<p>The points $A$, $B$ and $C$ lie on the circumference of a circle with center $O$ and diameter $AC$. It is given that&nbsp;$\overrightarrow{OA}=\mathbf{a}$ and  $\overrightarrow{OB}=\mathbf{b}$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Find $\overrightarrow{BC}$ in terms of $\mathbf{a}$ and $\mathbf{b}$. Hence show that&nbsp;$AB$ is perpendicular to $BC$.<span class="inline-marks">[3]</span></div></li></ol>
<p>The point $D$ is on $BC$ such that&nbsp;$BD:DC=3:2$.</p>
<ol class="i-list" start="2"><li><div class="list-item-paragraph">Show that the area of triangle $OCD$ can be written as&nbsp;$k\left| \mathbf{a}\times \mathbf{b} \right|$ where&nbsp;$k$ is a real constant to be found.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Given that angle $AOB={{120}^{\circ }}$, find $\overrightarrow{ON}$ in terms of $\mathbf{a}$ where&nbsp;$N$ is the foot of perpendicular of $D$ to $AC$.<span class="inline-marks">[4]</span></div></li></ol>

The points $A$, $B$ and $C$ lie on the circumference of a circle with center $O$ and diameter $AC$. It is given that $\overrightarrow{OA}=\mathbf{a}$ and  $\overrightarrow{OB}=\mathbf{b}$.

Find $\overrightarrow{BC}$ in terms of $\mathbf{a}$ and $\mathbf{b}$. Hence show that $AB$ is perpendicular to $BC$.

The point $D$ is on $BC$ such that $BD:DC=3:2$.

Show that the area of triangle $OCD$ can be written as $k\left| \mathbf{a}\times \mathbf{b} \right|$ where $k$ is a real constant to be found.

Given that angle $AOB={{120}^{\circ }}$, find $\overrightarrow{ON}$ in terms of $\mathbf{a}$ where $N$ is the foot of perpendicular of $D$ to $AC$.

2020 TJC Promo Q9

2020 TJC Promo Q9

Video Solution: