Vectors: MI 2024 Promo Q4

(i) [2]Method 1 Method 2

$AC:AB=1:3\Rightarrow AC:CB=1:2$.$AC:AB=1:3\Rightarrow3\overrightarrow{AC}=\overrightarrow{AB}$. $$\begin{aligned} \mathbf c&=\frac{(1)\mathbf b+(2)\mathbf a}{1+2}=\frac{2\mathbf a+\mathbf b}{3},\\ 3(\mathbf c-\mathbf a)&=\mathbf b-\mathbf a,\qquad \mathbf c=\frac{\mathbf b}{3}-\frac{\mathbf a}{3}+\mathbf a=\frac{2\mathbf a+\mathbf b}{3}. \end{aligned}$$

4(ii) [2]Method 1 $$\begin{aligned} \text{Area of triangle }OAC&=\frac13\text{ Area of triangle }OAB\\ &=\frac13\times\frac12|\overrightarrow{OA}\times\overrightarrow{OB}|\\ &=\frac16|\mathbf a\times\mathbf b|\quad\text{i.e. }k=\frac16. \end{aligned}$$

Method 2 $$\begin{aligned} \text{Area of triangle }OAC&=\frac12|\overrightarrow{OA}\times\overrightarrow{OC}|\\ &=\frac12\left|\mathbf a\times\left(\frac{2\mathbf a+\mathbf b}{3}\right)\right|\\ &=\frac16|2\mathbf a\times\mathbf a+\mathbf a\times\mathbf b|\\ &=\frac16|\mathbf a\times\mathbf b|\quad\text{i.e. }k=\frac16. \end{aligned}$$ 4(iii) [3] $$\begin{aligned} (2\mathbf a-\mathbf b)\cdot\mathbf c&=(2\mathbf a-\mathbf b)\cdot\left(\frac{2\mathbf a+\mathbf b}{3}\right)\\ &=\frac13(2\mathbf a-\mathbf b)\cdot(2\mathbf a+\mathbf b)\\ &=\frac13(4\mathbf a\cdot\mathbf a+2\mathbf a\cdot\mathbf b-2\mathbf b\cdot\mathbf a-\mathbf b\cdot\mathbf b)\\ &=\frac13(4|\mathbf a|^2-|\mathbf b|^2)=\frac13(4(1)^2-(1)^2)=1. \end{aligned}$$