Vector Algebra
Position Vectors
Grade 12
Question:
<p>If position vector of a point A is \(\vec{a} + 2\vec{b}\) and any point P(\(\vec{a}\)) divides AB in the ratio of 2:3, then position vector of B is</p>
<p>(a) \(2\vec{a} - \vec{b}\)</p>
<p>(b) \(\vec{b} - 2\vec{a}\)</p>
<p>(c) \(\vec{a} - 3\vec{b}\)</p>
<p>(d) \(\vec{b}\)</p>
Step-by-Step Solution
Key Concept: Use the section formula: if point P divides AB in ratio m:n, then position vector of P is (n·A + m·B)/(m+n). Here we know P's position vector and must solve for B.
Step 1: Write the section formula. If point P divides line segment AB in ratio 2:3 (internally), then:
$\vec{P} = \frac{3\vec{A} + 2\vec{B}}{2+3} = \frac{3\vec{A} + 2\vec{B}}{5}$
Step 2: Substitute known values. We are given:
- Position vector of A: $\vec{A} = \vec{a} + 2\vec{b}$
- Position vector of P: $\vec{P} = \vec{a}$
- Ratio AP:PB = 2:3
Step 3: Set up the equation.
$\vec{a} = \frac{3(\vec{a} + 2\vec{b}) + 2\vec{B}}{5}$
Step 4: Multiply both sides by 5.
$5\vec{a} = 3\vec{a} + 6\vec{b} + 2\vec{B}$
Step 5: Solve for position vector of B.
$2\vec{B} = 5\vec{a} - 3\vec{a} - 6\vec{b}$
$2\vec{B} = 2\vec{a} - 6\vec{b}$
$\vec{B} = \vec{a} - 3\vec{b}$
Step 6: Verify. Check using section formula:
$\vec{P} = \frac{3(\vec{a} + 2\vec{b}) + 2(\vec{a} - 3\vec{b})}{5} = \frac{3\vec{a} + 6\vec{b} + 2\vec{a} - 6\vec{b}}{5} = \frac{5\vec{a}}{5} = \vec{a}$ ✓
∴ Answer: A
Correct Answer: A