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.

Figure 8

Vector subtraction is similar to vector addition. To find u − v, 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.

Figure 9

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).

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.

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]

Figure 11

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]

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.

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].