You and your friends are at a ski weekend for your school. While skiing, you go down a hill that is rather steep. You decide to use a vector to represent your motion from the top of the hill going down. Counting the top of the hill as the origin, you ski down a slope and measure how far your "x" and "y" positions have changed. As it turns out, you can represent this displacement with the vector \begin{align*}(12, -256)\end{align*}

### Translating Vectors and Slopes

Vectors with the same magnitude and direction are equal. This means that the same ordered pair could represent many different vectors.

For instance, the ordered pair (4, 8) can represent a vector in standard position where the initial point is at the origin and the terminal point is at (4, 8). This vector could be thought of as the resultant of a horizontal vector with a magnitude or 4 units and a vertical vector with a magnitude of 8 units. Therefore, any vector with a horizontal component of 4 and vertical component of 8 could also be represented by the ordered pair (4, 8).

If you think back to Algebra, you know that the slope of a line is the change in \begin{align*}y\end{align*}

#### Consider the vector from (4, 7) to (12, 11). What would the representation of a vector that had 2.5 times the magnitude be?

Here, \begin{align*}k = 2.5\end{align*}

Mathematically, two vectors are equal if their direction and magnitude are the same. The positions of the vectors do not matter. This means that if we have a vector that is not in standard position, we can translate it to the origin. The initial point of \begin{align*}\vec{v}\end{align*}**translate** this to the origin, we would need to add (-4, -7) to both the initial and terminal points of the vector.

Initial point: \begin{align*}(4, 7) + (-4, -7) = (0,0)\end{align*}

Terminal point: \begin{align*}(12, 11) + (-4, -7) = (8, 4)\end{align*}

Now, to calculate \begin{align*}\vec{kv}\end{align*}

\begin{align*}\vec{kv} & = (2.5(8), 2.5(4)) \\
\vec{kv} & = (20,10)\end{align*}

The new coordinates of the directed segment are (0, 0) and (20, 10). To translate this back to our original terminal point:

Initial point: \begin{align*}(0, 0) + (4, 7) = (4, 7) \end{align*}

Terminal point: \begin{align*}(20, 10) + (4, 7) = (24, 17)\end{align*}

The new coordinates of the directed segment are (4, 7) and (24, 17).

#### What is the slope of a vector starting from the origin with terminal coordinates (5 , 7)?

Since the slope is defined as the change in "y" divided by the change in "x", we can find the slope of this vector:

\begin{align*}
\frac{\Delta y}{\Delta x} = \frac{7 - 0}{5 - 0} = \frac{7}{5} = 1.4
\end{align*}

#### Find the new coordinates of the vectors

A vector starts at the origin and has terminal coordinates (11 , 17). What would the new coordinates of the tail and tip of the vector be if the vector were shifted 15 units along the "x" axis?

The vector maintains the same orientation in space, it is just moved down the "x" axis. Therefore, only the "x" coordinates of the vector's tail and tip change.

So the new coordinates of the tail of the vector are:

\begin{align*}
(0 + 15, 0) = (15,0)
\end{align*}

And the new coordinates of the tip are:

\begin{align*}
(11 + 15, 17) = (26,17)
\end{align*}

### Examples

#### Example 1

Earlier, you were asked to calculate the incline (slope) of the hill you came down.

Since the slope is defined as the change in "y" divided by the change in "x", we can find the slope of the vector representing your trip down the hill:

\begin{align*}
\frac{\Delta y}{\Delta x} = \frac{-256 - 0}{12 - 0} = \frac{-256}{12} \approx -21.33
\end{align*}

This means that for every foot the hill changed in the "x" direction, it went down 21.33 feet in the "y" direction. That's a steep hill indeed!

#### Example 2

Find the magnitude of the horizontal and vertical components of the following vector given the following coordinates of their initial and terminal points.

\begin{align*}\text{initial} = (-3,8) \qquad \qquad \quad \text{terminal} = (2, -1)\end{align*}

The vector needs to be translated to (0,0). Also, recall that magnitudes are always positive.

\begin{align*}(-3, 8) + (3, -8) = (0, 0) \qquad \qquad (2, -1) + (3, -8) = (5, -9)\end{align*}

horizontal \begin{align*}= 5\end{align*}

#### Example 3

Find the magnitude of the horizontal and vertical components of the following vector given the following coordinates of their initial and terminal points.

\begin{align*}\text{initial} = (7, 13) \qquad \qquad \quad \ \text{terminal} = (11, 19)\end{align*}

The vector needs to be translated to (0,0). Also, recall that magnitudes are always positive.

\begin{align*}(7, 13) + (-7, -13) = (0, 0)\qquad \qquad (11, 19) + (-7, -13) = (4, 6)\end{align*}

horizontal \begin{align*}= 4\end{align*}

#### Example 4

Find the magnitude of the horizontal and vertical components of the following vector given the following coordinates of their initial and terminal points.

\begin{align*}\text{initial} = (4.2, -6.8) \qquad \quad \ \text{terminal} = (-1.3, -9.4)\end{align*}

The vector needs to be translated to (0,0). Also, recall that magnitudes are always positive.

\begin{align*}(4.2, -6.8) + (-4.2, 6.8) = (0, 0) \qquad \qquad (-1.3, -9.4) + (-4.2, 6.8) = (-5.5, -2.6)\end{align*}

horizontal \begin{align*}= 5.5\end{align*}

### Review

In each question below, the initial and terminal coordinates for a vector are given. If the vector is translated so that it is in standard position (with the initial point at the origin), what are the new terminal coordinates?

- initial (2, 5) and terminal (7, -1)
- initial (4, 3) and terminal (3, -5)
- initial (8, 1) and terminal (-4, 7)
- initial (-2, 7) and terminal (3, 5)
- initial (4, -3) and terminal (4, 3)
- initial (0, 2) and terminal (6, -4)

Find the slope of each vector below with the given terminal coordinates. Assume the vector is in standard position.

- terminal (6, 7)
- terminal (3, 6)
- terminal (-2, 4)
- terminal (5, 8)
- terminal (1, 3)

Find the magnitude of the horizontal and vertical components of each vector given the coordinates of their initial and terminal points.

- initial (1, 5) and terminal (1, -3)
- initial (4, 5) and terminal (6, -5)
- initial (6, 1) and terminal (-4, 4)
- initial (-2, 3) and terminal (2, 5)

### Review (Answers)

To see the Review answers, open this PDF file and look for section 5.18.