Performing Vector Addition and Scalar Multiplication

Now that we understand the properties of vectors, we can perform operations involving them. While it is convenient to think of the vector [latex]u[/latex] [latex]=\langle x,y\rangle[/latex] as an arrow or directed line segment from the origin to the point [latex]\left(x,y\right)[/latex], vectors can be situated anywhere in the plane. The sum of two vectors u and v, or vector addition, produces a third vector u+ v, the resultant vector.

To find u + v, we first draw the vector u, and from the terminal end of u, we drawn the vector v. In other words, we have the initial point of v meet the terminal end of u. This position corresponds to the notion that we move along the first vector and then, from its terminal point, we move along the second vector. The sum u + v is the resultant vector because it results from addition or subtraction of two vectors. The resultant vector travels directly from the beginning of u to the end of v in a straight path, as shown in Figure 8.

Diagrams of vector addition and subtraction.

Figure 8

Vector subtraction is similar to vector addition. To find uv, view it as u + (−v). Adding −v is reversing direction of v and adding it to the end of u. The new vector begins at the start of u and stops at the end point of −v. See Figure 9 for a visual that compares vector addition and vector subtraction using parallelograms.

Showing vector addition and subtraction with parallelograms. For addition, the base is u, the side is v, the diagonal connecting the start of the base to the end of the side is u+v. For subtraction, thetop is u, the side is -v, and the diagonal connecting the start of the top to the end of the side is u-v.

Figure 9

Example 5: Adding and Subtracting Vectors

Given [latex]u[/latex] [latex]=\langle 3,-2\rangle[/latex] and [latex]v[/latex] [latex]=\langle -1,4\rangle[/latex], find two new vectors u + v, and uv.

Solution

To find the sum of two vectors, we add the components. Thus,

[latex]\begin{array}{l}u+v=\langle 3,-2\rangle +\langle -1,4\rangle \hfill \\ =\langle 3+\left(-1\right),-2+4\rangle \hfill \\ =\langle 2,2\rangle \hfill \end{array}[/latex]

See Figure 10(a).

To find the difference of two vectors, add the negative components of [latex]v[/latex] to [latex]u[/latex]. Thus,

[latex]\begin{array}{l}u+\left(-v\right)=\langle 3,-2\rangle +\langle 1,-4\rangle \hfill \\ =\langle 3+1,-2+\left(-4\right)\rangle \hfill \\ =\langle 4,-6\rangle \hfill \end{array}[/latex]

See Figure 10(b).

Further diagrams of vector addition and subtraction.

Figure 10. (a) Sum of two vectors (b) Difference of two vectors

Multiplying By a Scalar

While adding and subtracting vectors gives us a new vector with a different magnitude and direction, the process of multiplying a vector by a scalar, a constant, changes only the magnitude of the vector or the length of the line. Scalar multiplication has no effect on the direction unless the scalar is negative, in which case the direction of the resulting vector is opposite the direction of the original vector.

A General Note: Scalar Multiplication

Scalar multiplication involves the product of a vector and a scalar. Each component of the vector is multiplied by the scalar. Thus, to multiply [latex]v[/latex] [latex]=\langle a,b\rangle[/latex] by [latex]k[/latex] , we have

[latex]kv=\langle ka,kb\rangle[/latex]

Only the magnitude changes, unless [latex]k[/latex] is negative, and then the vector reverses direction.

Example 6: Performing Scalar Multiplication

Given vector [latex]v[/latex] [latex]=\langle 3,1\rangle[/latex], find 3v, [latex]\frac{1}{2}[/latex] [latex]v[/latex], and −v.

Solution

See Figure 11 for a geometric interpretation. If [latex]v[/latex] [latex]=\langle 3,1\rangle[/latex], then

[latex]\begin{array}{l}3v=\langle 3\cdot 3,3\cdot 1\rangle \hfill \\ =\langle 9,3\rangle \hfill \\ \frac{1}{2}v=\langle \frac{1}{2}\cdot 3,\frac{1}{2}\cdot 1\rangle \hfill \\ =\langle \frac{3}{2},\frac{1}{2}\rangle \hfill \\ -v=\langle -3,-1\rangle \hfill \end{array}[/latex]
Showing the effect of scaling a vector: 3x, 1x, .5x, and -1x. The 3x is three times as long, the 1x stays the same, the .5x halves the length, and the -1x reverses the direction of the vector but keeps the length the same. The rest keep the same direction; only the magnitude changes.

