Adding Geometric Vectors Tail to Tail

Adding vectors results in a new Resultant Vector. Geometric and algebraic vectors can be added using different techniques.

Adding Geometric Vectors

Geometric vectors can be added by positioning the vectors Head to Tail or Tail to Tail. Remember, you can move a vector around in space as long as it maintains its magnitude and direction.

Adding Geometric Vectors Tail to Tail

To add vectors tail to tail, move one vector so that it's tail starts at the tail of the previous vector. Next, make a parallelogram. The Resultant Vector starts at the tail of the vectors and ends at the opposite side of the parallelogram.

Below are the steps to add vectors \(\vec{u}\) and \(\vec{v}\) Tail to Tail. Sometimes, this method is referred to the Parallelogram Law of Vector Addition.

Vectors u and v arranged tail to tail with resultant vector.

Notice that this method is very similar to the Triangle Law of Vector Addition. By moving vector \(\vec{v}\) over to the head of \(\vec{u}\) makes a triangle. However, sometimes the exact direction of the vectors is unknown. Instead, you may be provided information about the angle between the vectors. That is when the Parallelogram Law of Vector Addition is most useful.

Adding Geometric vectors involves solving triangles using trigonometry equations. Below is a summary:

Right Triangles
Pythagorean Theorem	\(c^2 = a^2 + b^2 \)
SOH CAH TOA	\(\sin(\theta) = \cfrac{\text{opposite}}{\text{hypotenuse}} = \cfrac{y}{r}\) \(\cos(\theta) = \cfrac{\text{adjacent}}{\text{hypotenuse}} = \cfrac{x}{r} \) \(\tan(\theta) = \cfrac{\text{opposite}}{\text{adjacent}} = \cfrac{y}{x}\)

Other Triangles
Cosine Law	\(c^2 = a^2 + b^2 - 2 a b \cos(\theta)\)
Sine Law	\(\cfrac{\sin(A)}{a} = \cfrac{\sin(B)}{b} = \cfrac{\sin(C)}{c} \)

Review these lessons:

Try these questions:

Frac and Michelle take their son Marco tobogganing. On the way to the slopes, Marco sits on the toboggan and Frac and Michelle each pull on a string attached to the front. Frac applies a force of \(45 \; [\text{N}]\) away from the center of the toboggan. Michelle applies a force of \(26 \; [\text{N}]\) away from the center of the toboggan in the other direction. The angle between the strings is \(60^\circ\). What is the resultant force applied to Marco on the toboggan?

Show Answer

Consider the top view of the toboggan with force vectors. Next, arrange the vectors tail to tail:

Force vectors arranged next to each other with magnitudes and angles.

Force vectors arranged tail to tail with magnitudes, angles, and resultant vector.

We need to know the angle between the two vectors. For a parallelogram, this is simply \(\textcolor{green}{180^\circ - 60^\circ = 120^\circ} \). We can use Sine Law to calculate the magnitude:

\(||\vec{R}||^2 = 45^2 + 26^2 - (2)(45)(26) \cos(120^\circ)\)

\( ||\vec{R}|| = \sqrt{3871} \approx 62 \; [N]\)

For this question, we don't have a lot of information of the direction of the specific vectors. We can assume the angle to the \(x\)-axis is equal (\(30^\circ\)). Use Sine Law to find the angle:

\(\cfrac{\sin(\theta)}{26} = \cfrac{\sin(120^\circ)}{62} \)

\(\theta = \sin^{-1}\left(\sin(120^\circ) \cfrac{45}{62}\right) \approx 21^\circ \)

The angle to the \(x\)-axis is \(30^\circ - 21^\circ = 9^\circ\)

\(\vec{R} = 62 \; [N] \; 9^\circ \)

Since we don't know which direction the vectors are pointing (only that there is between them), it would actually be more convienent for us if the \(x\)-axis was in the direction of one of the vectors. We can rotate the \(x\) and \(y\)-axes (yes! you actually can!) to make the \(x\)-axis parallel with the \(26 \; [\text{N}]\) vector.

Alternatively, we can rotate the vectors:

Force vectors arranged head to tail and rotated.

This is a handy trick that can simplify our work!

To find the direction of the Resultant Vector, use the sine law to find the angle of the left corner:

\(\cfrac{\sin(\theta)}{45} = \cfrac{\sin(120^\circ)}{62} \)

\(\theta = \sin^{-1}\left(\sin(120^\circ) \cfrac{45}{62}\right) \approx 39^\circ \)

\(\vec{R} = 62 \; [\text{N}] \; 39^\circ \)

Therefore, we can determine the Resultant Vector is \(\boldsymbol{62 \; [\textbf{N}] \; 39^\circ}\).