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A Geometric View of Vectors

A vector is a specific quantity drawn as a line segment with an arrowhead at one end. It has an initial point, where it begins, and a terminal point, where it ends. A vector is defined by its magnitude, or the length of the line, and its direction, indicated by an arrowhead at the terminal point. Thus, a vector is a directed line segment. There are various symbols that distinguish vectors from other quantities:

• Lower case, boldfaced type, with or without an arrow on top such as [latex]v,\,\,u,\,\,w,\,\,\stackrel{\to }{v},\,\,\stackrel{\to }{u},\,\stackrel{\to }{w}.[/latex]
• Given initial point[latex]\,P\,[/latex]and terminal point[latex]\,Q,\,[/latex]a vector can be represented as[latex]\,\stackrel{\to }{PQ}\,.\,\,[/latex]The arrowhead on top is what indicates that it is not just a line, but a directed line segment.
• Given an initial point of[latex]\,\left(0,0\right)\,[/latex]and terminal point[latex]\,\left(a,b\right),\,[/latex]a vector may be represented as[latex]〈a,b〉.[/latex]

This last symbol [latex]〈a,b〉[/latex] has special significance. It is called the standard position. The position vector has an initial point [latex]\left(0,0\right)\,[/latex]and a terminal point[latex]〈a,b〉.[/latex]To change any vector into the position vector, we think about the change in the x-coordinates and the change in the y-coordinates. Thus, if the initial point of a vector[latex]\,\stackrel{\to }{CD}\,[/latex]is[latex]\,C\left({x}_{1},{y}_{1}\right)\,[/latex]and the terminal point is[latex]\,D\left({x}_{2},{y}_{2}\right),\,[/latex]then the position vector is found by calculating

In (Figure), we see the original vector[latex]\,\stackrel{\to }{CD}\,[/latex]and the position vector[latex]\,\stackrel{\to }{AB}.[/latex]

Find the Position Vector

Consider the vector whose initial point is[latex]\,P\left(2,3\right)\,[/latex]and terminal point is[latex]\,Q\left(6,4\right).\,[/latex]Find the position vector.

Show Solution

The position vector is found by subtracting one x-coordinate from the other x-coordinate, and one y-coordinate from the other y-coordinate. Thus

[latex]\begin{array}{l}v=〈6-2,4-3〉\hfill \\ \,\,\,=〈4,1〉\hfill \end{array}[/latex]

The position vector begins at[latex]\,\left(0,0\right)\,[/latex]and terminates at[latex]\,\left(4,1\right).\,[/latex]The graphs of both vectors are shown in (Figure).

We see that the position vector is[latex]〈4,1〉.[/latex]

Drawing a Vector with the Given Criteria and Its Equivalent Position Vector

Find the position vector given that vector[latex]\,v\,[/latex]has an initial point at [latex]\,\left(-3,2\right)\,[/latex]and a terminal point at[latex]\,\left(4,5\right),\,[/latex]then graph both vectors in the same plane.

Show Solution

The position vector is found using the following calculation:

[latex]\begin{array}{l}v=〈4-\left(-3\right),5-2〉\hfill \\ \text{ }=〈7,3〉\hfill \end{array}[/latex]

Thus, the position vector begins at[latex]\,\left(0,0\right)\,[/latex]and terminates at[latex]\,\left(7,3\right).\,[/latex]See (Figure).

Finding Magnitude and Direction

To work with a vector, we need to be able to find its magnitude and its direction. We find its magnitude using the Pythagorean Theorem or the distance formula, and we find its direction using the inverse tangent function.

Magnitude and Direction of a Vector

Given a position vector[latex]\,v[/latex][latex]=〈a,b〉,[/latex]the magnitude is found by[latex]|v|=\sqrt{{a}^{2}+{b}^{2}}.[/latex]The direction is equal to the angle formed with the x-axis, or with the y-axis, depending on the application. For a position vector, the direction is found by[latex]\,\mathrm{tan}\,\theta =\left(\frac{b}{a}\right)⇒\theta ={\mathrm{tan}}^{-1}\left(\frac{b}{a}\right),\,[/latex]as illustrated in (Figure).

Two vectors v and u are considered equal if they have the same magnitude and the same direction. Additionally, if both vectors have the same position vector, they are equal.

Finding the Magnitude and Direction of a Vector

Find the magnitude and direction of the vector with initial point[latex]\,P\left(-8,1\right)\,[/latex]and terminal point[latex]\,Q\left(-2,-5\right).[/latex]Draw the vector.

Show Solution

First, find the position vector.