Figure 11

Analysis of the Solution

Notice that the vector 3v is three times the length of v, [latex]\frac{1}{2}[/latex] [latex]v[/latex] is half the length of v, and –v is the same length of v, but in the opposite direction.

Try It 2

Find the scalar multiple 3 [latex]u[/latex] given [latex]u[/latex] [latex]=\langle 5,4\rangle[/latex].

Solution

Example 7: Using Vector Addition and Scalar Multiplication to Find a New Vector

Given [latex]u=\langle 3,-2\rangle[/latex] and [latex]v=\langle -1,4\rangle[/latex], find a new vector w = 3u + 2v.

Solution

First, we must multiply each vector by the scalar.

[latex]\begin{array}{l}3u=3\langle 3,-2\rangle \hfill \\ =\langle 9,-6\rangle \hfill \\ 2v=2\langle -1,4\rangle \hfill \\ =\langle -2,8\rangle \hfill \end{array}[/latex]

Then, add the two together.

[latex]\begin{array}{l}w=3u+2v\hfill \\ =\langle 9,-6\rangle +\langle -2,8\rangle \hfill \\ =\langle 9 - 2,-6+8\rangle \hfill \\ =\langle 7,2\rangle \hfill \end{array}[/latex]

So, [latex]w=\langle 7,2\rangle[/latex].

Finding Component Form

In some applications involving vectors, it is helpful for us to be able to break a vector down into its components. Vectors are comprised of two components: the horizontal component is the [latex]x[/latex] direction, and the vertical component is the [latex]y[/latex] direction. For example, we can see in the graph in Figure 12 that the position vector [latex]\langle 2,3\rangle[/latex] comes from adding the vectors v1 and v2. We have v1 with initial point [latex]\left(0,0\right)[/latex] and terminal point [latex]\left(2,0\right)[/latex].

[latex]\begin{array}{l}{v}_{1}=\langle 2 - 0,0 - 0\rangle \hfill \\ =\langle 2,0\rangle \hfill \end{array}[/latex]

We also have v2 with initial point [latex]\left(0,0\right)[/latex] and terminal point [latex]\left(0,3\right)[/latex].

[latex]\begin{array}{l}{v}_{2}=\langle 0 - 0,3 - 0\rangle \hfill \\ =\langle 0,3\rangle \hfill \end{array}[/latex]

Therefore, the position vector is

[latex]\begin{array}{l}v=\langle 2+0,3+0\rangle \hfill \\ =\langle 2,3\rangle \hfill \end{array}[/latex]

Using the Pythagorean Theorem, the magnitude of v1 is 2, and the magnitude of v2 is 3. To find the magnitude of v, use the formula with the position vector.

[latex]\begin{array}{l}\hfill \\ \begin{array}{l}|v|=\sqrt{|{v}_{1}{|}^{2}+|{v}_{2}{|}^{2}}\hfill \\ \begin{array}{l}=\sqrt{{2}^{2}+{3}^{2}}\hfill \\ =\sqrt{13}\hfill \end{array}\hfill \end{array}\hfill \end{array}[/latex]

The magnitude of v is [latex]\sqrt{13}[/latex]. To find the direction, we use the tangent function [latex]\tan \theta =\frac{y}{x}[/latex].

[latex]\begin{array}{l}\tan \theta =\frac{{v}_{2}}{{v}_{1}}\hfill \\ \tan \theta =\frac{3}{2}\hfill \\ \theta ={\tan }^{-1}\left(\frac{3}{2}\right)=56.3^\circ \hfill \end{array}[/latex]
Diagram of a vector in root position with its horizontal and vertical components.

Figure 12

Thus, the magnitude of [latex]v[/latex] is [latex]\sqrt{13}[/latex] and the direction is [latex]{56.3}^{\circ }[/latex] off the horizontal.

Example 8: Finding the Components of the Vector

Find the components of the vector [latex]v[/latex] with initial point [latex]\left(3,2\right)[/latex] and terminal point [latex]\left(7,4\right)[/latex].

Solution

First find the standard position.

[latex]\begin{array}{l}v=\langle 7 - 3,4 - 2\rangle \hfill \\ =\langle 4,2\rangle \hfill \end{array}[/latex]

See the illustration in Figure 13.

Diagram of a vector in root position with its horizontal (4,0) and vertical (0,2) components.

Figure 13

The horizontal component is [latex]{v}_{1}=\langle 4,0\rangle[/latex] and the vertical component is [latex]{v}_{2}=\langle 0,2\rangle[/latex].