Vector Algebra
Vector Equations
Grade 12
Question:
<p>Let \(\overrightarrow{EB} = p\overrightarrow{AB}\) and \(\overrightarrow{CE} = q\overrightarrow{CD}\) in the given figure where \(0 < p\) and \(q \leq 1\). Given \(\overrightarrow{EB} + \overrightarrow{BC} + \overrightarrow{CE} = \vec{0}\) and using the vector equation \(pm(2\hat{i} - 6\hat{j} + 2\hat{k}) + (\hat{i} - 2\hat{j}) + qn(-6\hat{i} + 15\hat{j} - 3\hat{k}) = \vec{0}\), which of the following are the correct constraints on \(m\) and \(n\)?</p>
<p>\(m \geq \frac{1}{2}\) and \(n \geq \frac{1}{3}\)</p>
<p>\(m \leq \frac{1}{2}\) and \(n \leq \frac{1}{3}\)</p>
<p>\(m \geq 1\) and \(n \geq 1\)</p>
<p>\(m \leq 1\) and \(n \leq 1\)</p>
Step-by-Step Solution
Key Concept: Use the constraint $\overrightarrow{EB} + \overrightarrow{BC} + \overrightarrow{CE} = \vec{0}$ to express vectors in terms of position vectors, then apply linear independence of basis vectors (or compare coefficients) to find relationships between parameters p, q, m, and n.
Step 1: From the given vector equation, equate coefficients of $\hat{i}$, $\hat{j}$, and $\hat{k}$ to zero: Coefficient of $\hat{i}$: $2pm + 1 - 6qn = 0$ ... (1) Coefficient of $\hat{j}$: $-6pm - 2 + 15qn = 0$ ... (2) Coefficient of $\hat{k}$: $2pm - 3qn = 0$ ... (3) Step 2: From equation (3): $2pm = 3qn$ → $pm = \frac{3qn}{2}$ Step 3: Substitute into equation (1): $2 \cdot \frac{3qn}{2} + 1 - 6qn = 0$ $3qn + 1 - 6qn = 0$ $1 - 3qn = 0$ $qn = \frac{1}{3}$ Step 4: From $pm = \frac{3qn}{2} = \frac{3}{2} \cdot \frac{1}{3} = \frac{1}{2}$ Step 5: Verify with equation (2): $-6 \cdot \frac{1}{2} - 2 + 15 \cdot \frac{1}{3} = -3 - 2 + 5 = 0$ ✓ ∴ Answer: The constraints are $\boxed{pm = \frac{1}{2} \text{ and } qn = \frac{1}{3}}$ (or equivalent: $2pm = 1$ and $3qn = 1$)
Correct Answer: A