Review question

# How can we show that $P, Q$ and $R$ are collinear? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9575

## Solution

1. A vector of magnitude $OP$ in the direction from $O$ to $P$ is represented by $\mathbf{OP}$. If $\mathbf{OP}-3\mathbf{OQ}+2\mathbf{OR}=\mathbf{0}$, show that $P$,$Q$,$R$ are collinear.

Three points with position vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ are collinear if and only if the vectors $(\mathbf{a}-\mathbf{b})$ and $(\mathbf{a}-\mathbf{c})$ are parallel.

In other words, to prove collinearity, we would need to show $(\mathbf{a}-\mathbf{b})=k(\mathbf{a}-\mathbf{c})$ for some constant $k$.

For our example, we have $\mathbf{OP}-\mathbf{OQ}= 2(\mathbf{OQ}-\mathbf{OR})$, and so $\mathbf{OQ}-\mathbf{OP}= -2(\mathbf{OQ}-\mathbf{OR})$, telling us that $P, Q$ and $R$ are collinear.

1. A unit vector parallel to the $x$-axis is represented by $\mathbf{i}$ and a unit vector parallel to the $y$-axis by $\mathbf{j}$. If $\mathbf{OP}=a\mathbf{i}+s\mathbf{j}$ and $\mathbf{OQ}=-a\mathbf{i}+t\mathbf{j}$, where $a$ is a constant and $s$ and $t$ are variables, show that the loci of $P$ and $Q$ are parallel straight lines. In this case find $\mathbf{OQ}$ when $\mathbf{OP}=2\mathbf{i}+3\mathbf{j}$ and $OQ$ is perpendicular to $OP$.

The locus of $P$ will be the line $x = a$, while the locus of $Q$ will be $x = -a$. These are parallel straight lines.

The diagram shows the case $a = 2$. The point $P$ is at $(2,3)$, and $Q$ is at $(-2,k)$.

We are told that $OP$ and $OQ$ are perpendicular, so the gradients of $OP$ and $OQ$ must multiply to $-1$.

We could alternatively use that $\mathbf{OP}.\mathbf{OQ}=0$.

Thus $\dfrac{3}{2} \times \dfrac{k}{-2}=-1 \implies k = \dfrac{4}{3}$. Thus $\mathbf{OQ} = -2\mathbf{i}+\dfrac{4}{3}\mathbf{j}$.