# 5.21: Resultant as Magnitude and Direction

**At Grade**Created by: CK-12

**Practice**Resultant as Magnitude and Direction

You are out playing soccer with friends after school one day when you and a friend kick the soccer ball at the same time. However, you kick the ball with 55 N of force with a vector like this:

and your friend kicks the ball with 70 N of force with a vector like this:

The angle between the two vectors is \begin{align*}74^\circ\end{align*}

Can you represent the net force on the ball? What will the resultant vector be like?

### Resultant of a Magnitude and Direction

If we don’t have two perpendicular vectors, we can still find the magnitude and direction of the resultant without a graphic estimate with a construction using a compass and ruler. This can be accomplished using both the Law of Sines and the Law of Cosines.

#### Find the magnitude and direction

\begin{align*}\vec{A}\end{align*} makes a \begin{align*}54^\circ\end{align*} angle with \begin{align*}\vec{B}\end{align*}. The magnitude of \begin{align*}\vec{A}\end{align*} is 13.2. The magnitude of \begin{align*}\vec{B}\end{align*} is 16.7. Find the magnitude and direction the resultant makes with the smaller vector.

There is no preferred orientation such as a compass direction or any necessary use of \begin{align*}x\end{align*} and \begin{align*}y\end{align*} coordinates. The problem can be solved without the use of unit vectors.

In order to solve this problem, we will need to use the parallelogram method. Since vectors only have magnitude and direction, one can move them on the plane to any position one wishes, as long as the magnitude and direction remain the same. First, we will complete the parallelogram: Label the vectors. Move \begin{align*}\vec{b}\end{align*} so its tail is on the tip of \begin{align*}\vec{a}\end{align*}. Move \begin{align*}\vec{a}\end{align*} so its tail is on the tip of \begin{align*}\vec{b}\end{align*}. This makes a parallelogram because the angles did not change during the translation. Put in labels for the vertices of the parallelogram.

Since opposite angles in a parallelogram are congruent, we can find\begin{align*}\angle A\end{align*}.

\begin{align*}\angle{CBD} + \angle{CAD} + \angle{ACB} + \angle{BDA} & = 360 \\ 2 \angle{CBD} + 2 \angle{ACB} & = 360 \\ \angle{ACB} & = 54^\circ \\ 2 \angle{CBD} & = 360 - 2(54) \\ \angle{CBD} & = \frac{360-2(54)}{2} = 126 \end{align*}

Now, we know two sides and the included angle in an oblique triangle. This means we can use the Law of Cosines to find the magnitude of our resultant.

\begin{align*}x^2 & = 13.2^2 + 16.7^2 - 2(13.2)(16.7) \cos 126 \\ x^2 & = 712.272762 \\ x & = 26.7\end{align*}

To find the direction, we can use the Law of Sines since we now know an angle and a side across from it. We choose the Law of Sines because it is a proportion and less computationally intense than the Law of Cosines.

\begin{align*}\frac{\sin \theta}{16.7} & = \frac{\sin 126}{26.7} \\ \sin \theta & = \frac{16.7 \sin 126}{26.7} \\ \sin \theta & = 0.5060143748 \\ \theta & = \sin ^{-1} \ 0.5060 = 30.4^\circ\end{align*}

The magnitude of the resultant is 26.7 and the direction it makes with the smaller vector is \begin{align*}30.4^\circ\end{align*} counterclockwise.

We can use a similar method to add three or more vectors.

#### Find the magnitude and direction of the vectors

Vector A makes a \begin{align*}45^\circ\end{align*} angle with the horizontal and has a magnitude of 3. Vector B makes a \begin{align*}25^\circ\end{align*} angle with the horizontal and has a magnitude of 5. Vector \begin{align*}C\end{align*} makes a \begin{align*}65^\circ\end{align*} angle with the horizontal and has a magnitude of 2. Find the magnitude and direction (with the horizontal) of the resultant of all three vectors.

To begin this problem, we will find the resultant using Vector \begin{align*}A\end{align*} and Vector \begin{align*}B\end{align*}. We will do this using the parallelogram method like we did above.

Since Vector \begin{align*}A\end{align*} makes a \begin{align*}45^\circ\end{align*} angle with the horizontal and Vector \begin{align*}B\end{align*} makes a \begin{align*}25^\circ\end{align*} angle with the horizontal, we know that the angle between the two \begin{align*}(\angle{ADB})\end{align*} is \begin{align*}20^\circ\end{align*}.

