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Vectors: MI 2024 Promo Q4

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Millennia Institute2024 PromoPromo Q4●●○ Standard7 marks

What this question tests

Vectors (Products).

Question

. Referred to the origin $O$ , the points $A$ , $B$ and $C$ have position vectors $\mathbf{a}$ , $\mathbf{b}$ and $\mathbf{c}$ . $C$ lies on $AB$ such that $AC : AB = 1 : 3$ .

Find $\mathbf{c}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ . [2]
Show that the area of triangle $OAC$ is given by $k\,|\mathbf{a} \times \mathbf{b}|$ , where $k$ is a constant to be determined. [2]

It is now given that $\mathbf{a}$ and $\mathbf{b}$ are unit vectors.

Using the result in part (i), find the value of $(2\mathbf{a} - \mathbf{b}) \cdot \mathbf{c}$ . [3]

Show full worked solution▾

(i) [2]Method 1 Method 2

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

$Millennia Institute 2024 Promo Q4: solution diagram$

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

$Millennia Institute 2024 Promo Q4: solution diagram$

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] $\begin{aligned} (2\mathbf a-\mathbf b)\cdot\mathbf c&=(2\mathbf a-\mathbf b)\cdot\left(\frac{2\mathbf a+\mathbf b}{3}\right)\\ &=\frac13(2\mathbf a-\mathbf b)\cdot(2\mathbf a+\mathbf b)\\ &=\frac13(4\mathbf a\cdot\mathbf a+2\mathbf a\cdot\mathbf b-2\mathbf b\cdot\mathbf a-\mathbf b\cdot\mathbf b)\\ &=\frac13(4|\mathbf a|^2-|\mathbf b|^2)=\frac13(4(1)^2-(1)^2)=1. \end{aligned}$

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