ACJC/2021/JC1/Promo/P1/Q05

Referred to the origin $O$, points $A$ and $B$ have position vectors a and $\mathbf{b}$ respectively. The modulus of $\mathbf{a}$ is 2 and $\mathbf{b}$ is a unit vector. The angle between $\mathbf{a}$ and $\mathbf{b}$ is $60^{\circ}$. Point $C$ lies on $A B$, between $A$ and $B$, such that $A C=k C B$, where $0<k<1$.
(i) Express $\overrightarrow{O C}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.
(ii) Show that the length of projection of $\overrightarrow{O C}$ on $\overrightarrow{O A}$ is given by $\frac{k+4}{2(k+1)}$.

[3]

(iii) Find, in terms of $k$, the area of triangle $O A C$.

[3]

(i) Using Ratio Theorem,

$\overrightarrow{O C}=\frac{1}{k+1}(\mathbf{a}+k \mathbf{b})$

(ii) Length of projection of $\overrightarrow{O C}$ onto $\overrightarrow{O A}$

$=\frac{1}{2(k+1)}|(\mathbf{a}+k \mathbf{b}) \cdot \mathbf{a}| \quad$ since $0<k<1$

$=\frac{1}{2(k+1)}|(\mathbf{a} \cdot \mathbf{a}+k \mathbf{b} \cdot \mathbf{a})|$

$=\left.\frac{1}{2(k+1)}|| \mathbf{a}\right|^2+k \mathbf{b} \cdot \mathbf{a} \mid$

$=\frac{1}{2(k+1)}|4+k| \mathbf{a}|| \mathbf{b}\left|\cos \left(60^{\circ}\right)\right|$

$=\frac{1}{2(k+1)}\left|4+k(2)(1)\left(\frac{1}{2}\right)\right|=\frac{k + 4}{2(k+1)}$ (proved)

(iii)

Area of triangle $O A C$
$=\frac{1}{2}|\mathbf{a} \times \mathbf{c}| =\frac{1}{2}\left|\mathbf{a} \times \frac{1}{k+1}(\mathbf{a}+k \mathbf{b})\right| $
$=\frac{1}{2(k+1)}|(\mathbf{a} \times \mathbf{a})+k(\mathbf{a} \times \mathbf{b})| \quad \mid \text { since } 0<k<1 $
$=\frac{1}{2(k+1)} \mid(k(\mathbf{a} \times \mathbf{b}) \mid \text { since } \mathbf{a} \times \mathbf{a}=0$
$=\frac{k}{2(k +1)}|| \mathbf{a}|| \mathbf{b}\left|\sin \left(60^{\circ}\right)\right|=\frac{k}{2(k + 1)}\left|2(1)\frac{\sqrt{3}}{2} \right|$

$=\frac{\sqrt{3} k}{2(k+1)}$