[latex]\begin{array}{l}u=〈-2,-\left(-8\right),-5-1〉\hfill \\ \text{ }=〈6,-6〉\hfill \end{array}[/latex]

We use the Pythagorean Theorem to find the magnitude.

[latex]\begin{array}{l}|u|=\sqrt{{\left(6\right)}^{2}+{\left(-6\right)}^{2}}\hfill \\ \,\,\,\,\,\,=\sqrt{72}\hfill \\ \,\,\,\,\,\,=6\sqrt{2}\hfill \end{array}[/latex]

The direction is given as

[latex]\begin{array}{l}\mathrm{tan}\,\theta =\frac{-6}{6}=-1⇒\theta ={\mathrm{tan}}^{-1}\left(-1\right)\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,=-45°\hfill \end{array}[/latex]

However, the angle terminates in the fourth quadrant, so we add 360° to obtain a positive angle. Thus,[latex]\,-45°+360°=315°.\,[/latex]See (Figure).

Showing That Two Vectors Are Equal

Show that vector v with initial point at[latex]\,\left(5,-3\right)\,[/latex]and terminal point at[latex]\,\left(-1,2\right)\,[/latex]is equal to vector u with initial point at[latex]\,\left(-1,-3\right)\,[/latex]and terminal point at[latex]\,\left(-7,2\right).\,[/latex]Draw the position vector on the same grid as v and u. Next, find the magnitude and direction of each vector.

Show Solution

As shown in (Figure), draw the vector[latex]\,v\,[/latex]starting at initial[latex]\,\left(5,-3\right)\,[/latex]and terminal point[latex]\,\left(-1,2\right).\,[/latex]Draw the vector[latex]\,u\,[/latex]with initial point[latex]\,\left(-1,-3\right)\,[/latex]and terminal point[latex]\,\left(-7,2\right).\,[/latex]Find the standard position for each.

Next, find and sketch the position vector for v and u. We have

[latex]\begin{array}{l}v=〈-1-5,2-\left(-3\right)〉\hfill \\ \text{ }=〈-6,5〉\hfill \\ \hfill \\ u=〈-7-\left(-1\right),2-\left(-3\right)〉\hfill \\ \text{ }=〈-6,5〉\hfill \end{array}[/latex]

Since the position vectors are the same, v and u are the same.

An alternative way to check for vector equality is to show that the magnitude and direction are the same for both vectors. To show that the magnitudes are equal, use the Pythagorean Theorem.

[latex]\begin{array}{l}|v|=\sqrt{{\left(-1-5\right)}^{2}+{\left(2-\left(-3\right)\right)}^{2}}\hfill \\ \,\,\,\,\,\,=\sqrt{{\left(-6\right)}^{2}+{\left(5\right)}^{2}}\hfill \\ \,\,\,\,\,\,=\sqrt{36+25}\hfill \\ \,\,\,\,\,\,=\sqrt{61}\hfill \\ |u|=\sqrt{{\left(-7-\left(-1\right)\right)}^{2}+{\left(2-\left(-3\right)\right)}^{2}}\hfill \\ \,\,\,\,\,\,=\sqrt{{\left(-6\right)}^{2}+{\left(5\right)}^{2}}\hfill \\ \,\,\,\,\,\,=\sqrt{36+25}\hfill \\ \,\,\,\,\,\,=\sqrt{61}\hfill \end{array}[/latex]

As the magnitudes are equal, we now need to verify the direction. Using the tangent function with the position vector gives

[latex]\begin{array}{l}\mathrm{tan}\,\theta =-\frac{5}{6}⇒\theta ={\mathrm{tan}}^{-1}\left(-\frac{5}{6}\right)\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,=-39.8°\hfill \end{array}[/latex]

However, we can see that the position vector terminates in the second quadrant, so we add[latex]\,180°.\,[/latex]Thus, the direction is[latex]\,-39.8°+180°=140.2°.[/latex]

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]=〈x,y〉[/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).

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) for a visual that compares vector addition and vector subtraction using parallelograms.

Adding and Subtracting Vectors

Given [latex]u[/latex][latex]=〈3,-2〉[/latex]and[latex]v[/latex][latex]=〈-1,4〉,[/latex]find two new vectors u + v, and u − v.

Show Solution

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

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

See (Figure)(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)=〈3,-2〉+〈1,-4〉\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=〈3+1,-2+\left(-4\right)〉\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=〈4,-6〉\hfill \end{array}[/latex]

See (Figure)(b).

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.

