Given three vectors, can we find a point equidistant from them all?

Given three vectors, can we find a point equidistant from them all?

Question

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 $\mathbf{OP}$.

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

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

Suggestion

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 $P$?

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?

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 $Oyz$?

Solution

Let $A'$ be a point on $OA$ such that $A'P$ is perpendicular to $OA$ and let $B'$ and $C'$ be the equivalent points on $OB$ and $OC$ respectively.

If we think about the three right angled triangles $A'OP$, $B'OP$ and $C'OP$ we have hypotenuse $OP$ common to each and $P$ was defined to be equidistant from the lines $OA$, $OB$ and $OC$. So $OA'=OB'=OC'$ making the triangles congruent (RHS).

The question asks about the cosine of the angle $\alpha$, which suggests using the scalar product. Let the vector $OP$ be $\mathbf{p} = \begin{pmatrix}p_i \\ p_j \\ p_k\end{pmatrix}$, so we have $\mathbf{p}.\mathbf{a} = 3p_i+4p_j+5p_k$.

But $\mathbf{p}.\mathbf{a} = |\mathbf{p}||\mathbf{a}|\cos \alpha$, so $\cos \alpha = \dfrac{\mathbf{p}.\mathbf{a}}{|\mathbf{p}|\sqrt{50}}=\dfrac{3p_i+4p_j+5p_k}{|\mathbf{p}|\sqrt{50}}$. We have used the fact that $|\mathbf{a}|=\sqrt{3^2+4^2+5^2} = \sqrt{50}$.

Since the three triangles are congruent, they each contain the same angle $\alpha$. Also notice that $|\mathbf{b}|=|\mathbf{c}|=\sqrt{50}$.

So for vector $\mathbf{b}$ we have $\cos \alpha= \dfrac{7p_i-p_k}{|\mathbf{p}|\sqrt{50}}$, and for $\mathbf{c}$ we have $\cos \alpha = \dfrac{5p_i+5p_j}{|\mathbf{p}|\sqrt{50}}$.

Combining these we can write $3p_i+4p_j+5p_k = 7p_i-p_k= 5p_i+5p_j$.

Using the second and third parts gives $p_k = 2p_i - 5p_j$, and substituting this into the first two parts gives $3p_i+4p_j+5(2p_i - 5p_j) = 7p_i-(2p_i - 5p_j)$.

Thus $p_j = \dfrac{4}{13} p_i$, and $p_k = \dfrac{6}{13} p_i$, and so the vector $\mathbf{p}$ is $t \begin{pmatrix}13 \\ 4 \\ 6\end{pmatrix}$ for some scalar $t$.

To find the unit vector in this direction, we divide by the magnitude $\sqrt{13^2+4^2+6^2} = \sqrt{221}$, and so the unit vector in the direction of $\mathbf{OP}$ is $\hat{\mathbf{p}} = \frac{13}{\sqrt{221}} \mathbf{i} + \frac{4}{\sqrt{221}}\mathbf{j} + \frac{6}{\sqrt{221}}\mathbf{k}.$

We can define the plane $ABC$ from the three position vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ via $\mathbf{r} = \mathbf{a} + \lambda(\mathbf{b}-\mathbf{a}) + \mu(\mathbf{c}-\mathbf{a}),$ where $\lambda$ and $\mu$ are scalar parameters.

If we start at the origin, $\mathbf{a}$ takes us onto the plane. The vectors $(\mathbf{b}-\mathbf{a})$ and $(\mathbf{c}-\mathbf{a})$ are distinct vectors lying in the plane, so a linear combination of these will take us anywhere on the plane from $\mathbf{a}$.

Thus the plane $ABC$ is $\mathbf{r} = (3+4\lambda+2\mu)\mathbf{i} + (4-4\lambda+\mu) \mathbf{j} + (5-6\lambda-5\mu)\mathbf{k}.$

The line $OP$ is given by $\mathbf{r} = t (13\mathbf{i}+ 4\mathbf{j}+ 6\mathbf{k})$. To find the intersection we set these equal, then the three components give us three equations in three unknowns, $\lambda$, $\mu$ and $t$.

Solving these simultaneously gives

$t = \dfrac{5}{13}$

and so

$\mathbf{OQ}$

=

$5\mathbf{i} + \frac{20}{13}\mathbf{j} +\frac{30}{13}\mathbf{k}$

.

The position vector of the centroid $G$ of a triangle whose vertices have position vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ is given by $\mathbf{g}=\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}.$ In our case $\mathbf{g}=5 \mathbf{i} + 3 \mathbf{j} + \frac{4}{3} \mathbf{k}.$

The direction of the line $QG$ is given by the vector $\mathbf{g}-\mathbf{OQ}$.

If this direction vector is parallel to the $Oyz$ plane, its $x$-component will be zero. Hence we have to show that $q_i = g_i$.

The $x$-components of $\mathbf{g}$ and $\mathbf{OQ}$ are both $5$, so $QG$ is indeed parallel to the plane.