gamma\) are the three roots of the quadratic equation \((a-2)t^2 + (b-3)t + c = 0\). Step 4: But a quadratic can have at most 2 roots. For three distinct roots to exist, the coefficient of \(t^2\) must be zero: \[a - 2 = 0 \Rightarrow a = 2\] \[b - 3 = 0 \Rightarrow b = 3\] \[c = 0\] Step 5: Therefore, \(a^2 + b^2 + c^2 - 4 = 4 + 9 + 0 - 4 = 9\) and checking: \(a^2 + b^2 + c^2 = 4 + 9 + 0 = 13\) ∴ Answer is 13
Correct Answer: 13