# 5.20: Resultant as the Sum of Two Components

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

**Practice**Resultant as the Sum of Two Components

You are working in science class on a "weather unit". As part of this class, you are tasked with going out and checking the wind speed each day at a meter behind your school. The wind speed you record for the day is 20 mph at a \begin{align*}E50^\circ N\end{align*} trajectory. This means that the wind is blowing at an angle \begin{align*}50^\circ\end{align*} taken from the direction that would be due East. At the conclusion of each day, you are supposed to break the wind speed (which is a vector) into two components: the portion that is in a North/South direction and the portion that is in an East/West direction. Can you figure out how to do this?

### Resultant of the Sum of Two Components

We can look at any vector as the resultant of two perpendicular components. If we generalize some vector \begin{align*}\vec{q}\end{align*} into perpendicular components, \begin{align*}|\vec{r}|\hat{i}\end{align*} is the horizontal component of a vector \begin{align*}\vec{q}\end{align*} and \begin{align*}|\vec{s}| \hat{j}\end{align*} is the vertical component of \begin{align*}\vec{q}\end{align*}. Therefore \begin{align*}\vec{r}\end{align*} is a magnitude, \begin{align*}|\vec{r}|\end{align*}, times the unit vector in the \begin{align*}x\end{align*} direction and \begin{align*}\vec{s}\end{align*} is its magnitude, \begin{align*}|\vec{s}|\end{align*}, times the unit vector in the \begin{align*}y\end{align*} direction. The sum of \begin{align*}\vec{r}\end{align*} plus \begin{align*}\vec{s}\end{align*} is: \begin{align*}\vec{r} + \vec{s} = \vec{q}\end{align*}. This addition can also be written as \begin{align*}|\vec{r}|\hat{i} + |\vec{s}| \hat{j} = \vec{q}\end{align*}.

If we are given the vector \begin{align*}\vec{q}\end{align*}, we can find the components of \begin{align*}\vec{q},\vec{r}\end{align*}, and \begin{align*}\vec{s}\end{align*} using trigonometric ratios if we know the magnitude and direction of \begin{align*}\vec{q}\end{align*}.

This is accomplished by taking the magnitude of the vector times the cosine of the vector's angle to find the horizontal component, and the magnitude of the vector times the sine of the vector's angle to find the vertical component.

#### Find the horizontal and vertical components.

If \begin{align*}|\vec{q}| = 19.6\end{align*} and its direction is \begin{align*}73^\circ\end{align*}, find the horizontal and vertical components.

If we know an angle and a side of a right triangle, we can find the other remaining sides using trigonometric ratios. In this case, \begin{align*}\vec{q}\end{align*} is the hypotenuse of our triangle, \begin{align*}\vec{r}\end{align*} is the side adjacent to our \begin{align*}73^\circ\end{align*} angle, \begin{align*}\vec{s}\end{align*} is the side opposite our \begin{align*}73^\circ\end{align*} angle, and \begin{align*}\vec{r}\end{align*} is directed along the \begin{align*}x-\end{align*}axis.

To find \begin{align*}\vec{r}\end{align*}, we will use cosine and to find \begin{align*}\vec{s}\end{align*} we will use sine. Notice this is a scalar equation so all quantities are just numbers. It is written as the quotient of the magnitudes, not the vectors.

\begin{align*}\cos 73 & = \frac{|\vec{r}|}{|\vec{q}|} = \frac{r}{q} && \sin 73 = \frac{|\vec{s}|}{|\vec{q}|} = \frac{s}{q} \\ \cos 73 & = \frac{r}{19.6} && \sin 73 = \frac{s}{19.6} \\ r & = 19.6 \cos 73 && \qquad \ \ s = 19.6 \sin 73 \\ r & = 5.7 && \qquad \ \ s = 18.7\end{align*}

The horizontal component is 5.7 and the vertical component is 18.7. One can rewrite this in vector notation as \begin{align*}5.7\hat{i} + 18.7\hat{j} = \vec{q}\end{align*}. The components can also be written \begin{align*}\vec{q} = \big \langle{5.7, 18.7}\big\rangle\end{align*}, with the horizontal component first, followed by the vertical component. Be careful not to confuse this with the notation for plotted points.

#### Find the horizontal and vertical components.

If \begin{align*}|\vec{m}| = 12.1\end{align*} and its direction is \begin{align*}31^\circ\end{align*}, find the horizontal and vertical components.

To find \begin{align*}\vec{r}\end{align*}, we will use cosine and to find \begin{align*}\vec{s}\end{align*} we will use sine. Notice this is a scalar equation so all quantities are just numbers. It is written as the quotient of the magnitudes, not the vectors.

\begin{align*}\cos 31 & = \frac{|\vec{r}|}{|\vec{m}|} = \frac{r}{m} && \sin 31 = \frac{|\vec{s}|}{|\vec{m}|} = \frac{s}{m} \\ \cos 31 & = \frac{r}{12.1} && \sin 31 = \frac{s}{12.1} \\ r & = 12.1 \cos 31 && \qquad \ \ s = 12.1 \sin 31 \\ r & = 10.37 && \qquad \ \ s = 6.23\end{align*}

#### Find the resultant vector length and angle.

If \begin{align*}|\vec{r}| = 15\end{align*} and \begin{align*}|\vec{s}| = 11\end{align*}, find the resultant vector length and angle.