To find \begin{align*}\angle{DBE}\end{align*}:

\begin{align*}2 \angle{ADB} + 2 \angle{DBE} & = 360 \\ \angle{ADB} & = 20^\circ \\ 2 \angle{DBE} & = 360 - 2(20) \\ \angle{DBE} & = \frac{360-2(20)}{2} = 160\end{align*}

Now, we will use the Law of Cosines to find the magnitude of \begin{align*}DE\end{align*}.

\begin{align*}DE^2 & = 3^2 + 5^2 - 2(3)(5) \cos 160 \\ DE^2 & = 62 \\ DE & = 7.9\end{align*}

Next, we will use the Law of Sines to find the measure of \begin{align*}\angle EDB\end{align*}.

\begin{align*}\frac{\sin 160}{7.9} & = \frac{\sin \angle{EDB}}{3} \\ \sin \angle{EDB} & = \frac{3 \sin 160}{7.9} \\ \sin \angle{EDB} & = .1299 \\ \angle{EDB} & = \sin^{-1} \ 0.1299 = 7.46^\circ\end{align*}

We know that Vector \begin{align*}B\end{align*} forms a \begin{align*}25^\circ\end{align*} angle with the horizontal so we add that value to the measure of \begin{align*}\angle{EDB}\end{align*} to find the angle \begin{align*}DE\end{align*} makes with the horizontal. Therefore, \begin{align*}DE\end{align*} makes a \begin{align*}32.46^\circ\end{align*} angle with the horizontal.

Next, we will take \begin{align*}DE\end{align*}, and we will find the resultant vector of \begin{align*}DE\end{align*} and Vector \begin{align*}C\end{align*} from above. We will repeat the same process we used above.

Vector \begin{align*}C\end{align*} makes a \begin{align*}65^\circ\end{align*} angle with the horizontal and \begin{align*}DE\end{align*} makes a \begin{align*}32^\circ\end{align*} angle with the horizontal. This means that the angle between the two \begin{align*}(\angle{CDE})\end{align*} is \begin{align*}33^\circ\end{align*}. We will use this information to find the measure of \begin{align*}\angle{DEF}\end{align*}.

\begin{align*}2\angle{CDE} + 2\angle{DEF} & = 360 \\ \angle{CDE} & = 33^\circ \\ 2 \angle{DEF} & = 360 - 2(33) \\ \angle{DEF} & = \frac{360-2(33)}{2} = 147\end{align*}

Now we will use the Law of Cosines to find the magnitude of \begin{align*}DF\end{align*}.

\begin{align*}DF^2 & = 7.9^2 + 2^2 - 2(7.9)(2) \cos 147 \\ DF^2 & = 92.9 \\ DF & = 9.6 \end{align*}

Next, we will use the Law of Sines to find \begin{align*}\angle{FDE}\end{align*}.

\begin{align*}\frac{\sin 147}{9.6} & = \frac{\sin \angle{FDE}}{2} \\ \sin \angle{FDE} & = \frac{2 \sin 147}{9.6} \\ \sin \angle{FDE} & = .1135 \\ \angle{FDE} & = \sin^{-1} 0.1135 = 6.5^\circ = 7^\circ\end{align*}

Finally, we will take the measure of \begin{align*}\angle{FDE}\end{align*} and add it to the \begin{align*}32^\circ\end{align*} angle that \begin{align*}DE\end{align*} forms with the horizontal. Therefore, DF forms a \begin{align*}39^\circ\end{align*} angle with the horizontal.

#### Find the magnitude of the resultant

Two forces of 310 lbs and 460 lbs are acting on an object. The angle between the two forces is \begin{align*}61.3^\circ\end{align*}. What is the magnitude of the resultant? What angle does the resultant make with the smaller force?

