If $\overrightarrow{OM}$ is the projection of $\overrightarrow{OA}$ on $m$, where is $M$?

If $\overrightarrow{OM}$ is the projection of $\overrightarrow{OA}$ on $m$, where is $M$?

Question

The straight lines $l$ and $m$ intersect at the origin $O$, and their directions are parallel to $\mathbf{i}$ and $\mathbf{i}+2\mathbf{j}$ respectively. The point $A$ has position vector $\overrightarrow{OA}$ given by $2\mathbf{i}+3\mathbf{j}+5\mathbf{k}$. The point $L$ on $l$ is such that $\overrightarrow{OL}$ is the projection of $\overrightarrow{OA}$ on $l$, and the point $M$ on $m$ is such that $\overrightarrow{OM}$ is the projection of $\overrightarrow{OA}$ on $m$. State the position vector of $L$, and show that $\overrightarrow{OM} = \dfrac{8}{5}\mathbf{i}+ \dfrac{16}{5}\mathbf{j}$.

Calculate the angle $MAL$, giving your answer correct to the nearest degree.

The point $N$, where $\overrightarrow{ON} = 2\mathbf{i} + 3\mathbf{j}$, is the foot of the perpendicular from $A$ on to the plane $OLM$. The lines $ON$ and $LM$ intersect at $X$. Show that $\dfrac{OX}{ON} = \dfrac{16}{19}$.

Suggestion

Drawing a diagram should help to clarify what is being described here.

If $\overrightarrow{OL}$ is the projection of $\overrightarrow{OA}$ on $l$ it means that $\overrightarrow{OL}$ is the component of $\overrightarrow{OA}$ in the direction of $l$. So we can find $L$ by dropping a perpendicular…

To find the angle between vectors we can often use the dot product.

Can we find the position vector of $N$? From there, can we find the position vector of $X$?

Solution

Since $l$ is parallel to $\mathbf{i}$ (or the $x$-axis), the vector $\overrightarrow{OL}$ is the $\mathbf{i}$-component of $\overrightarrow{OA}$. So the position vector of $L$ is $2\mathbf{i}$.

The point

$M$

is on

$m$

, so its position vector can be written as

$\mu(\mathbf{i}+2\mathbf{j})$

for some number

$\mu$

. Since it is the projection, it follows that

$m$

and

$\overrightarrow{AM}$

are perpendicular. In other words, their

dot product

is zero.

$$\begin{align*} (\mathbf{i}+2\mathbf{j}) . ((\mu-2)\mathbf{i}+(2\mu-3)\mathbf{j}-5\mathbf{k}) &= 0 \\ \implies\quad (\mu-2)+2(2\mu-3) &= 0 \\ \implies\quad \mu &= \frac{8}{5} \end{align*}$$

So $\overrightarrow{OM} = \dfrac{8}{5}\mathbf{i}+ \dfrac{16}{5}\mathbf{j}$ as required.

We require the angle between the vectors $\overrightarrow{AM}$ and $\overrightarrow{AL}$ and we can find this using the dot product.

$$\begin{align*} \overrightarrow{AM} &= \overrightarrow{OM} -\overrightarrow{OA} = \dfrac{8}{5}\mathbf{i}+ \dfrac{16}{5}\mathbf{j} - (2\mathbf{i}+3\mathbf{j}+5\mathbf{k}) = -\frac{2}{5}\mathbf{i} + \frac{1}{5}\mathbf{j} - 5\mathbf{k} \\ \overrightarrow{AL} &= \overrightarrow{OL} -\overrightarrow{OA} = 2\mathbf{i} - (2\mathbf{i}+3\mathbf{j}+5\mathbf{k}) = -3\mathbf{j}-5\mathbf{k} \end{align*}$$

So now,

$$\begin{align*} \overrightarrow{AM} . \overrightarrow{AL} &= 0-\frac{3}{5}+25 = \frac{122}{5} \\ &= \big\vert AM\big\vert\times\big\vert AL\big\vert\cos\theta \\ &= \sqrt{\left(\tfrac{2}{5}\right)^2 + \left(\tfrac{1}{5}\right)^2 +5^2}\sqrt{3^2+5^2}\cos\theta \\ &= \sqrt{\tfrac{126}{5}}\sqrt{34}\cos\theta \\ \implies\quad \cos\theta &=\frac{122/5}{\sqrt{126/5}\sqrt{34}}\approx0.8336 \\ \implies\quad \theta &\approx33.53^\circ \end{align*}$$

The angle $MAL$ is $34^\circ$.

We can now work in two dimensions. We have that $\overrightarrow{OL} = 2\mathbf{i}$, $\overrightarrow{OM} = \frac{8}{5}\mathbf{i}+\frac{16}{5}\mathbf{j}$ and $\overrightarrow{ON} = 2\mathbf{i}+3\mathbf{j}$. Let $\overrightarrow{OX} = \alpha (2\mathbf{i}+3\mathbf{j})$ so that $\alpha$ is the required ratio.

The vector $\overrightarrow{LM} = \frac{8}{5}\mathbf{i}+\frac{16}{5}\mathbf{j} - 2\mathbf{i} = -\frac{2}{5}\mathbf{i}+\frac{16}{5}\mathbf{j}$.

The line

$LM$

is

$\mathbf{r} = 2\mathbf{i} + \beta(-\frac{2}{5}\mathbf{i}+\frac{16}{5}\mathbf{j})$

. So at

$X$

,

$$\alpha (2\mathbf{i}+3\mathbf{j}) = 2\mathbf{i} + \beta\left(-\tfrac{2}{5}\mathbf{i}+\tfrac{16}{5}\mathbf{j}\right)$$

Comparing coefficients, we have

$$\begin{align*} 2\alpha &= 2-\tfrac{2}{5}\beta \text{ and } 3\alpha = \tfrac{16}{5}\beta \\ \implies\quad \alpha &= \frac{16}{19} \end{align*}$$

as required.

The line $LM$ is $\mathbf{r} = \begin{pmatrix}2\\0\end{pmatrix} + \mu \begin{pmatrix}-2/5\\16/5\end{pmatrix}$, and so $\lambda \begin{pmatrix}2\\3\end{pmatrix} = \begin{pmatrix}2\\0\end{pmatrix} + \mu \begin{pmatrix}-2/5\\16/5\end{pmatrix}$ for some $\mu$. Thus $2\lambda = 2 + -\dfrac{2}{5}\mu$, and $3\lambda = \dfrac{16}{5}\mu \implies \mu = \dfrac{15}{16}\lambda \implies 2 \lambda = 2 + -\dfrac{2}{5}\times \dfrac{15}{16}\lambda$, which gives us $\lambda = \dfrac{16}{19}$. So $\dfrac{OX}{ON} = \dfrac{16}{19}$, as required.