3D Geometry
Planes and determinants
Grade 12
Question:
<p>Consider the determinant \(\Delta = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = \frac{1}{2}(a + b + c)[(a-b)^2 + (b-c)^2 + (c-a)^2]\).</p><p>Which of the following is correct?</p>
<p>(A) If \(a + b + c \neq 0\) and \(a^2 + b^2 + c^2 = ab + bc + ca\), then \(\Delta = 0\) and \(a = b = c \neq 0\). The equations represent identical planes.</p>
<p>(B) If \(a + b + c = 0\) and \(a^2 + b^2 + c^2 \neq ab + bc + ca\), then \(\Delta = 0\). The equations have infinitely many solutions.</p>
<p>(C) If \(a + b + c \neq 0\) and \(a^2 + b^2 + c^2 \neq ab + bc + ca\), then \(\Delta \neq 0\). The equations represent planes meeting at only one point.</p>
<p>(D) If \(a + b + c = 0\) and \(a^2 + b^2 + c^2 = ab + bc + ca\), then \(\Delta = 0\).</p>
Step-by-Step Solution
Key Concept: The determinant's value determines whether the system of plane equations has a unique solution, infinitely many solutions, or no solution.
Analysis: From the given determinant formula: \(\Delta = \frac{1}{2}(a + b + c)[(a-b)^2 + (b-c)^2 + (c-a)^2]\) (A) If \(a^2 + b^2 + c^2 = ab + bc + ca\), then \((a-b)^2 + (b-c)^2 + (c-a)^2 = 0\), so \(a = b = c\). If \(a + b + c \neq 0\), then \(\Delta = 0\). ✓ (B) If \(a + b + c = 0\), then \(\Delta = 0\) regardless of the second condition. ✓ (C) If \(a + b + c \neq 0\) and \(a^2 + b^2 + c^2 \neq ab + bc + ca\), then \((a-b)^2 + (b-c)^2 + (c-a)^2 \neq 0\), so \(\Delta \neq 0\). The system has a unique solution (planes meet at one point). ✓ (D) If both conditions hold, \(\Delta = \frac{1}{2}(0)[0] = 0\). ✓ Option C correctly characterizes the case where \(\Delta \neq 0\).
Correct Answer: C