Hello maybealways Originally Posted by

**maybealways** I'm having troubles with constructing vectors. I understand and can apply the basics of vectors: dot products, null vector, unit vector, adding and subtracting vectors, scalar multiplication, etc... Which is fine and all, but when I'm faced with something relatively simple, like...

$\displaystyle

If \

\vec{AB} = \begin{pmatrix} 3 \\ -1 \end{pmatrix} and\

\vec{CA} = \begin{pmatrix} -1 \\ 2 \end{pmatrix},\

find\ CB

$

or

$\displaystyle

If \

\vec{PQ} = \begin{pmatrix} -4 \\ 1 \end{pmatrix} and\

\vec{RQ} = \begin{pmatrix} -1 \\ 2 \end{pmatrix} and\

\vec{RS} = \begin{pmatrix} 2 \\ -3 \end{pmatrix},\

find\ SP

$

^When I see those types of questions, my head becomes confused. It's not that I don't know how to go about answering them, but more like, I can't get my head around how the vectors are constructed so I can't use/find the right numbers to find the vector (I hope that made sense).

I've searched around numerous sites/textbooks, but nothing that illuminates, or what I'm looking for... What I really want is a book/website that has a lot of questions and examples about the construction of these vectors (geometrically and algebraically), but that probably isn't going to happen.

Anyway, I'd love any sort of enlightenment on the construction of vectors, whether algebraically or geometrically! I'd appreciate any help!

You can use simple words to translate these things. Use the following 'vocabulary':(1) The vector $\displaystyle \vec{AB}$ translates as "a movement from $\displaystyle A$ to $\displaystyle B$"

(2) $\displaystyle -\vec{AB}$ translates as "the opposite of $\displaystyle \vec{AB}$" or "and then the opposite of $\displaystyle \vec{AB}$"; in other words "a movement from $\displaystyle B$ to $\displaystyle A$".

(3) $\displaystyle =$ translates as "is the same as" (No surprise there!)

(4) $\displaystyle +$ translates as "and then" or "followed by"

So we can write the vector Law of Addition like this:$\displaystyle \underbrace{\vec{AC}}_{\text{a movement from A to C ...}} \underbrace{=}_{\text{ is the same as ...}}\underbrace{\vec{AB}}_{\text{ a movement from A to B ...}}\underbrace{+}_{\text{and then ...}}\underbrace{\vec{BC}}_{\text{ a movement from B to C }}$

And, using (2) above, we can subtract the vector $\displaystyle \vec{BC}$ from both sides of this equation to get:$\displaystyle \underbrace{\vec{AC}}_{\text{a movement from A to C ...}} \underbrace{-}_{\text{ and then the opposite of ...}}\underbrace{\vec{BC}}_{\text{ a movement from B to C ...}}$$\displaystyle \underbrace{=}_{\text{is the same as ...}}\underbrace{\vec{AB}}_{\text{ a movement from A to B}}$

We can now begin to solve the problems above. In the first one we want $\displaystyle \vec{CB}$ - that's the movement from $\displaystyle C$ to $\displaystyle B$. If we go from $\displaystyle C$ to $\displaystyle B$ via $\displaystyle A$, we get:$\displaystyle \underbrace{\vec{CB}}_{\text{a movement from C to B ...}} \underbrace{=}_{\text{ is the same as ...}}\underbrace{\vec{CA}}_{\text{ a movement from C to A ...}}\underbrace{+}_{\text{followed by ...}}\underbrace{\vec{AB}}_{\text{ a movement from A to B }}$

$\displaystyle \Rightarrow \vec{CB}=\binom{-1}{2}+\binom{3}{-1}$ $\displaystyle =\binom21$

In the second example above, you have to find a 'route' from $\displaystyle S$ to $\displaystyle P$ via $\displaystyle R$ and $\displaystyle Q$, using the movements you've been given: that is $\displaystyle \vec{PQ}$, $\displaystyle \vec{RQ}$ and $\displaystyle \vec{RS}$. Bearing in mind that we can reverse the directions of any of these movements by using a minus sign, we want:$\displaystyle \underbrace{\vec{SP}}_{\text{a movement from S to P ...}}\underbrace{=}_{\text{is the same as ...}} \underbrace{\vec{SR}}_{\text{ a movement from S to R ...}}$$\displaystyle \underbrace{+}_{\text{ and then ...}}\underbrace{\vec{RQ}}_{\text{ a movement from R to Q }}\underbrace{+}_{\text{ and then ...}}\underbrace{\vec{QP}}_{\text{ a movement from Q to P }}$

Reversing the signs where necessary (to take account of the vectors we've been given in the question) this is:$\displaystyle \vec{SP}=-\vec{RS}+\vec{RQ}-\vec{PQ}$$\displaystyle =-\binom{2}{-3}+\binom{-1}{2}-\binom{-4}{1}$

$\displaystyle =\binom{-2-1+4}{3+2-1}$

$\displaystyle = \binom{1}{4}$

I hope that helps to make things clearer.

Grandad