# Alternative approaches to solving advanced level vector problems

Trying to find more than one solution for a given problem can be an effective way to convert routine exercises or exam questions into exploratory `research’ type activities which are more open-ended and fun. Getting students to generate, analyse and compare alternative solutions to problems has long been recommended in the mathematics education literature (see, e.g., Cai, J., Brook, M., 2006, Looking back in problem solving, Mathematics Teaching Incorporating Micromath, 196, pp. 42-45). I recently explored the potential for this approach in the context of advanced level vector problems and want to document some alternative approaches I found for answering some of these problems here. A typical advanced level vector problem – in fact, an exam question – is as follows (the reader should attempt this question before reading on):

With respect to a fixed origin $O$, the line $l$ has equation

$\mathbf{r} = \begin{pmatrix}13\\8\\1\end{pmatrix} + \lambda\begin{pmatrix}2\\2\\-1\end{pmatrix}$

where $\lambda$ is a scalar parameter. The point $A$ lies on $l$ and has coordinates $(3, -2, 6)$.
The point $P$ has position vector $(-p \mathbf{i} + 2p \mathbf{k})$ relative to $O$ where $p$ is a constant.
Given that vector $\overrightarrow{PA}$ is perpendicular to $l$,
(a) find the value of $p$.

Given also that $B$ is a point on $l$ such that $\angle BPA = 45^{\circ}$,
(b) find the coordinates of the two possible positions of $B$.

The following is a sketch of the scenario.

Let $\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OB} = \mathbf{b}$, and $\overrightarrow{OP} = \mathbf{p}$. The interesting part of this question is (b), but with regard to (a) we have

$\overrightarrow{PA} = \mathbf{a} - \mathbf{p} = \begin{pmatrix}3\\-2\\6\end{pmatrix} - \begin{pmatrix}-p\\\text{0}\\2p\end{pmatrix} = \begin{pmatrix}3+p\\-2\\6-2p\end{pmatrix}$

Since $\overrightarrow{PA}$ is perpendicular to $l$, the dot product with the direction vector of $l$ must be zero, so we must have

$\begin{pmatrix}3+p\\-2\\6-2p\end{pmatrix} \cdot \begin{pmatrix}2\\2\\-1\end{pmatrix} = 6 + 2p - 4 - 6 + 2p = 0$

so $p = 1$. It follows that

$\mathbf{p} = \begin{pmatrix}-p\\\text{0}\\2p\end{pmatrix} = \begin{pmatrix}-1\\\text{0}\\2\end{pmatrix}$

and

$\overrightarrow{PA} = \begin{pmatrix}3+p\\-2\\6-2p\end{pmatrix} = \begin{pmatrix}4\\-2\\4\end{pmatrix}$

With regard to part (b), the most straightforward approach is to observe that since $\overrightarrow{AB}$ is collinear with $l$, the coordinates of $B$ with respect to $O$ must be given by

$\mathbf{b} = \mathbf{a} + \overrightarrow{AB} = \begin{pmatrix}3\\-2\\6\end{pmatrix} + \mu\begin{pmatrix}2\\2\\-1\end{pmatrix}$

where $\mu$ is a parameter to be found, and where the length of $\overrightarrow{AB}$ is equal to the length of $\mathbf{b} - \mathbf{a} = \mu\begin{pmatrix}2\\2\\-1\end{pmatrix}$, which is $|\mu| \sqrt{2^2 + 2^2 + (-1)^2} = 3|\mu|$. Now, $|\overrightarrow{PA}| = \sqrt{4^2 + (-2)^2 + 4^2} = \sqrt{36} = 6$ and since the right-angled triangle in the sketch above is isosceles we must also have $|\overrightarrow{AB}| = 3|\mu| = 6$ and so $|\mu| = 2$. Substituting $\mu = 2$ into the expression for $\mathbf{b}$ above we get the coordinates of $B$ in the sketch above to be $(7, 2, 4)$. However, the sketch above also shows another possible position for $B$, to the left of $A$. By symmetry, the coordinates of this other possible position for $B$ can be found by setting $\mu = -2$ (since in this case the vector $\overrightarrow{AB}$ would point to the left of $A$ rather than to the right). Substituting $\mu = -2$ into the expression for $\mathbf{b}$ above we get the other possible coordinates of $B$ to be $(-1, -6, 8)$. This solves part (b).

