Review question

# If $\overrightarrow{OM}$ is the projection of $\overrightarrow{OA}$ on $m$, where is $M$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7272

## Suggestion

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

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…

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

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

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 $\dfrac{OX}{ON} = \dfrac{16}{19}$.

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