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 \(\lambda\)
Path from O to R

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 \(\mu\).
Different path from O to R

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 \[\overrightarrow{OR} = \frac{3}{2}\mathbf{q}+\frac{1}{2}\mathbf{p}.\]