Vector Algebra
Collinearity Condition
Grade 12
Question:
<p>If the position vectors of the points <span class="math">\(A\)</span>, <span class="math">\(B\)</span> and <span class="math">\(C\)</span> be <span class="math">\(\vec{i} + \vec{j}\)</span>, <span class="math">\(\vec{i} - \vec{j}\)</span> and <span class="math">\(a\vec{i} + b\vec{j} + c\vec{k}\)</span> respectively, then the points <span class="math">\(A\)</span>, <span class="math">\(B\)</span> and <span class="math">\(C\)</span> are collinear, if</p>
<p>(a) <span class="math">\(a = b = c = 1\)</span></p>
<p>(b) <span class="math">\(a = 1\)</span>, <span class="math">\(b\)</span> and <span class="math">\(c\)</span> are arbitrary scalars</p>
<p>(c) <span class="math">\(a = b = c = 0\)</span></p>
<p>(d) <span class="math">\(c = 0\)</span>, <span class="math">\(a = 1\)</span> and <span class="math">\(b\)</span> is arbitrary scalar</p>
Step-by-Step Solution
Key Concept: Three points are collinear if and only if one of the vectors connecting them is a scalar multiple of the other. Use component-wise comparison.
Step 1: Calculate \(\vec{AB}\) and \(\vec{BC}\) : \(\vec{AB} = (\hat{i} - \hat{j}) - (\hat{i} + \hat{j}) = -2\hat{j}\) \(\vec{BC} = (a\hat{i} + b\hat{j} + c\hat{k}) - (\hat{i} - \hat{j}) = (a-1)\hat{i} + (b+1)\hat{j} + c\hat{k}\) Step 2: For collinearity, \(\vec{AB} = k\vec{BC}\) for some scalar \(k\) : \(-2\hat{j} = k\{(a-1)\hat{i} + (b+1)\hat{j} + c\hat{k}\}\) Step 3: Comparing coefficients of \(\hat{i}\) , \(\hat{j}\) , and \(\hat{k}\) : Coefficient of \(\hat{i}\) : \(k(a-1) = 0 \Rightarrow a = 1\) Coefficient of \(\hat{k}\) : \(kc = 0 \Rightarrow c = 0\) Coefficient of \(\hat{j}\) : \(k(b+1) = -2\) (satisfied for any \(b\) ) Step 4: Hence, \(c = 0\) , \(a = 1\) and \(b\) is arbitrary scalar. ∴ Answer is (d).
Correct Answer: D