The question now arises as to whether there is any other way of solving part (b) of this problem? I found the following alternative solution approach which I think is instructive (although algebraically more complicated, perhaps) in that it uses the cosine expression for the scalar product. Since the angle between $\overrightarrow{PA}$ and $\overrightarrow{PB}$ is $45^{\circ}$, we must have

$\overrightarrow{PA} \cdot \overrightarrow{PB} = |\overrightarrow{PA}||\overrightarrow{PB}|\cos 45^{\circ} = \frac{1}{\sqrt{2}}|\overrightarrow{PA}||\overrightarrow{PB}|$

Since $B$ is on the line $l$, it must have coordinates of the form $(13+2\theta, 8+2\theta, 1-\theta)$. As we found in part (a) above, we have (in row vector form)

$\overrightarrow{PA} = (4, -2, 4)$

and

$\overrightarrow{PB} = (13+2\theta, 8+2\theta, 1-\theta) - (-1, 0, 2) = (14+2\theta, 8+2\theta, -1-\theta)$

Therefore

$\overrightarrow{PA} \cdot \overrightarrow{PB} = 56 + 8\theta - 16 - 4\theta - 4 - 4\theta = 36$

but also

$\overrightarrow{PA} \cdot \overrightarrow{PB} = \frac{1}{\sqrt{2}}|\overrightarrow{PA}||\overrightarrow{PB}| = \frac{1}{\sqrt{2}}\cdot6\sqrt{(14+2\theta)^2 + (8+2\theta)^2 + (1+\theta)^2} = 3\sqrt{2}\sqrt{9\theta^2 + 90\theta + 261}$

Equating these two we get

$3\sqrt{2}\sqrt{9\theta^2 + 90\theta + 261} = 36$

which simplifies to

$\theta^2 + 10\theta + 21 = 0$

or

$(\theta + 3)(\theta + 7) = 0$

The solutions to the quadratic are therefore $\theta = -3$ and $\theta = -7$. Since $B$ must have coordinates of the form $(13+2\theta, 8+2\theta, 1-\theta)$, substituting the two possible solutions for $\theta$ gives us the two possible coordinates of $B$ as $(7, 2, 4)$ and $(-1, -6, 8)$ as before.

I performed similar exercises with other advanced level vector problems and was able to find alternative solution approaches which were usually instructive in some way, although also usually more algebraically demanding than the most straightforward approach available. Another example I want to record here concerns the following advanced level vector problem – also an exam question (again, the reader should attempt to solve this problem before reading my discussion of it below):

With respect to a fixed origin $O$, the lines $l_1$ and $l_2$ are given by the equations

$l_1: \mathbf{r} = \begin{pmatrix}6\\-3\\-2\end{pmatrix} + \lambda\begin{pmatrix}-1\\2\\3\end{pmatrix}$

and

$l_2: \mathbf{r} = \begin{pmatrix}-5\\15\\3\end{pmatrix} + \mu\begin{pmatrix}2\\-3\\1\end{pmatrix}$

where $\lambda$ and $\mu$ are scalar parameters.

(a) Show that $l_1$ and $l_2$ meet and find the position vector of their point of intersection $A$.

(b) Find, to the nearest $0.1^{\circ}$, the acute angle between $l_1$ and $l_2$.

The point $B$ has position vector $\begin{pmatrix}5\\-1\\1\end{pmatrix}$

(c) Show that $B$ lies on $l_1$.

(d) Find the shortest distance from B to the line $l_2$.

