Vector Algebra
Dot Product and Angles
Grade 12
Question:
<p>Let the cosine of angle between the vectors <strong>p</strong> and <strong>q</strong> be \(λ\) such that \(2\mathbf{p} + \mathbf{q} = \mathbf{i} + \mathbf{j}\) and \(\mathbf{p} + 2\mathbf{q} = \mathbf{i} - \mathbf{j}\), then \(λ\) is equal to</p>
<p>(a) \(\frac{5}{9}\)</p>
<p>(b) \(-\frac{4}{5}\)</p>
<p>(c) \(\frac{3}{9}\)</p>
<p>(d) \(\frac{7}{9}\)</p>
Step-by-Step Solution
Key Concept: Solve the system of vector equations to find <strong>p</strong> and <strong>q</strong>, then use the formula for cosine of angle between two vectors.
Step 1: From \(2\mathbf{p} + \mathbf{q} = \mathbf{i} + \mathbf{j}\) and \(\mathbf{p} + 2\mathbf{q} = \mathbf{i} - \mathbf{j}\) Step 2: Solving these equations: \(\mathbf{p} = \frac{1}{3}\mathbf{i} + \mathbf{j}\) and \(\mathbf{q} = \frac{1}{3}\mathbf{i} - \mathbf{j}\) Step 3: Calculate \(\mathbf{p} ⋅ \mathbf{q} = \frac{1}{9} - 1 = -\frac{8}{9}\) Step 4: Calculate \(|\mathbf{p}| = \sqrt{\frac{1}{9} + 1} = \frac{\sqrt{10}}{3}\) and \(|\mathbf{q}| = \sqrt{\frac{1}{9} + 1} = \frac{\sqrt{10}}{3}\) Step 5: \(\cos θ = \frac{\mathbf{p} ⋅ \mathbf{q}}{|\mathbf{p}||\mathbf{q}|} = \frac{-8/9}{10/9} = -\frac{4}{5}\) ∴ Answer is (b): \(λ = -\frac{4}{5}\)
Correct Answer: b