Adding Geometric Vectors Head to Tail

Zaina's displacement is the Resultant Vector of her travels. It includes both the distance and direction.

First, let's arrange the vectors head to tail. The Resultant Vector will start at the tail of the first vector and end at the head of the second:

Since these vectors are perpendicular (\(90^{\circ}\)) this is a right angle triangle and we can use Pythagorean theorem to find the magnitude of \(R\):

\( \vec{R} = 5 \; [\text{km}] \; N + 2 \; [\text{km}] \; E \)

\( ||\vec{R}||^2 = 5^2 + 2^2\)

\( ||\vec{R}|| = \sqrt{5^2 + 2^2}\)

\( ||\vec{R}|| = \sqrt{25 + 4}\)

\( ||\vec{R}|| = \sqrt{29} \approx 5.4 \; [\text{km}]\)

Next, we need to find the direction. Solve for \( \theta \):

\(\tan\theta = \cfrac{2}{5}\)

\(\theta = \tan^{-1}\left({\cfrac{2}{5}}\right)\)

\(\theta \approx 22^\circ\)

Zaina's displacement is:

Overall, Zaina travels \(5.4 \; [\text{km}]\) away from her starting point. Notice that this is different from the distance she travelled:

Sometimes it is convienient to describe the final angle in relation to an original vector. For example, if Zaina walked \(NE\) and then \(SE\) then the vectors would look like this:

Let's define \(\vec{u} = 5 \; [\text{km}] \; N \theta E \) and \(\vec{v} = 2 \; [\text{km}] \; E \phi S \). Where \( \theta \) and \( \phi \) are angles. Then:

\(\vec{R} = 5.4 \; [\text{km}] \; 22^\circ\) clockwise from \(\vec{u}\)

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

We need to know the angle between the two vectors. By drawing another \(x\) and \(y\)-axis we can see that the angle made by the first vector and the \(y\)-axis is \( \textcolor{green}{60^\circ} \) and the second vector and the \(y\)-axis is \( \textcolor{red}{30^\circ} \). The sum is \(90^\circ\) so we can use Pythagorean Theorem:

\( ||\vec{R}||^2 = 45^2 + 26^2\)

\( ||\vec{R}|| = \sqrt{45^2 + 26^2}\)

\( ||\vec{R}|| = \sqrt{2025 + 676}\)

\( ||\vec{R}|| = \sqrt{2701} \approx 52 \; [\text{N}]\)

The direction can be found by finding the angle in the left corner:

\(\tan\theta = \cfrac{26}{45}\)

\(\theta = \tan^{-1}\left(\cfrac{26}{45}\right)\)

\(\theta \approx 30^\circ\)

This means that the resulant vector right along the \(x\)-axis. Thus, the toboggan will travel in a straight line!

Let's see what happens when the angles change.

Now the angle between the vectors is \(120^\circ \). We can use Sine Law to calculate the magnitude:

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

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

We can already see by drawing the Resultant Vector that it is not in the same direction as the \(x\)-axis. Use Sine Law to find the angle of the left corner:

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

\(\theta = \sin^{-1}\left(\sin(120^\circ) \cfrac{26}{62}\right) \approx 21^\circ\)

The angle between the \(x\)-axis and Resultant Vector is \(30^\circ - 21^\circ = 9^\circ \). In general, the angles are referenced clockwise from the positive \(x\)-axis.

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

We can move the order of addition (Associative Law of Addition) to allow us to use the collapsing shortcut:

\(=\vec{AB} + \vec{BD} + \vec{DE} \)

\(=\vec{A \require{cancel}\cancel {B}} + \vec{\cancel{B}D} + \vec{DE} \)

\(=\vec{AD} + \vec{DE} \)

\(=\vec{A\cancel{D}} + \vec{\cancel{D}E} \)

\(=\vec{AE} \)