Vector Pathways (SQA National 5 Maths): Revision Note

In the exam, different correct pathways will earn full marks, as long as the final answer is fully simplified.

Check for symmetries in the diagram to see if the vectors given can be used anywhere else.

The following diagram consists of a grid of identical parallelograms.

Vectors a and b are defined by $\mathbf{a}=\overrightarrow{\mathrm{AB}}$ and $\mathbf{b}=\overrightarrow{\mathrm{AF}}$.

Write the following vectors in terms of a and b.

a) $\overrightarrow{\mathrm{AE}}$

b) $\overrightarrow{\mathrm{GT}}$

c) $\overrightarrow{\mathrm{EK}}$

Answer:

Part (a)

To get from A to E you need to follow vector $\mathbf{a}$ four times to the right 

$\begin{array}{rcl}\overrightarrow{\mathrm{AE}}& =& \overrightarrow{\mathrm{AB}}+\hspace{0.17em}\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CD}}+\overrightarrow{\mathrm{DE}}\\ & =& \mathbf{a}+\mathbf{a}+\mathbf{a}+\mathbf{a}\end{array}$

$\overrightarrow{\mathrm{AE}}=4\mathbf{a}$ 

Part (b)

There are many ways to get from G to T

One option is to go from G to Q ($\mathbf{b}$ twice), and then from Q to T ($\mathbf{a}$ three times) 

$\begin{array}{rcl}\overrightarrow{\mathrm{GT}}& =& \overrightarrow{\mathrm{GL}}+\hspace{0.17em}\overrightarrow{\mathrm{LQ}}+\overrightarrow{\mathrm{QR}}+\overrightarrow{\mathrm{RS}}+\overrightarrow{\mathrm{ST}}\\ & =& \mathbf{b}+\mathbf{b}+\mathbf{a}+\mathbf{a}+\mathbf{a}\end{array}$

$\overrightarrow{\mathrm{GT}}=2\mathbf{b}+3\mathbf{a}\left(\mathrm{or}3\mathbf{a}+2\mathbf{b}\right)$

Part (c)

There are many ways to get from E to K
One option is to go from E to O ($\mathbf{b}$ twice), and then from O to K ( $-\mathbf{a}$ four times)

$\begin{array}{rcl}\overrightarrow{\mathrm{EK}}& =& \overrightarrow{\mathrm{EJ}}+\hspace{0.17em}\overrightarrow{\mathrm{JO}}+\overrightarrow{\mathrm{ON}}+\overrightarrow{\mathrm{NM}}+\overrightarrow{\mathrm{ML}}+\overrightarrow{\mathrm{LK}}\\ & =& \mathbf{b}+\mathbf{b}-\mathbf{a}-\mathbf{a}-\mathbf{a}-\mathbf{a}\end{array}$

$\overrightarrow{\mathrm{EK}}=2\mathbf{b}-4\mathbf{a}\left(\mathrm{or}-4\mathbf{a}+2\mathbf{b}\right)$

The diagram shows a parallelogram $\mathrm{PQRS}$ with a diagonal $\mathrm{QS}$ drawn.

$\overrightarrow{\mathrm{QR}}$ represents vector $\mathbf{a}$ and $\overrightarrow{\mathrm{QS}}$ represents vector $\mathbf{b}$.

a) Express $\overrightarrow{\mathrm{RS}}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

$\mathrm{T}$ is the point such that $\mathrm{PT}=\frac{1}{3}\mathrm{PS}$.

b) Express $\overrightarrow{\mathrm{RT}}$ in terms of $\mathbf{a}$ and $\mathbf{b}$. Give your answer in simplest form.

