## Understanding the situation

Let's see how the addition of multiple vectors works in an specific example.

## Presenting a sample problem

Given that

$$\overrightarrow{a} = \hat{i} + 2\hat{j},\;\;\; \overrightarrow{b}=-3\hat{j}, \;\;\;\overrightarrow{c}=4\hat{i}$$

For the following vector operation, find the results both algebraically and show the meaning geometrically.

$$2\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}$$

## Solving this problem

#### An algebraic solution

To add vectors algebraically, we just replace the vectors by how they are expressed in terms of the unit vectors and then rearrange to collect terms.

$$2\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = 2 \times(1\hat{i}+2\hat{j}) + (-3\hat{j}) + (4\hat{i})$$

$$2\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = (2\times 1 + 0 + 4)\hat{i} + (2\times 2 -3 + 0) \hat{j}$$

$$2\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} =6\hat{i} + 1 \hat{j} = 6\hat{i} + \hat{j}$$.

Note two tricky bits that are obvious once you see them but can be confusing when you are getting started:

When we have just a unit vector by itself ($\hat{i}$ or $\hat{j}$) we mean "1" times that vector. If we're adding we have to put it in.

When we have no component in a vector, it corresponds to 0 amount of that vector.

These are perhaps more obvious when we use the parenthesis notation that is commonly used when you are doing geometry in an algebra class: $(x\hat{i} + y\hat{j}) \rightarrow (x,y)$. This is only convenient if we are never going to change what we mean by $\hat{i}$ or $\hat{j}$. Since in this class we do want to be free to use different choices of directions as our unit vectors, we will mostly use $\hat{i}$ and $\hat{j}$

#### A geometric solution

Now let's see what it looks like geometrically.

Conceptually, the idea is to use the displacement metaphor: The idea of vectors is built on the physical idea of displacement. You start from the tail and move to the head.

Adding vectors is like doing the displacements one after the other. You put the tail of the second at the head of the first and follow it to the second head. The sum is the displacement from the first tail to the second head.

In our case, we'll do it in two steps since we have three vectors to add. The geometry looks like this:

In the left panel above we've added $2\overrightarrow{a} + \overrightarrow{b}$. We've drawn the original vectors $\overrightarrow{a}$ and $\overrightarrow{a}$ in lightly in blue and red respectively both starting at the origin. We've then doubled $\overrightarrow{a}$ (deep blue) and shifted $\overrightarrow{b}$ so that its tail starts at the head of $2\overrightarrow{a}$ — just as if you had walked from the origin to the head of the vector $2\overrightarrow{a}$ and then started to walk the vector $\overrightarrow{b}$. The result, shown in green, is $2\overrightarrow{a} + \overrightarrow{b}$.

Next we add the vector $\overrightarrow{c}$ to that result. Again, we draw the original vector in lightly (in purple) and shift it so it follows on as a tail-to-head displacement beginning at the end of $2\overrightarrow{a} + \overrightarrow{b}$. The result, shown in black, is found by counting across the x axis to get the x component: 6; and counting up the y axis to get the y component: 1. So the result is $6\hat{i} + \hat{j}$, the same as we got from our algebraic analysis.

Joe Redish 12/26/14

Article 499