While at summer camp, you are enjoying a tug of war with your friends. You are on one side of the rope, pulling with a force of 200 N. The vector for this force can be represented like this:

You decide to really go for the win and pull as hard as you possibly can. As it turns out, you are pulling with twice the force you were before. Do you know how you can represent the vector for this?

### Multiplying Vectors and Scalars

In working with vectors there are two kinds of quantities employed. The first is the vector, a quantity that has both magnitude and direction. The second quantity is a scalar. Scalars are just numbers. The magnitude of a vector is a scalar quantity. A vector can be multiplied by a real number. This real number is called a **scalar**. The product of a vector \begin{align*}\vec{a}\end{align*} and a scalar \begin{align*}k\end{align*} is a vector, written \begin{align*}\vec{ka}\end{align*}. It has the same direction as \begin{align*}\vec{a}\end{align*} with a magnitude of \begin{align*}k|\vec{a}|\end{align*} if \begin{align*}k > 0\end{align*}. If \begin{align*}k < 0\end{align*}, the vector has the opposite direction of \begin{align*}\vec{a}\end{align*} and a magnitude of \begin{align*}k |\vec{a}|\end{align*}.

#### Find the current vector for wind velocity.

The speed of the wind before a hurricane arrived was 20 mph from the SSE (\begin{align*}N 22.5^\circ W\end{align*}). It quadrupled when the hurricane arrived. What is the current vector for wind velocity?

The wind is coming now at 80 mph from the same direction.

#### Find the current velocity vector of the ship.

A sailboat was traveling at 15 knots due north. After realizing he had overshot his destination, the captain turned the boat around and began traveling twice as fast due south. What is the current velocity vector of the ship?

The ship is traveling at 30 knots in the opposite direction.

If the vector is expressed in coordinates with the starting end of the vector at the origin, this is called standard form. To perform a scalar multiplication, we multiply our scalar by both the coordinates of our vector. The word scalar comes from “scale.” Multiplying by a scalar just makes the vectors longer or shorter, but doesn't change their direction.

#### Multiply the vector and scalar together

Consider the vector from the origin to (4, 6). What would the representation of a vector that had three times the magnitude be?

Here \begin{align*}k = 3\end{align*} and \begin{align*}\vec{v}\end{align*} is the directed segment from (0,0) to (4, 6).

Multiply each of the components in the vector by 3.

\begin{align*}\vec{kv} & = (0,0)\ to\ (12,18)\end{align*}

The new coordinates of the directed segment are (0, 0), (12, 18).

### Examples

#### Example 1

Earlier, you were asked to represent a vector.

Since the original vector looked like this:

The new vector is equal to 2 times the old vector. As we found in this Concept, multiplication of a vector by a scalar doesn't change the direction of the vector, only its magnitude. Therefore, the new vector looks like this:

Its length is twice the length of the original vector, and its direction is unchanged.

#### Example 2

Find the resulting ordered pair that represents \begin{align*}\vec{a}\end{align*} in each equation if you are given \begin{align*}\vec{b} = (0,0)\ to \ (5,4)\end{align*} and \begin{align*}\vec{a} = 2 \vec{b}\end{align*}.

\begin{align*}2 \vec{b} = 2 \big \langle{5, 4}\big\rangle = \big \langle{10, 8}\big\rangle = 10 \hat{i} + 8 \hat{j}\end{align*}

#### Example 3

Find the resulting ordered pair that represents \begin{align*}\vec{a}\end{align*} in each equation if you are given \begin{align*}\vec{b} = (0,0)\ to \ (5,4)\end{align*} and \begin{align*}\vec{a} = - \frac{1}{2} \vec{b}\end{align*}.

\begin{align*}- \frac{1}{2} \vec{c} = - \frac{1}{2} \big \langle{-3, 7}\big\rangle = \big \langle{1.5, -3.5}\big\rangle = 1.5 \hat{i} - 3.5 \hat{j}\end{align*}

#### Example 4

Find the resulting ordered pair that represents \begin{align*}\vec{a}\end{align*} in each equation if you are given \begin{align*}\vec{b} = (0,0)\ to \ (5,4)\end{align*} and \begin{align*}\vec{a} = 0.6\vec{b}\end{align*}.

\begin{align*}0.6\vec{b} = 0.6 \big \langle{5, 4}\big\rangle = \big \langle{3,2.4}\big\rangle = 3 \hat{i} + 2.4 \hat{j}\end{align*}

### Review

\begin{align*}\vec{a}\end{align*} is in standard position with terminal point (1, 5) and \begin{align*}\vec{b}\end{align*} is in standard position with terminal point (4, 2).

- Find the coordinates of the terminal point of \begin{align*}2\vec{a}\end{align*}.
- Find the coordinates of the terminal point of \begin{align*}\frac{1}{2}\vec{b}\end{align*}.
- Find the coordinates of the terminal point of \begin{align*}6\vec{a}\end{align*}.

\begin{align*}\vec{c}\end{align*} is in standard position with terminal point (4, 3) and \begin{align*}\vec{d}\end{align*} is in standard position with terminal point (2, 2).

- Find the coordinates of the terminal point of \begin{align*}3\vec{c} + 2\vec{d}\end{align*}.
- Find the coordinates of the terminal point of \begin{align*}4\vec{c} - 0.3\vec{d}\end{align*}.
- Find the coordinates of the terminal point of \begin{align*}\vec{c} - 3\vec{d}\end{align*}.

\begin{align*}\vec{e}\end{align*} is in standard position with terminal point (3, 2) and \begin{align*}\vec{f}\end{align*} is in standard position with terminal point (-1, 2).

- Find the coordinates of the terminal point of \begin{align*}4\vec{e} + 5\vec{f}\end{align*}.
- Find the coordinates of the terminal point of \begin{align*}3\vec{e} - 3\vec{f}\end{align*}.
- Find the coordinates of the terminal point of \begin{align*}5\vec{e} + \frac{3}{4}\vec{f}\end{align*}.

\begin{align*}\vec{g}\end{align*} is in standard position with terminal point (5, 5) and \begin{align*}\vec{h}\end{align*} is in standard position with terminal point (4, 2).

- Find the coordinates of the terminal point of \begin{align*}\vec{g} + 4\vec{h}\end{align*}.
- Find the coordinates of the terminal point of \begin{align*}5\vec{g} - 2\vec{h}\end{align*}.
- Find the coordinates of the terminal point of \begin{align*}2\vec{g} - 3\vec{h}\end{align*}.

\begin{align*}\vec{i}\end{align*} is in standard position with terminal point (1, 5) and \begin{align*}\vec{j}\end{align*} is in standard position with terminal point (-3, 1).

- Find the coordinates of the terminal point of \begin{align*}3\vec{i} - \vec{j}\end{align*}.
- Find the coordinates of the terminal point of \begin{align*}0.5\vec{i} - 0.6\vec{j}\end{align*}.
- Find the coordinates of the terminal point of \begin{align*}6\vec{i} + 1.2\vec{j}\end{align*}.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 5.17.