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]=〈a,b〉[/latex] by [latex]k[/latex], we have

[latex]kv=〈ka,kb〉[/latex]

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

Performing Scalar Multiplication

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

Show Solution

See (Figure) for a geometric interpretation. If[latex]\,v[/latex][latex]=〈3,1〉,[/latex]then

[latex]\begin{array}{l}\,\,3v=〈3\cdot 3,3\cdot 1〉\hfill \\ \,\,\,\,\,\,\,\,=〈9,3〉\hfill \\ \,\frac{1}{2}v=〈\frac{1}{2}\cdot 3,\frac{1}{2}\cdot 1〉\hfill \\ \,\,\,\,\,\,\,\,=〈\frac{3}{2},\frac{1}{2}〉\hfill \\ -v=〈-3,-1〉\hfill \end{array}[/latex]

Analysis

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.

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) that the position vector[latex]〈2,3〉[/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]

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

Therefore, the position vector is

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.

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

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

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]

Show Solution

First find the standard position.

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

See the illustration in (Figure).

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

Finding the Unit Vector in the Direction of v

In addition to finding a vector’s components, it is also useful in solving problems to find a vector in the same direction as the given vector, but of magnitude 1. We call a vector with a magnitude of 1 a unit vector. We can then preserve the direction of the original vector while simplifying calculations.

Unit vectors are defined in terms of components. The horizontal unit vector is written as [latex]i[/latex][latex]=〈1,0〉[/latex]and is directed along the positive horizontal axis. The vertical unit vector is written as[latex]j[/latex][latex]=〈0,1〉[/latex]and is directed along the positive vertical axis. See (Figure).

Finding the Unit Vector in the Direction of v

Find a unit vector in the same direction as [latex]v[/latex][latex]=〈-5,12〉.[/latex]

Show Solution

First, we will find the magnitude.

[latex]\begin{array}{l}|v|=\sqrt{{\left(-5\right)}^{2}+{\left(12\right)}^{2}}\hfill \\ \,\,\,\,\,\,=\sqrt{25+144}\hfill \\ \,\,\,\,\,\,=\sqrt{169}\hfill \\ \,\,\,\,\,\,=13\hfill \end{array}[/latex]

Then we divide each component by[latex]\,|v|,\,[/latex]which gives a unit vector in the same direction as v:

[latex]\frac{v}{|v|}=-\frac{5}{13}i+\frac{12}{13}j[/latex]

or, in component form

[latex]\frac{v}{|v|}=〈-\frac{5}{13},\frac{12}{13}〉[/latex]

See (Figure).

Verify that the magnitude of the unit vector equals 1. The magnitude of[latex]\,-\frac{5}{13}i+\frac{12}{13}j\,[/latex]is given as

[latex]\begin{array}{l}\sqrt{{\left(-\frac{5}{13}\right)}^{2}+{\left(\frac{12}{13}\right)}^{2}}=\sqrt{\frac{25}{169}+\frac{144}{169}}\hfill \\ \text{ }=\sqrt{\frac{169}{169}}=1\hfill \end{array}[/latex]

The vector u[latex]=\frac{5}{13}[/latex]i[latex]+\frac{12}{13}[/latex]j is the unit vector in the same direction as v[latex]=〈-5,12〉.[/latex]

Performing Operations with Vectors in Terms of i and j

So far, we have investigated the basics of vectors: magnitude and direction, vector addition and subtraction, scalar multiplication, the components of vectors, and the representation of vectors geometrically. Now that we are familiar with the general strategies used in working with vectors, we will represent vectors in rectangular coordinates in terms of i and j.

Vectors in the Rectangular Plane

Given a vector[latex]\,v\,[/latex]with initial point[latex]\,P=\left({x}_{1},{y}_{1}\right)\,[/latex]and terminal point [latex]Q=\left({x}_{2},{y}_{2}\right),[/latex] v is written as

[latex]v=\left({x}_{2}-{x}_{1}\right)i+\left({y}_{2}-{y}_{1}\right)j[/latex]

The position vector from[latex]\,\left(0,0\right)\,[/latex]to[latex]\,\left(a,b\right),\,[/latex]where[latex]\,\left({x}_{2}-{x}_{1}\right)=a\,[/latex]and[latex]\,\left({y}_{2}-{y}_{1}\right)=b,\,[/latex]is written as v = ai + bj. This vector sum is called a linear combination of the vectors i and j.

The magnitude of v = ai + bj is given as[latex]\,|v|=\sqrt{{a}^{2}+{b}^{2}}.\,[/latex]See (Figure).