Answer:

Part (a)

To get from R to S you can go

• the 'wrong way' down vector $\mathbf{a}$ to Q

• then the 'right way' down vector $\mathbf{b}$ to S

$\overrightarrow{\mathrm{RS}}=-\mathbf{a}+\mathbf{b}\left(\mathrm{or}\mathbf{b}-\mathbf{a}\right)$

Part (b)

Write $\overrightarrow{\mathrm{RT}}$ as a vector pathway; one possibility is

$\overrightarrow{\mathrm{RT}}=\overrightarrow{\mathrm{RQ}}+\overrightarrow{\mathrm{QP}}+\overrightarrow{\mathrm{PT}}$

Use what you know about those vectors

• $\overrightarrow{\mathrm{RQ}}=-\overrightarrow{\mathrm{QR}}=-\mathbf{a}$

• Because PQRS is a parallelogram, sides QP and RS are parallel and the same length

  • Therefore $\overrightarrow{\mathrm{QP}}=\overrightarrow{\mathrm{RS}}=-\mathbf{a}+\mathbf{b}$ (using the answer from part (a) )

• Also because PQRS is a parallelogram, sides PS and QR are parallel and the same length

  • And $\mathrm{PT}=\frac{1}{3}\mathrm{PS}$

  • So $\overrightarrow{\mathrm{PT}}=\frac{1}{3}\overrightarrow{\mathrm{PS}}=\frac{1}{3}\overrightarrow{\mathrm{QR}}=\frac{1}{3}\mathbf{a}$

$\overrightarrow{\mathrm{RT}}=-\mathbf{a}+\left(-\mathbf{a}+\mathbf{b}\right)+\frac{1}{3}\mathbf{a}$

Collect like terms and simplify

$\overrightarrow{\mathrm{RT}}=-\frac{5}{3}\mathbf{a}+\mathbf{b}\left(\mathrm{or}\mathbf{b}-\frac{5}{3}\mathbf{a}\right)$

In triangle $\mathrm{ABC}$, $\overrightarrow{\mathrm{AC}}=\left(\begin{array}{c}8\\ -2\end{array}\right)$ and $\overrightarrow{\mathrm{CB}}=\left(\begin{array}{c}-1\\ 7\end{array}\right)$.

a) Express $\overrightarrow{\mathrm{AB}}$ in component form.

$\mathrm{M}$ is the midpoint of $\mathrm{AC}$.

b) Express $\overrightarrow{\mathrm{MB}}$ in component form.

Answer:

Part (a)

Find a path from A to B using vectors whose components you know

• You can get from A to B by going from A to C, then from C to B

$\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{CB}}$

Substitute in the components from the question

$\overrightarrow{\mathrm{AB}}=\left(\begin{array}{c}8\\ -2\end{array}\right)+\left(\begin{array}{c}-1\\ 7\end{array}\right)$

Add the components

$\overrightarrow{\mathrm{AB}}=\left(\begin{array}{c}8+\left(-1\right)\\ -2+7\end{array}\right)$

$\overrightarrow{\mathrm{AB}}=\left(\begin{array}{c}7\\ 5\end{array}\right)$

Part (b)

Write $\overrightarrow{\mathrm{MB}}$ as a vector pathway; one possibility is

$\overrightarrow{\mathrm{MB}}=\overrightarrow{\mathrm{MC}}+\overrightarrow{\mathrm{CB}}$

$\mathrm{M}$ is the midpoint of $\mathrm{AC}$

• That means that $\overrightarrow{\mathrm{MC}}=\frac{1}{2}\overrightarrow{\mathrm{AC}}$

$\overrightarrow{\mathrm{MB}}=\frac{1}{2}\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{CB}}$

Substitute in the components from the question

$\overrightarrow{\mathrm{MB}}=\frac{1}{2}\left(\begin{array}{c}8\\ -2\end{array}\right)+\left(\begin{array}{c}-1\\ 7\end{array}\right)$

Multiply the components of $\left(\begin{array}{c}8\\ -2\end{array}\right)$ by $\frac{1}{2}$

$\overrightarrow{\mathrm{MB}}=\left(\begin{array}{c}4\\ -1\end{array}\right)+\left(\begin{array}{c}-1\\ 7\end{array}\right)$

Add the components

$\overrightarrow{\mathrm{MB}}=\left(\begin{array}{c}4+\left(-1\right)\\ -1+7\end{array}\right)$

$\overrightarrow{\mathrm{MB}}=\left(\begin{array}{c}3\\ 6\end{array}\right)$