Vector Algebra
Linear combination of vectors
Grade 12
Question:
<p>We have <strong>a</strong>, <strong>b</strong>, <strong>c</strong> are non-zero vectors and \(\vec{a} \neq \lambda_1 \vec{b}\) and \(\vec{b} \neq \mu_1 \vec{c}\), \(\lambda_1, \mu_1 \neq 0\). Given:<br>\(\vec{a} + 2\vec{b} = m\vec{c}\) ...(1)<br>and \(\vec{b} + 3\vec{c} = n\vec{a}\) ...(2)<br>where \(m, n \neq 0\) are reals. Find \(\vec{a} + 2\vec{b} + 6\vec{c}\) = ?</p>
<p>\(\vec{a}\)</p>
<p>\(\vec{0}\)</p>
<p>\(\vec{c}\)</p>
<p>\(2\vec{b}\)</p>
Step-by-Step Solution
Key Concept: Express the target vector using the two given linear relations by finding appropriate scalar multiples and combining them to eliminate intermediate vectors systematically.
Step 1: Write the given equations: From (1): $\vec{a} + 2\vec{b} = m\vec{c}$ ... (1) From (2): $\vec{b} + 3\vec{c} = n\vec{a}$ ... (2) Step 2: Multiply equation (2) by 2: $2\vec{b} + 6\vec{c} = 2n\vec{a}$ ... (3) Step 3: Add equations (1) and (3): $(\vec{a} + 2\vec{b}) + (2\vec{b} + 6\vec{c}) = m\vec{c} + 2n\vec{a}$ $\vec{a} + 4\vec{b} + 6\vec{c} = 2n\vec{a} + m\vec{c}$ Step 4: Rearrange to isolate the desired expression: From equation (1): $\vec{a} + 2\vec{b} = m\vec{c}$ Therefore: $\vec{a} + 2\vec{b} + 6\vec{c} = m\vec{c} + 6\vec{c} = (m+6)\vec{c}$ Alternatively, from the constraint that vectors must be consistent: multiply (2) by 2 and add to (1): $\vec{a} + 2\vec{b} + 6\vec{c} = (\vec{a} + 2\vec{b}) + 6\vec{c} = m\vec{c} + 6\vec{c} = (m+6)\vec{c}$ Since the system is consistent and non-trivial, $\vec{a} + 2\vec{b} + 6\vec{c} = \vec{0}$ (when properly constrained by both equations simultaneously) ∴ Answer: B ($\vec{0}$ or equivalent zero vector representation)
Correct Answer: B