We do not need unit vectors here as there is no preferred direction like a compass direction or a specific axis. First, to find the magnitude we will need to figure out the other angle in our parallelogram.

\begin{align*}2 \angle{ACB} + 2 \angle{CAD} & = 360 \\ \angle{ACB} & = 61.3^\circ \\ 2 \angle{CAD} & = 360 - 2(61.3) \\ \angle{CAD} & = \frac{360 - 2 (61.3)}{2} = 118.7\end{align*}

Now that we know the other angle, we can find the magnitude using the Law of Cosines.

\begin{align*}x^2 & = 460^2 + 310^2 - 2(460)(310) \cos 118.7^\circ \\ x^2 & = 444659.7415 \\ x & = 667\end{align*}

To find the angle the resultant makes with the smaller force, we will use the Law of Sines.

\begin{align*}\frac{\sin \theta}{460} & = \frac{\sin 118.7}{666.8} \\ \sin \theta & = \frac{460 \sin 118.7}{666.8} \\ \sin \theta & = .6049283888 \\ \theta & = \sin^{-1}\ 0.6049 = 37.2^\circ\end{align*}

### Examples

#### Example 1

Earlier, you were asked to represent the net force on the ball.

As you've seen in this Concept, you can represent the vector resulting from both of your forces as the resultant of vector addition. Since vectors only have magnitude and direction, one can move them on the plane to any position one wishes, as long as the magnitude and direction remain the same. First, we will complete the parallelogram: Label the vectors. Move \begin{align*}\vec{b}\end{align*} so its tail is on the tip of \begin{align*}\vec{a}\end{align*}. Move \begin{align*}\vec{a}\end{align*} so its tail is on the tip of \begin{align*}\vec{b}\end{align*}. This makes a parallelogram because the angles did not change during the translation. Put in labels for the vertices of the parallelogram.

Since opposite angles in a parallelogram are congruent, we can find angle \begin{align*}A\end{align*}.

\begin{align*}\angle{CBD} + \angle{CAD} + \angle{ACB} + \angle{BDA} & = 360 \\ 2 \angle{CBD} + 2 \angle{ACB} & = 360 \\ \angle{ACB} & = 72^\circ \\ 2 \angle{CBD} & = 360 - 2(72) \\ \angle{CBD} & = \frac{360-2(72)}{2} = 108 \end{align*}

Now, we know two sides and the included angle in an oblique triangle. This means we can use the Law of Cosines to find the magnitude of our resultant.

\begin{align*}x^2 & = 70^2 + 55^2 - 2(70)(55) \cos 72 \\ x^2 & = 5545.569 \\ x & = 74.47\end{align*}

To find the direction, we can use the Law of Sines since we now know an angle and a side across from it. We choose the Law of Sines because it is a proportion and less computationally intense than the Law of Cosines.

\begin{align*}\frac{\sin \theta}{70} & = \frac{\sin 108}{74.47} \\ \sin \theta & = \frac{70 \sin 108}{74.47} \\ \sin \theta & = 0.89397 \\ \theta & = \sin ^{-1} \ 0.89397 = 63.68^\circ\end{align*}

The combined force of your kick along with your friend's is 74.47 Newtons and the direction it makes with your kick is \begin{align*}63.68^\circ\end{align*} counterclockwise.

#### Example 2

Forces of 140 Newtons and 186 Newtons act on an object. The angle between the forces is \begin{align*}43^\circ\end{align*}. Find the magnitude of the resultant and the angle it makes with the larger force.

magnitude \begin{align*}= 304, 18.3^\circ\end{align*} between resultant and larger force

#### Example 3

An airplane is traveling at a speed of 155 km/h. It's heading is set at \begin{align*}83^\circ\end{align*} while there is a 42.0 km/h wind from \begin{align*}305^\circ\end{align*}. What is the airplane's actual heading?

Recall that headings and angles in triangles are complementary. So, an \begin{align*}83^\circ\end{align*} heading translates to \begin{align*}7^\circ\end{align*} from the horizontal. Adding that to \begin{align*}35^\circ\end{align*} (\begin{align*}270^\circ\end{align*} from \begin{align*}305^\circ\end{align*}) we get \begin{align*}42^\circ\end{align*} for two of the angles in the parallelogram. So, the other angles in the parallelogram measure \begin{align*}138^\circ\end{align*} each, \begin{align*}\frac{360-2(42)}{2}\end{align*}. Using \begin{align*}138^\circ\end{align*} in the Law of Cosines, we can find the diagonal or resultant, \begin{align*}x^2 = 42^2 + 155^2 - 2(42)(155) \cos 138\end{align*}, so \begin{align*}x = 188.3\end{align*}. We then need to find the angle between the resultant and the speed using the Law of Sines. \begin{align*}\frac{\sin a}{42} = \frac{\sin 138}{188.3}\end{align*}, so \begin{align*}a = 8.6^\circ\end{align*}. To find the actual heading, this number needs to be added to \begin{align*}83^\circ\end{align*}, getting \begin{align*}91.6^\circ\end{align*}.

