Review question

# Given three vectors, can we find a point equidistant from them all? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5813

## Suggestion

The position vectors of points $A$, $B$ and $C$, referred to an origin $O$, are $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ respectively, where

\begin{align*} \mathbf{a} &= 3\mathbf{i} + 4 \mathbf{j} + 5\mathbf{k}, \\ \mathbf{b} &= 7\mathbf{i} - \mathbf{k}, \\ \mathbf{c} &= 5\mathbf{i} + 5 \mathbf{j}. \\ \end{align*}

$P$ is a point which is equidistant from the lines $OA$, $OB$ and $OC$. Write down expressions for the cosine of the angle between $OP$ and $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ respectively and hence deduce an expression for a unit vector in the direction of $\mathbf{OP}$.

We want to find the cosine of an angle between vectors. What does this suggest?

We are also interested in the distance of $P$ from the lines. Can you draw a diagram showing the point on $OA$ closest to $P$?

Given that the line $OP$ meets the plane $ABC$ at $Q$, find $\mathbf{OQ}$.

We will need equations for both the line $OP$ and the plane $ABC$ to find the intersection point.

Can you write the equation of the plane in a helpful form?

Given that $G$ is the centroid of the triangle $ABC$, show that $QG$ is parallel to the coordinate plane $Oyz$.

Can we define the position vector of $G$ in terms of the those of the three corners?

Which vectors are parallel to the coordinate plane $Oyz$?