The most interesting part of this question for me is part (d), but for part (a) we observe that since the two lines meet, it must be the case that $6 - \lambda = -5 + 2\mu$ or equivalently

$2\mu + \lambda = 11$

and also $-3 + 2\lambda = 15 - 3\mu$ or equivalently

$3\mu + 2\lambda = 18$

Solving these two simultaneously gives $\lambda = 3$ and $\mu = 4$. If the two lines do indeed meet, these values for $\lambda$ and $\mu$ will be consistent with equality of the third coordinates of $l_1$ and $l_2$ at the point of intersection. Using $\lambda = 3$ to find the third coordinate for $l_1$ and using $\mu = 4$ to find the third coordinate for $l_2$, we find that the third coordinate is 7 in both cases, so the two lines do indeed meet. The position vector of their point of intersection $A$ is then

$\begin{pmatrix}6\\-3\\-2\end{pmatrix} + 3\begin{pmatrix}-1\\2\\3\end{pmatrix} = \begin{pmatrix}3\\3\\7\end{pmatrix}$

or equivalently

$\begin{pmatrix}-5\\15\\3\end{pmatrix} + 4\begin{pmatrix}2\\-3\\1\end{pmatrix} = \begin{pmatrix}3\\3\\7\end{pmatrix}$

For part (b) we find the dot product of the direction vectors of the two lines. We get

$(-1, 2, 3) \cdot (2, -3, 1) = -5$

but also

$(-1, 2, 3) \cdot (2, -3, 1) = |(-1, 2, 3)||(2, -3, 1)|\cos\theta = 14\cos\theta$

Equating the two gives

$\cos\theta = \frac{-5}{14}$

so $\theta = \arccos(\frac{-5}{14}) = 110.9248324^{\circ}$, and therefore the acute angle must be $180^{\circ} - \theta = 69.1^{\circ}$ (to nearest $0.1^{\circ}$).

For part (c) we observe that if $B$ lies on $l_1$, there is a $\lambda$ such that

$\begin{pmatrix}5\\-1\\1\end{pmatrix} = \begin{pmatrix}6-\lambda\\-3+2\lambda\\-2+3\lambda\end{pmatrix}$

By inspection, the required value is $\lambda = 1$.

Finally, for part (d) it is helpful to draw the following sketch of the situation:

We know that point $A$ has coordinates $(3, 3, 7)$ and point $B$ has coordinates $(5, -1, 1)$, and therefore the length of the line from $B$ to $A$ is

$|\overrightarrow{BA}| = |(3, 3, 7) - (5, -1, 1)| = |(-2, 4, 6)| = \sqrt{(-2)^2 + 4^2 + 6^2} = \sqrt{56}$

The shortest distance from $B$ to the line $l_2$ is the length of the perpendicular from $B$ which is shown in the sketch as intersecting line $l_2$ at a point $C$. Therefore the required shortest distance is $|\overrightarrow{BC}|$ which can be obtained from simple trigonometry as

$|\overrightarrow{BC}| = |\overrightarrow{BA}|\sin69.1^{\circ} = \sqrt{56}\sin69.1^{\circ} = 6.99$

This solves part (d). Again, we ask if there is some alternative way to obtain this shortest distance? I found the following alternative approach which avoids using trigonometry (at the expense of being algebraically more cumbersome). A point on the line $l_2$ must have coordinates of the form

$(-5 + 2\mu, 15 - 3\mu, 3 + \mu)$

The distance between $B$ and any such point is given by

$|(5, -1, 1) - (-5 + 2\mu, 15 - 3\mu, 3 + \mu)| = |(10 - 2\mu, -16 + 3\mu, -2 - \mu)|$

$= \sqrt{(10 - 2\mu)^2 + (-16 + 3\mu)^2 + (-2 - \mu)^2}$

$= \sqrt{14\mu^2 - 132\mu + 360}$

$= \sqrt{14(\mu - \frac{33}{7})^2 + \frac{342}{7}}$

Therefore the shortest distance is when $\mu = \frac{33}{7}$ and is $\sqrt{\frac{342}{7}} = 6.99$, which agrees with the previous result.