Q11 (e) (iii)

Let point $P$ [be the] intersection of the perpendicular [with] $OP$.

Method 1 Find $OP$, the projection of $OM$ on to $ON$, then use Pythagoras on $riangle OPM$.

$$\begin{aligned} \overrightarrow{OP}&=\operatorname{proj}_{\overrightarrow{ON}}\overrightarrow{OM} =\frac{\overrightarrow{OM}\cdot\overrightarrow{ON}}{|\overrightarrow{ON}|^2}\overrightarrow{ON}\\ &=\frac{0+6+0}{3^2+1^2}\overrightarrow{ON}\\ &=\frac{6}{10}\overrightarrow{ON} \end{aligned}$$ $$|\overrightarrow{OP}|=\frac35|\overrightarrow{ON}|=\frac{3\sqrt{10}}5$$ $$\begin{aligned} |PM|^2&=OM^2-OP^2\quad(\text{Pythagoras})\\ &=8-\frac{9}{25}\\ &=\frac{110}{25}\\ &=\frac{22}{5} \end{aligned}$$ $$\therefore |PM|=\sqrt{\frac{22}{5}}\text{ units.}$$

(N.B. The diagram was useful to see what was required.)

Method 2 Find the coordinates of $P$, then calculate the length of the resulting vector $\overrightarrow{MP}$.

$$\overrightarrow{OP}=\lambda\overrightarrow{ON}\quad\text{for some }\lambda\in\mathbb{R}$$ $$\overrightarrow{OP}=3\lambda\,\vec{j}+\lambda\,\vec{k}$$

Now $\overrightarrow{OP}\cdot\overrightarrow{PM}=0$ (perpendicular)

$$\begin{aligned} (3\lambda\vec{j}+\lambda\vec{k})\cdot((2-3\lambda)\vec{j}-\lambda\vec{k})&=0\\ 3\lambda(2-3\lambda)-\lambda^2&=0\\ 6\lambda-9\lambda^2-\lambda^2&=0\\ 10\lambda^2&=6\lambda\\ \lambda&=\frac35\quad(\lambda\ne0) \end{aligned}$$ $$\overrightarrow{OP}=\frac95\vec{j}+\frac35\vec{k}$$ $$\overrightarrow{PM}=2\vec{i}+\frac15\vec{j}-\frac35\vec{k}$$ $$\begin{aligned} |PM|&=\sqrt{4+\frac1{25}+\frac9{25}}\\ &=\sqrt{\frac{100+1+9}{25}}\\ &=\sqrt{\frac{110}{25}}\\ &=\sqrt{\frac{22}{5}}\text{ units.} \end{aligned}$$