Review question

# What's the position vector of the reflection of $O$ in the plane $ABC$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9794

## Solution

The position vectors of the points $A, B$ and $C$ with respect to an origin $O$ are the unit vectors $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ respectively. It is given that $OB$ is perpendicular to $OC$, $m = \cos \widehat{AOC}$, $n = \cos \widehat{AOB}$, and $m+n \neq 1$. The vector $\mathbf{r}$, where $\mathbf{r} = \mathbf{a} + \lambda \mathbf{b} + \mu \mathbf{c}$, is perpendicular to the plane $ABC$. Show that

$\lambda = \dfrac{(1-m)(1+m-n)}{1-m-n}$

and find a similar expression for $\mu$.

We know (where $a$ is the magnitude of $\mathbf{a}$) that $\mathbf{a}.\mathbf{b} = abn = n$, and similarly, $\mathbf{a}.\mathbf{c} = acm=m.$

Since $\mathbf{r}$ is perpendicular to $ABC$, we have that $\mathbf{r}.(\mathbf{a}-\mathbf{b})=0$, and so $(\mathbf{a} + \lambda \mathbf{b} + \mu \mathbf{c}).(\mathbf{a}-\mathbf{b})=0,$ and since the distributive law holds for the scalar product, and since $\mathbf{b}.\mathbf{c}=0$, \begin{align} 0 &= a^2 + \lambda \mathbf{a}.\mathbf{b} + \mu \mathbf{a}.\mathbf{c} -\mathbf{a}.\mathbf{b}-\lambda b^2 \notag \\ &= 1+\lambda n + \mu m -n-\lambda.\label{eq:1} \end{align} Similarly we know that $\mathbf{r}.(\mathbf{a}-\mathbf{c}) = 0$, which gives us that $$$0 = 1 + \lambda n + \mu m -m -\mu. \label{eq:2}$$$ Subtracting $\eqref{eq:2}$ from $\eqref{eq:1}$ gives $n+\lambda=m+\mu$. Then substituting $\mu$ in $\eqref{eq:2}$ we get \begin{align*} 0 &= 1+\lambda n+(\lambda+n-m)m-m-(\lambda+n-m) \\ \implies\quad \lambda(n+m-1) &= m^2+n-mn-1 \\ \implies\quad \lambda &= \frac{(m-1)(m-n+1)}{n+m-1} = \frac{(1-m)(1+m-n)}{1-m-n} \end{align*} as required.

In dividing here, we have used the given fact that $m+n \neq 1$.

Similarly, substituting $\lambda$ gives us $\begin{equation*} \mu = \frac{(1-n)(1+n-m)}{1-m-n}. \end{equation*}$

Notice how the expressions for $\lambda$ and $\mu$ are symmetrical in $m$ and $n$.

It is further given that $\widehat{AOB}=\widehat{AOC}=120^\circ$. The point $N$ is the foot of the perpendicular from $O$ on the plane $ABC$. Find the position vector of $N$, and deduce that the position vector of the reflection of $O$ in the plane $ABC$ is

$\dfrac{4}{5}\mathbf{a}+ \dfrac{3}{5}\mathbf{b} + \dfrac{3}{5}\mathbf{c}.$

We are given now that $m = n = \cos 120^\circ = -\dfrac{1}{2}$, which means on substitution that $\lambda = \mu = \dfrac{3}{4}$, giving $\mathbf{r} = \mathbf{a} + \dfrac{3}{4}\mathbf{b}+\dfrac{3}{4}\mathbf{c}$.

Let $\mathbf{n}$ be the position vector of $N$ (note that $\mathbf{n}$ is a vector, which distinguishes it from $n$).

Since $\mathbf{r}$ is perpendicular to the plane, we know that $\mathbf{n}$ is a multiple of $\mathbf{r}$, so $\mathbf{n} = \gamma \mathbf{a}+ \dfrac{3}{4}\gamma \mathbf{b} + \dfrac{3}{4}\gamma \mathbf{c}$ for some $\gamma$.

But $N$ is also on the plane $ABC$, which means that $\mathbf{n} = \mathbf{a} + \alpha(\mathbf{b}-\mathbf{a}) + \beta (\mathbf{c}-\mathbf{a})$, and so

$\mathbf{n} = (1-\alpha-\beta)\mathbf{a}+\alpha \mathbf{b} + \beta \mathbf{c}.$

Now $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ are independent vectors (we cannot write one as the sum of multiples of the other two) which means that we can compare coefficients here.

So the equation $(1-\alpha-\beta)\mathbf{a}+\alpha \mathbf{b} + \beta \mathbf{c} = \gamma \mathbf{a}+ \dfrac{3}{4}\gamma \mathbf{b} + \dfrac{3}{4}\gamma \mathbf{c}$ leads to $(1-\alpha-\beta)=\gamma$, $\alpha = \dfrac{3}{4}\gamma$, $\beta = \dfrac{3}{4}\gamma$.

This gives us that $\gamma = \dfrac{2}{5}$, and so $\alpha = \beta = \dfrac{3}{10}$.

Thus $\mathbf{n} = \dfrac{2}{5}\mathbf{a}+ \dfrac{3}{10}\mathbf{b} + \dfrac{3}{10}\mathbf{c}$ and if we reflect the point $O$ in the plane $ABC$, it takes on the position vector $2\mathbf{n}= \dfrac{4}{5}\mathbf{a}+ \dfrac{3}{5}\mathbf{b} + \dfrac{3}{5}\mathbf{c},$ as required.