Review question

# If $A$, $P$ and $B$ are collinear, what's the value of $\lambda$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7285

## Solution

Points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively relative to a point $O$. Given that $L$ is the midpoint of $OA$, and that $M$ is the point on $OB$ produced such that $OM = 3OB$, express $\overrightarrow{LM}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

Given further that $P$ is the point on $LM$ such that $LP = \lambda LM$, express $\overrightarrow{AP}$ in terms of $\mathbf{a, b}$ and $\lambda$.

We have that $\overrightarrow{OM} = 3\mathbf{b}, \overrightarrow{OL} = \dfrac{1}{2}\mathbf{a}$, so $\overrightarrow{LM}= 3\mathbf{b}- \dfrac{1}{2}\mathbf{a}.$

The vector \begin{align*} \overrightarrow{AP} &= \overrightarrow{OP} - \overrightarrow{OA} \\ &= (\overrightarrow{OL}+\lambda \overrightarrow{LM}) - \mathbf{a} \\ &= \dfrac{1}{2}\mathbf{a}+\lambda\left( 3\mathbf{b}- \dfrac{1}{2}\mathbf{a}\right) -\mathbf{a}\\ &=3\lambda \mathbf{b} - \dfrac{1}{2}\mathbf{a}(1+\lambda). \end{align*}

In the case where $A, P$ and $B$ are collinear, calculate the value of

1. $\lambda$
2. $\dfrac{AP}{PB}$.

If $P$ is on $AB$, then $\overrightarrow{AP} = \mu\overrightarrow{AB}$ for some scalar $\mu$. This means that $3\lambda \mathbf{b} - \dfrac{1}{2}\mathbf{a}(1+\lambda) = \mu(\mathbf{b}-\mathbf{a}).$

Now any vector in the plane can be written uniquely as $\alpha\, \mathbf{a} + \beta\, \mathbf{b}$, so we can compare the coefficients of $\mathbf{a}$ and $\mathbf{b}$ here, giving $3\lambda = \mu \quad\text{and}\quad \dfrac{1}{2} + \dfrac{\lambda}{2} = \mu,$ which solve to give $\lambda = \dfrac{1}{5}, \mu = \dfrac{3}{5}.$

Thus

1. $\lambda = \dfrac{1}{5}$

2. $\dfrac{AP}{PB} = \dfrac{\mu}{1-\mu} = \dfrac{3/5}{2/5} = \dfrac{3}{2}$.