Vector Algebra
Vector Triple Product and Coplanar Vectors
Grade 12
Question:
<p>Let <strong>a</strong> = \(2\mathbf{i} + \mathbf{j} + \mathbf{k}\), <strong>b</strong> = \(\mathbf{i} + 2\mathbf{j} - \mathbf{k}\) and <strong>c</strong> is a unit vector coplanar to them. If <strong>c</strong> is perpendicular to <strong>a</strong>, then <strong>c</strong> is equal to</p>
<p>(a) \(\frac{1}{2}(-\mathbf{j} + \mathbf{k})\)</p>
<p>(b) \(-\frac{1}{3}(\mathbf{i} + \mathbf{j} + \mathbf{k})\)</p>
<p>(c) \(\frac{1}{5}(\mathbf{i} - 2\mathbf{j})\)</p>
<p>(d) \(\frac{1}{3}(\mathbf{i} + \mathbf{j} + \mathbf{k})\)</p>
Step-by-Step Solution
Key Concept: Use the vector triple product formula and the coplanarity condition to find the vector perpendicular to a and coplanar with a and b, then normalize to unit length.
Step 1: Since c is coplanar to a and b , and perpendicular to a , we use: \(\mathbf{a} \times (\mathbf{a} \times \mathbf{b}) = (\mathbf{a} \cdot \mathbf{b})\mathbf{a} - (\mathbf{a} \cdot \mathbf{a})\mathbf{b}\) Step 2: Calculate \(\mathbf{a} \cdot \mathbf{b} = 2(1) + 1(2) + 1(-1) = 3\) \(\mathbf{a} \cdot \mathbf{a} = 4 + 1 + 1 = 6\) Step 3: \(\mathbf{a} \times (\mathbf{a} \times \mathbf{b}) = 3(2\mathbf{i} + \mathbf{j} + \mathbf{k}) - 6(\mathbf{i} + 2\mathbf{j} - \mathbf{k})\) \(= 6\mathbf{i} + 3\mathbf{j} + 3\mathbf{k} - 6\mathbf{i} - 12\mathbf{j} + 6\mathbf{k} = -9\mathbf{j} + 9\mathbf{k}\) Step 4: The unit vector is: \(\mathbf{c} = \pm\frac{1}{\sqrt{81 + 81}}(-9\mathbf{j} + 9\mathbf{k}) = \pm\frac{1}{2}(-\mathbf{j} + \mathbf{k})\) ∴ Answer is (a).
Correct Answer: A