Imagine you are out for a walk. Initially you walk along one street for some distance, but then you turn onto another street and walk along it for a ways. Both of these movements are different vectors! You move some distance in one direction, then some distance in a different direction. Now suppose you ask, what is the overall distance and direction you travelled? This is a vector addition problem. Adding these vectors produces the net displacement you travel on your walk.
To add vectors graphically, we place them tip to tail, as though you travel along one vector, and then turn to travel along the other. Here is an example:
Above, to add \( \vec{u} \) and \(\vec{v}\), we place the tail of \(\vec{v}\) on the tip of \(\vec{u}\), and the red vector is \( \vec{u} + \vec{v}\).
In practice, this visual method not precise enough to calculate with. Instead, we split the vectors into components to perform the addition.
Let's look back at the vectors we want to add. This time, we can draw on the x component for both vectors (these are the blue vectors along the x axis).
Notice how when we add \(\vec{u}_x\) to \(\vec{v}_x\) by placing them tip to tail, we get the x-component of \(\vec{u} + \vec{v}\). Breaking \(\vec{u}\) and \( \vec{v}\) into x-components can give us the exact x-component of \(\vec{u} + \vec{v}\).
Naturally, we can do the same thing for the y-components to get the y-component of \(\vec{u} + \vec{v}\). This method of adding corresponding components is what we use to calculate with.
Example 1
Given that \(\vec{u} = (1.44, 2)\) and \(\vec{v} = (1.06, -1)\), calculate \(\vec{u} + \vec{v}\).
Solution
Let \(\vec{w} = \vec{u} + \vec{v}\). To find \(\vec{w}_x\) we add \(\vec{u}_x = 1.44\) and \(\vec{v}_x = 1.06\).
\[ \vec{w}_x = 1.44 + 1.06 = 2.5\]
We can find \(\vec{w}_y\) similarly.
\[\vec{w}_x = 2 + -1 = 1 \]
So \(\vec{w} = (2.5, 1)\).
Example 2
Suppose \(\vec{A} = (-3, 5)\), \(\vec{B} = (1.5, 7\)\), and \(\vec{C} = (4, 0.5)\). What is the length of \(\vec{A} + \vec{B} + \vec{C} \)?
Solution
Let \(\vec{D} = \vec{A} + \vec{B} + \vec{C}\). Then \(\vec{D}_x = -3 + 1.5 + 4= 2.5\), \(\vec{C}_y = 5 + 7 + 0.5 = 12.5\). To calculate the length of \(\vec{C}\), we use the same method as covered in the Vector Components page and apply the Pythagorean theorem. So the length is \(\sqrt{(2.5)^2 + 12.5^2} = 12.75\). Thus the length of \(\vec{A} + \vec{B} + \vec{C} = 12.75\).
Note:
In both of these examples, the vector components were given to you already. If they are not given, you will need to calculate the components first and then perform the addition.
Example 3
Suppose \(\vec{V}\) has length 10 at an angle of 20° and \(\vec{W}\) has length 6 at an angle of 290°. What is the magnitude and direction of their sum?
Solution
Let \(\vec{Z}\) be the resulting vector. We are not given the x and y components for \(\vec{V}\) and \(\vec{W}\), so we need to calculate them.
\[ \vec{V}_x = 10 \cos 20° = 9.40 \]
\[ \vec{V}_y = 10 \sin 20° = 3.42 \]
\[ \vec{W}_x = 6 \cos 290° = 2.05 \]
\[ \vec{W}_y = 6 \sin 290° = -5.63 \]
Now we add:
\[ \vec{Z}_x = \vec{V}_x + \vec{W}_x = 9.40 + 3.42 = 12.82 \]
\[ \vec{Z}_y = \vec{V}_y + \vec{W}_y = 2.05 - 5.63 = -3.88 \]
To calculate the magnitude, we use the Pythagorean theorem:
\[ | \vec{Z} | = \sqrt{12.82^2 + (-3.88)^2} = 13.39 \]
Now we need to find the angle of the resulting vector. Since we have the x and y components for \(\vec{Z}\), we can use the tan inverse to get the angle:
\[ \tan^-1\left(\frac{-3.88}{12.82}\right) = -16.8° = -16.8° + 360° = 343.2° \]
Thus \(\vec{Z}\) has magnitude 13.39 at an angle of 343.2°.