Vector Algebra
Position Vectors and Linear Combinations
Grade 12
Question:
<p><strong>Ex. 106:</strong> Find <strong>AN + BP + CM</strong> where points M, N and P are taken on sides AB, BC and CA respectively, such that <math>\frac{AM}{AB} = \frac{BN}{BC} = \frac{CP}{CA} = \alpha</math>, with A at the origin, B and C having position vectors <i>b</i> and <i>c</i> respectively.</p>
<p>(a) <math>3\alpha(\mathbf{b} + \mathbf{c})</math></p>
<p>(b) <math>\alpha(\mathbf{b} + \mathbf{c})</math></p>
<p>(c) <math>(1 - \alpha)(\mathbf{b} + \mathbf{c})</math></p>
<p>(d) <math>0</math></p>
Step-by-Step Solution
Key Concept: Use position vectors to express points M, N, P on the sides, then compute the vector sum using the given ratios.
Step 1: Since <math>\frac{AM}{AB} = \alpha</math>, the position vector of M is <math>\mathbf{m} = \alpha \mathbf{b}</math> Step 2: Since <math>\frac{BN}{BC} = \alpha</math>, we have <math>\frac{BN}{NC} = \frac{\alpha}{1-\alpha}</math> Step 3: Position vector of N is <math>\mathbf{n} = (1-\alpha)\mathbf{b} + \alpha\mathbf{c}</math> Step 4: By similar analysis for P on CA, position vector of P is <math>\mathbf{p} = (1-\alpha)\mathbf{c}</math> Step 5: Therefore <math>\overrightarrow{AN} + \overrightarrow{BP} + \overrightarrow{CM} = [(1-\alpha)\mathbf{b} + \alpha\mathbf{c}] + [(1-\alpha)\mathbf{c} - \mathbf{b}] + [\alpha\mathbf{b} - \mathbf{c}] = \mathbf{0}</math> ∴ Answer is D .
Correct Answer: D