We can view each of these vectors on the coordinate system here:

Each of these vectors then serves as sides in a right triangle. So we can use the Pythagorean Theorem to find the length of the resultant:

\begin{align*} c^2 = a^2 + b^2\\ c^2 = 15^2 + 11^2\\ c^2 = 225 + 121\\ c^2 = 346\\ c = \sqrt{346} \approx 18.60 \end{align*}

The angle of rotation that the vector makes with the "x" axis can be found using the tangent function:

\begin{align*} \tan \theta = \frac{opposite}{adjacent}\\ \tan \theta = \frac{11}{15}\\ \tan \theta = .73\\ \theta = \tan^{-1}.73\\ \theta \approx 36.13 \end{align*}

### Examples

#### Example 1

Earlier, you were asked to break down vectors into their individual components.

From this Concept you've learned how to take a vector and break it into components using trig functions. If you draw the wind speed you recorded as a vector:

You can find the "x" and "y" components. These are the same as the part of the wind that is blowing to the East and the part of the wind that is blowing to the North.

East component: \begin{align*}\cos 50^\circ = \frac{x}{20}\end{align*}

\begin{align*}20 \cos 50^\circ = x\end{align*}

x = 12.86 mph

North component: \begin{align*}\sin 50^\circ = \frac{y}{20}\end{align*}

\begin{align*}20 \sin 50^\circ = y\end{align*}

y = 15.32 mph

#### Example 2

Find the magnitude of the horizontal and vertical components of the following vector if the resultant vector’s magnitude and direction are given as \begin{align*}\text{magnitude} = 75 \qquad \qquad \quad \text{direction} = 35^\circ\end{align*}.

\begin{align*}\cos 35^\circ = \frac{x}{75}, \sin 35^\circ = \frac{y}{75}, x = 61.4, y = 43\end{align*}

#### Example 3

Find the magnitude of the horizontal and vertical components of the following vector if the resultant vector’s magnitude and direction are given as \begin{align*}\text{magnitude} = 3.4 \qquad \qquad \quad \text{direction} = 162^\circ\end{align*}.

\begin{align*}\cos 162^\circ = \frac{x}{3.4}, \sin 162^\circ = \frac{y}{3.4}, x = 3.2, y = 1.1\end{align*}

#### Example 4

Find the magnitude of the horizontal and vertical components of the following vector if the resultant vector’s magnitude and direction are given as \begin{align*}\text{magnitude} = 15.9 \qquad \qquad \ \text{direction} = 12^\circ\end{align*}.

\begin{align*}\cos 12^\circ = \frac{x}{15.9}, \sin 12^\circ = \frac{y}{15.9}, x = 15.6, y = 3.3\end{align*}

### Review

Find the horizontal and vertical components of the following vectors given the resultant vector’s magnitude and direction.

- \begin{align*}\text{magnitude} = 65 \quad \text{direction} = 22^\circ\end{align*}.
- \begin{align*}\text{magnitude} = 34 \quad \text{direction} = 15^\circ\end{align*}.
- \begin{align*}\text{magnitude} = 29 \quad \text{direction} = 160^\circ\end{align*}.
- \begin{align*}\text{magnitude} = 100 \quad \text{direction} = 320^\circ\end{align*}.
- \begin{align*}\text{magnitude} = 320 \quad \text{direction} = 200^\circ\end{align*}.
- \begin{align*}\text{magnitude} = 15 \quad \text{direction} = 110^\circ\end{align*}.
- \begin{align*}\text{magnitude} = 10 \quad \text{direction} = 80^\circ\end{align*}.
- \begin{align*}\text{magnitude} = 90 \quad \text{direction} = 290^\circ\end{align*}.
- \begin{align*}\text{magnitude} = 87 \quad \text{direction} = 10^\circ\end{align*}.
- \begin{align*}\text{magnitude} = 42 \quad \text{direction} = 150^\circ\end{align*}.

- If \begin{align*}|\vec{r}| = 12\end{align*} and \begin{align*}|\vec{s}| = 8\end{align*}, find the resultant vector magnitude and angle.
- If \begin{align*}|\vec{r}| = 14\end{align*} and \begin{align*}|\vec{s}| = 6\end{align*}, find the resultant vector magnitude and angle.
- If \begin{align*}|\vec{r}| = 9\end{align*} and \begin{align*}|\vec{s}| = 24\end{align*}, find the resultant vector magnitude and angle.
- Will cosine always be used to find the horizontal component of a vector?
- If you know the component form of a vector, how can you find its magnitude and direction?

### Review (Answers)

To see the Review answers, open this PDF file and look for section 5.20.

### Image Attributions

Here you'll learn how to express a vector as the sum of two component vectors.

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