Vector Algebra
Coplanar Vectors
Grade 12
Question:
<p>The points <span>A(2-x, 2, 2)</span>, <span>B(2, 2-y, 2)</span>, <span>C(2, 2, 2-z)</span> and <span>D(1, 1, 1)</span> are coplanar. Find the locus of <span>P(x, y, z)</span>.</p>
<p>(a) <span>\(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1\)</span></p>
<p>(b) <span>\(x + y + z = 1\)</span></p>
<p>(c) <span>\(\frac{1}{1-x} + \frac{1}{1-y} + \frac{1}{1-z} = 1\)</span></p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Four points are coplanar if and only if the scalar triple product of vectors from one point to the other three points equals zero.
Step 1: Find vectors from D to A, B, C: \(\overrightarrow{AB} = x\hat{i} - y\hat{j}\) \(\overrightarrow{AC} = x\hat{i} - z\hat{k}\) \(\overrightarrow{AD} = (x-1)\hat{i} - \hat{j} - \hat{k}\) Step 2: For coplanar vectors, scalar triple product equals zero: \(\begin{vmatrix} x & -y & 0 \\ x & 0 & -z \\ x-1 & -1 & -1 \end{vmatrix} = 0\) Step 3: Expanding the determinant: \(1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1\) ∴ Locus is \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1\) , Answer is (a).
Correct Answer: A