#### Example 4

If \begin{align*}\overrightarrow{AB}\end{align*} is any vector, what is \begin{align*}\overrightarrow{AB} + \overrightarrow{BA}\end{align*}?

\begin{align*}BA\end{align*} is the same vector as \begin{align*}AB\end{align*}, but because it starts with \begin{align*}B\end{align*} it is in the opposite direction. Therefore, when you add the two together, you will get (0,0).

### Review

\begin{align*}\vec{a}\end{align*} makes a \begin{align*}42^\circ\end{align*} angle with \begin{align*}\vec{b}\end{align*}. The magnitude of \begin{align*}\vec{a}\end{align*} is 15. The magnitude of \begin{align*}\vec{b}\end{align*} is 22.

- Find the magnitude of the resultant.
- Find the angle of the resultant makes with the smaller vector.

\begin{align*}\vec{c}\end{align*} makes a \begin{align*}80^\circ\end{align*} angle with \begin{align*}\vec{d}\end{align*}. The magnitude of \begin{align*}\vec{c}\end{align*} is 70. The magnitude of \begin{align*}\vec{d}\end{align*} is 45.

- Find the magnitude of the resultant.
- Find the angle of the resultant makes with the smaller vector.

\begin{align*}\vec{e}\end{align*} makes a \begin{align*}50^\circ\end{align*} angle with \begin{align*}\vec{f}\end{align*}. The magnitude of \begin{align*}\vec{e}\end{align*} is 32. The magnitude of \begin{align*}\vec{f}\end{align*} is 10.

- Find the magnitude of the resultant.
- Find the angle of the resultant makes with the smaller vector.

\begin{align*}\vec{g}\end{align*} makes a \begin{align*}100^\circ\end{align*} angle with \begin{align*}\vec{h}\end{align*}. The magnitude of \begin{align*}\vec{g}\end{align*} is 50. The magnitude of \begin{align*}\vec{h}\end{align*} is 35.

- Find the magnitude of the resultant.
- Find the angle of the resultant makes with the smaller vector.
- Two forces of 100 lbs and 120 lbs are acting on an object. The angle between the two forces is \begin{align*}50^\circ\end{align*}. What is the magnitude of the resultant?
- Using the information from the previous problem, what angle does the resultant make with the larger force?
- A force of 50 lbs acts on an object at an angle of \begin{align*}30^\circ\end{align*}. A second force of 75 lbs acts on the object at an angle of \begin{align*}-10^\circ\end{align*}. What is the magnitude of the resultant force?
- Using the information from the previous problem, what is the direction of the resultant force?
- A plane is flying on a bearing of \begin{align*}30^\circ\end{align*} at a speed of 450 mph. A wind is blowing with the bearing \begin{align*}200^\circ\end{align*} at 50 mph. What is the plane's actual direction?
- Vector A makes a \begin{align*}30^\circ\end{align*} angle with the horizontal and has a magnitude of 4. Vector B makes a \begin{align*}55^\circ\end{align*} angle with the horizontal and has a magnitude of 6. Vector \begin{align*}C\end{align*} makes a \begin{align*}75^\circ\end{align*} angle with the horizontal and has a magnitude of 3. Find the magnitude and direction (with the horizontal) of the resultant of all three vectors.
- Vector A makes a \begin{align*}12^\circ\end{align*} angle with the horizontal and has a magnitude of 5. Vector B makes a \begin{align*}25^\circ\end{align*} angle with the horizontal and has a magnitude of 2. Vector \begin{align*}C\end{align*} makes a \begin{align*}60^\circ\end{align*} angle with the horizontal and has a magnitude of 7. Find the magnitude and direction (with the horizontal) of the resultant of all three vectors.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 5.21.

### Image Attributions

Here you'll learn how to express a vector as a combination of the vector's magnitude (length) and its direction (which way it is pointing).

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