Question

Let \( ABC \) be a triangle and \( \vec{a}, \vec{b}, \vec{c} \) be the position vectors of \( A, B, C \) respectively. Let \( D \) divide \( BC \) in the ratio \( 3:1 \) internally and \( E \) divide \( AD \) in the ratio \( 4:1 \) internally. Let \( BE \) meet \( AC \) in \( F \). If \( E \) divides \( BF \) in the ratio \( 3:2 \) internally then the position vector of \( F \) is:

• A. \( \frac{\vec{a} + \vec{b} + \vec{c}}{3} \)
• B. \( \frac{\vec{a} - 2\vec{b} + 3\vec{c}}{2} \)
• C. \( \frac{\vec{a} + 2\vec{b} + 3\vec{c}}{2} \)
• D. \( \frac{\vec{a} - \vec{b} + 3\vec{c}}{3} \)

Hint:

Remember to use the section formula for internal division to find the position vectors in geometric vector problems.

Answer 1

The Correct Option is D

Given vectors: \[ \vec{OA} = \vec{a}, \quad \vec{OB} = \vec{b}, \quad \vec{OC} = \vec{c} \] Now, the Position Vector (PV) of point \( D \) is: \[ \vec{OD} = \frac{1}{1+3}(\vec{OB} + 3\vec{OC}) = \frac{1}{4}(\vec{b} + 3\vec{c}) \] The Position Vector of point \( E \) is: \[ \vec{OE} = \frac{4\vec{OD} + \vec{OA}}{4+1} = \frac{1}{5}(4(\frac{1}{4}(\vec{b} + 3\vec{c})) + \vec{a}) = \frac{1}{5}(\vec{a} + \vec{b} + 3\vec{c}) \] Now, the Position Vector of point \( F \) is calculated by: \[ \vec{OF} = \frac{20\vec{OB} + 30\vec{OC}}{2+3} = \frac{50\vec{OE} - 20\vec{OB}}{3} = \frac{50(\frac{1}{5}(\vec{a} + \vec{b} + 3\vec{c})) - 20\vec{b}}{3} \] \[ \vec{OF} = \frac{10(\vec{a} + \vec{b} + 3\vec{c}) - 20\vec{b}}{3} = \frac{10\vec{a} - 10\vec{b} + 30\vec{c}}{3} = \frac{10(\vec{a} - \vec{b} + 3\vec{c})}{3} \] \[ \vec{OF} = \frac{\vec{a} - \vec{b} + 3\vec{c}}{3} \] Hence, the Position Vector of \( F \) is: \[ \vec{OF} = \frac{1}{3} (\vec{a} - \vec{b} + 3\vec{c}) \]

Answer 2

Step 1: Understanding the Problem
We are given a triangle \( ABC \), and the position vectors of points \( A, B, C \) are represented by \( \vec{a}, \vec{b}, \vec{c} \) respectively. The following information is provided:
- Point \( D \) divides the line segment \( BC \) in the ratio \( 3:1 \) internally.
- Point \( E \) divides the line segment \( AD \) in the ratio \( 4:1 \) internally.
- Point \( BE \) intersects \( AC \) at point \( F \).
- Point \( E \) divides \( BF \) in the ratio \( 3:2 \) internally.
We are tasked with finding the position vector of point \( F \). 
Step 2: Find the Position Vector of Point \( D \)
Point \( D \) divides \( BC \) in the ratio \( 3:1 \) internally. The position vector of point \( D \), denoted as \( \vec{d} \), is calculated using the section formula: \[ \vec{d} = \frac{3\vec{c} + 1\vec{b}}{3 + 1} = \frac{3\vec{c} + \vec{b}}{4} \] 
Step 3: Find the Position Vector of Point \( E \)
Point \( E \) divides \( AD \) in the ratio \( 4:1 \) internally. The position vector of point \( E \), denoted as \( \vec{e} \), is calculated as: \[ \vec{e} = \frac{4\vec{d} + 1\vec{a}}{4 + 1} = \frac{4\left(\frac{3\vec{c} + \vec{b}}{4}\right) + \vec{a}}{5} \] Simplifying this expression: \[ \vec{e} = \frac{4 \times \frac{3\vec{c} + \vec{b}}{4} + \vec{a}}{5} = \frac{3\vec{c} + \vec{b} + \vec{a}}{5} \] 
Step 4: Equation of Line \( BE \)
The position vector of any point on the line \( BE \) can be expressed as a linear combination of \( \vec{b} \) and \( \vec{e} \). Let \( F \) divide \( BE \) in the ratio \( t:1 \), where \( t \) is a scalar. Then, the position vector of point \( F \), denoted as \( \vec{f} \), is given by: \[ \vec{f} = \frac{t \vec{e} + \vec{b}}{t + 1} \] Substituting \( \vec{e} = \frac{3\vec{c} + \vec{b} + \vec{a}}{5} \) into this equation: \[ \vec{f} = \frac{t \left( \frac{3\vec{c} + \vec{b} + \vec{a}}{5} \right) + \vec{b}}{t + 1} \] Simplifying: \[ \vec{f} = \frac{\frac{t(3\vec{c} + \vec{b} + \vec{a})}{5} + \vec{b}}{t + 1} \] 
Step 5: Using the Ratio \( E \) Divides \( BF \) in \( 3:2 \)
We are given that point \( E \) divides \( BF \) in the ratio \( 3:2 \) internally. This implies that the position vector of \( E \) divides the line segment \( BF \) in the ratio \( 3:2 \), so we can write: \[ \vec{e} = \frac{3\vec{f} + 2\vec{b}}{3 + 2} \] Substitute \( \vec{e} = \frac{3\vec{c} + \vec{b} + \vec{a}}{5} \) into this equation: \[ \frac{3\vec{c} + \vec{b} + \vec{a}}{5} = \frac{3\vec{f} + 2\vec{b}}{5} \] Simplifying: \[ 3\vec{c} + \vec{b} + \vec{a} = 3\vec{f} + 2\vec{b} \] Rearranging: \[ 3\vec{f} = 3\vec{c} + \vec{a} - \vec{b} \] \[ \vec{f} = \frac{\vec{a} - \vec{b} + 3\vec{c}}{3} \]