Review question

# Where do $ST$ and $AB$ cross? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8456

## Solution

1. Given that $\overrightarrow{OM}=\mathbf{i}+3\mathbf{j}$ and $\overrightarrow{ON}=\mathbf{i}+\mathbf{j}$, evaluate $\overrightarrow{OM}.\overrightarrow{ON}$ and hence calculate $\widehat{MON}$ to the nearest degree.

Given two vectors $\mathbf{f} = \begin{pmatrix}a\\b\end{pmatrix}$ and $\mathbf{g} = \begin{pmatrix}c\\d\end{pmatrix}$, the dot product is defined as $\mathbf{f.g} = ac+bd = \big \vert \mathbf{f} \big\vert \big\vert \mathbf{g} \big \vert \cos \theta,$ where $\theta$ is the angle between $\mathbf{f}$ and $\mathbf{g}$, and where $\big \vert \mathbf{f} \big\vert$ is the modulus of the vector $\mathbf{f}$.

Here we have $\overrightarrow{OM}.\overrightarrow{ON} = \begin{pmatrix}1\\3\end{pmatrix}.\begin{pmatrix}1\\1\end{pmatrix} = 1 + 3 = 4$.

And we have $\overrightarrow{OM}.\overrightarrow{ON}=\sqrt{1^2+3^2}\sqrt{1^2+1^2} \cos \widehat{MON}$, which yields $\cos\widehat{MON} = \frac{4}{\sqrt{10}\sqrt{2}} = \frac{2}{\sqrt{5}}.$

Substituting in the values and rounding to the nearest degree, we find $\widehat{MON} =27^\circ.$

1. The position vectors, relative to an Origin $O$, of two points $S$ and $T$ are $2\mathbf{p}$ and $2\mathbf{q}$ respectively. The point $A$ lies on $OS$ and is such that $OA=AS$. The point $B$ lies on $OT$ produced and is such that $OT=2TB$. The lines $ST$ and $AB$ intersect at $R$.

Given that $\overrightarrow{AR}=\lambda \overrightarrow{AB}$ and that $\overrightarrow{SR}=\mu \overrightarrow{ST}$, express $\overrightarrow{OR}$

1. in terms of $\mathbf{p}$, $\mathbf{q}$ and $\lambda$

The diagram shows the positions of $O, A, B, R, S$, and $T$. Travelling from $O$ to $R$ along the red line gives $\overrightarrow{OR} = \overrightarrow{OA} + \lambda \overrightarrow{AB} = \mathbf{p} + \lambda (3\mathbf{q}-\mathbf{p})$.

1. in terms of $\mathbf{p}$, $\mathbf{q}$ and $\mu$.

Following the new red path, we have $\overrightarrow{OR} = 2\mathbf{q}+(1-\mu)(2\mathbf{p}-2\mathbf{q})=2\mathbf{p}+2\mu(\mathbf{q}-\mathbf{p}).$

Hence evaluate $\lambda$ and $\mu$ and express $\overrightarrow{OR}$ in terms of $\mathbf{p}$ and $\mathbf{q}$.

The two expressions we derived for $\overrightarrow{OR}$ have to be equal. Hence we know that $\mathbf{p}+\lambda(3\mathbf{q}-\mathbf{p})=2\mathbf{p}+2\mu(\mathbf{q}-\mathbf{p}).$

Rearranging this equation, we find $(3\lambda-2\mu)\mathbf{q}-(\lambda+1-2\mu)\mathbf{p}=0.$

Assuming that vectors $\mathbf{p}$ and $\mathbf{q}$ are not parallel or zero, then no linear combination of them can vanish unless both coefficients vanish.

In other words, if $a\mathbf{p} + b\mathbf{q} = 0$ for constants $a$ and $b$, then $a = b = 0$.

Hence we have the pair of simultaneous equations

\begin{align*} 3\lambda-2\mu &= 0,\\ \lambda+1-2\mu &=0, \end{align*}

from which we conclude $\lambda=\dfrac{1}{2}$ and $\mu=\dfrac{3}{4}$, and so $\overrightarrow{OR} = \frac{3}{2}\mathbf{q}+\frac{1}{2}\mathbf{p}.$