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: \[ \vec{OF} = \frac{1}{3} (\vec{a} - \vec{b} + 3\vec{c}) \]