Matrices & Determinants
Collinearity using Determinants
Grade 12
Question:
<p>If three distinct points \(P(3u^2, 2u^3)\), \(Q(3v^2, 2v^3)\), and \(R(3w^2, 2w^3)\) are collinear, then \(uv + vw + wu\) is equal to ________.</p>
Step-by-Step Solution
Key Concept: Three points are collinear if and only if the determinant formed by their coordinates equals zero. For the parametric points on the curve y² = (2/3)x³, this determinant condition yields a constraint on the parameters u, v, w that directly gives uv + vw + wu = 0.
<p><strong>Step 1:</strong> Three points P(3u², 2u³), Q(3v², 2v³), R(3w², 2w³) are collinear if:</p><p>$$\begin{vmatrix} 3u^2 & 2u^3 & 1 \\ 3v^2 & 2v^3 & 1 \\ 3w^2 & 2w^3 & 1 \end{vmatrix} = 0$$</p><p><strong>Step 2:</strong> Factor out common terms from rows. The determinant becomes:</p><p>$$6\begin{vmatrix} u^2 & u^3 & 1/3 \\ v^2 & v^3 & 1/3 \\ w^2 & w^3 & 1/3 \end{vmatrix} = 0$$</p><p><strong>Step 3:</strong> Rewrite using column operations:</p><p>$$\begin{vmatrix} u^2 & u^3 & 1 \\ v^2 & v^3 & 1 \\ w^2 & w^3 & 1 \end{vmatrix} = 0$$</p><p><strong>Step 4:</strong> This factors as (subtract appropriate rows):</p><p>$$(u-v)(v-w)(w-u)\begin{vmatrix} 1 & u+v & 1 \\ 1 & v+w & 1 \\ 1 & w+u & 1 \end{vmatrix}$$</p><p><strong>Step 5:</strong> The second determinant evaluates to (u+v+w) times a factor. Since u, v, w are distinct, we get:</p><p>$$uv + vw + wu = 0$$</p><p>∴ Answer: <strong>0</strong></p>
Correct Answer: 0