Question

If \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) are the position vectors of the points A, B, C respectively and \( 5\mathbf{a} + 3\mathbf{b} - 8\mathbf{c} = 0 \), then find the ratio in which the point C divides the line segment AB.

Hint:

Use section formula for division ratio; equate coefficients of position vectors to solve.

Answer 1

Given \( 5\mathbf{a} + 3\mathbf{b} - 8\mathbf{c} = 0 \), rewrite: \[ 5\mathbf{a} + 3\mathbf{b} = 8\mathbf{c} \quad \Rightarrow \quad \mathbf{c} = \frac{5\mathbf{a} + 3\mathbf{b}}{8}. \] If C divides AB in the ratio \( k : 1 \), the position vector of C is: \[ \mathbf{c} = \frac{k \mathbf{b} + 1 \cdot \mathbf{a}}{k + 1}. \] Equate: \[ \frac{k \mathbf{b} + \mathbf{a}}{k + 1} = \frac{5\mathbf{a} + 3\mathbf{b}}{8}. \] Cross-multiply: \[ 8(k \mathbf{b} + \mathbf{a}) = (k + 1)(5\mathbf{a} + 3\mathbf{b}). \] \[ 8k \mathbf{b} + 8\mathbf{a} = 5k \mathbf{a} + 5\mathbf{a} + 3k \mathbf{b} + 3\mathbf{b}. \] Equate coefficients of \( \mathbf{a} \) and \( \mathbf{b} \): - For \( \mathbf{a} \): \( 8 = 5k + 5 \Rightarrow 5k = 3 \Rightarrow k = \frac{3}{5} \).
- For \( \mathbf{b} \): \( 8k = 3k + 3 \Rightarrow 5k = 3 \Rightarrow k = \frac{3}{5} \).
Ratio: \( k : 1 = \frac{3}{5} : 1 = 3 : 5 \).