1. **Problem statement:** Given triangle OAB with position vectors of points A and B from O as $\mathbf{a}$ and $\mathbf{b}$ respectively. Given ratios: $OC : CA = \frac{2}{3}$ and $AD : DB = 1 : 3$. Points D and C lie on AB and OA respectively, and lines OD and BC meet at P. We need to show that vectors $\mathbf{OP}$ and $\mathbf{BP}$ are not perpendicular. 2. **Step 1: Find position vectors of points C and D.** - Since $OC : CA = \frac{2}{3}$, point C divides OA in ratio 2:3 starting from O to A. Position vector of C: $$\mathbf{c} = \frac{3\mathbf{0} + 2\mathbf{a}}{2+3} = \frac{2}{5}\mathbf{a}$$ - Since $AD : DB = 1 : 3$, point D divides AB in ratio 1:3 starting from A to B. Position vector of D: $$\mathbf{d} = \frac{3\mathbf{a} + 1\mathbf{b}}{1+3} = \frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}$$ 3. **Step 2: Parametric forms of lines OD and BC.** - Line OD: from O (0) to D ($\mathbf{d}$), any point is $$\mathbf{r} = t\mathbf{d} = t\left(\frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}\right)$$ - Line BC: from B ($\mathbf{b}$) to C ($\mathbf{c}$), any point is $$\mathbf{r} = \mathbf{b} + s(\mathbf{c} - \mathbf{b}) = \mathbf{b} + s\left(\frac{2}{5}\mathbf{a} - \mathbf{b}\right) = s\frac{2}{5}\mathbf{a} + (1 - s)\mathbf{b}$$ 4. **Step 3: Find intersection point P by equating the two parametric forms:** $$t\left(\frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}\right) = s\frac{2}{5}\mathbf{a} + (1 - s)\mathbf{b}$$ Equate coefficients of $\mathbf{a}$ and $\mathbf{b}$: - For $\mathbf{a}$: $$t \cdot \frac{3}{4} = s \cdot \frac{2}{5} \implies \frac{3}{4}t = \frac{2}{5}s$$ - For $\mathbf{b}$: $$t \cdot \frac{1}{4} = 1 - s \implies \frac{1}{4}t = 1 - s$$ 5. **Step 4: Solve the system for $t$ and $s$.** From the first equation: $$s = \frac{3}{4}t \cdot \frac{5}{2} = \frac{15}{8}t$$ Substitute into second equation: $$\frac{1}{4}t = 1 - \frac{15}{8}t$$ Multiply both sides by 8: $$2t = 8 - 15t$$ $$2t + 15t = 8$$ $$17t = 8 \implies t = \frac{8}{17}$$ Then, $$s = \frac{15}{8} \times \frac{8}{17} = \frac{15}{17}$$ 6. **Step 5: Find position vector of P.** Using line OD: $$\mathbf{OP} = t\mathbf{d} = \frac{8}{17} \left( \frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b} \right) = \frac{8}{17} \times \frac{3}{4} \mathbf{a} + \frac{8}{17} \times \frac{1}{4} \mathbf{b} = \frac{6}{17}\mathbf{a} + \frac{2}{17}\mathbf{b}$$ 7. **Step 6: Find vector $\mathbf{BP} = \mathbf{OP} - \mathbf{OB} = \mathbf{OP} - \mathbf{b}$.** $$\mathbf{BP} = \left( \frac{6}{17}\mathbf{a} + \frac{2}{17}\mathbf{b} \right) - \mathbf{b} = \frac{6}{17}\mathbf{a} - \frac{15}{17}\mathbf{b}$$ 8. **Step 7: Check if $\mathbf{OP}$ and $\mathbf{BP}$ are perpendicular by dot product:** $$\mathbf{OP} \cdot \mathbf{BP} = \left( \frac{6}{17}\mathbf{a} + \frac{2}{17}\mathbf{b} \right) \cdot \left( \frac{6}{17}\mathbf{a} - \frac{15}{17}\mathbf{b} \right)$$ $$= \frac{36}{289} \mathbf{a} \cdot \mathbf{a} - \frac{90}{289} \mathbf{a} \cdot \mathbf{b} + \frac{12}{289} \mathbf{b} \cdot \mathbf{a} - \frac{30}{289} \mathbf{b} \cdot \mathbf{b}$$ Since $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$, combine terms: $$= \frac{36}{289} |\mathbf{a}|^2 - \frac{78}{289} \mathbf{a} \cdot \mathbf{b} - \frac{30}{289} |\mathbf{b}|^2$$ This expression is generally not zero unless specific conditions on $\mathbf{a}$ and $\mathbf{b}$ hold. Therefore, $\mathbf{OP}$ and $\mathbf{BP}$ are not perpendicular in general. **Final answer:** Vectors $\mathbf{OP}$ and $\mathbf{BP}$ are not perpendicular.