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Vectors

View vectors geometrically.

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 \(v,\) \(u,\) \(w,\) \(\overset{\to}{v},\) \(\overset{\to}{u},\ \overset{\to}{w}.\)
Given initial point \(P\) and terminal point \(Q,\) a vector can be represented as \(\overset{\to}{PQ}.\) The arrowhead on top is what indicates that it is not just a line, but a directed line segment.
Given an initial point of \((0,0)\) and terminal point \((ab),\) a vector may be represented as \(〈a,b〉.\)

This last symbol \(〈a,b〉\) has special significance. It is called the standard position. The position vector has an initial point \((0,0)\) and a terminal point \((a,b).\) 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 \(\overset{\to}{CD}\) is \(C({x}_{1},{y}_{1})\) and the terminal point is \(D({x}_{2},{y}_{2}),\) then the position vector is found by calculating

\[\begin{array}{l}\overset{\to}{AB}=〈{x}_{2}-{x}_{1},{y}_{2}-{y}_{1}〉 \\ =〈a,b〉\end{array}\]

In , we see the original vector \(\overset{\to}{CD}\) and the position vector \(\overset{\to}{AB}.\)

Example

Try it.

Consider the vector whose initial point is \(P(2,3)\) and terminal point is \(Q(6,4).\) Find the position vector.

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

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

The position vector begins at \((0,0)\) and terminates at \((4,1).\) The graphs of both vectors are shown in .

We see that the position vector is \(〈4,1〉.\)

Condensed — the full section is in OpenStax Algebra and Trigonometry 2e.

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.

Example

Try it.

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

Solution

First, find the position vector.

\[\begin{array}{l}u=〈-2,-(-8),-5-1〉 \\ =〈6,-6〉\end{array}\]

We use the Pythagorean Theorem to find the magnitude.

\[\begin{array}{l}|u|=\sqrt{{(6)}^{2}+{(-6)}^{2}} \\ =\sqrt{72} \\ =6\sqrt{2}\end{array}\]

The direction is given as

\[\begin{array}{l}\tan \ \theta =\frac{-6}{6}=-1⇒\theta ={\tan }^{-1}(-1) \\ =-45^{\circ}\end{array}\]

However, the angle terminates in the fourth quadrant, so we add 360° to obtain a positive angle. Thus, \(-45^{\circ}+360^{\circ}=315^{\circ}.\) See .

Condensed — the full section is in OpenStax Algebra and Trigonometry 2e.

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 \(u\) \(=〈x,y〉\) as an arrow or directed line segment from the origin to the point \((x,y),\) 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 .

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

Example

Try it.

Given \(u\) \(=〈3,-2〉\) and \(v\) \(=〈-1,4〉,\) find two new vectors u + v, and u − v.

Solution

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

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

See (a).

To find the difference of two vectors, add the negative components of \(v\) to \(u.\) Thus,

\[\begin{array}{l}u+(-v)=〈3,-2〉+〈1,-4〉 \\ =〈3+1,-2+(-4)〉 \\ =〈4,-6〉\end{array}\]

See (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.

Example

Try it.

Given vector \(v\) \(=〈3,1〉,\) find 3v, \(\frac{1}{2}\) \(v,\) and −v.

Solution

See for a geometric interpretation. If \(v\) \(=〈3,1〉,\) then

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

Example

Try it.

Given \(u\) \(=〈3,-2〉\) and \(v\) \(=〈-1,4〉,\) find a new vector w = 3u + 2v.

Solution

First, we must multiply each vector by the scalar.

\[\begin{array}{l}3u=3〈3,-2〉 \\ =〈9,-6〉 \\ 2v=2〈-1,4〉 \\ =〈-2,8〉\end{array}\]

Then, add the two together.

\[\begin{array}{l}w=3u+2v \\ =〈9,-6〉+〈-2,8〉 \\ =〈9-2,-6+8〉 \\ =〈7,2〉\end{array}\]

So, \(w\) \(=〈7,2〉.\)

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 \(x\) direction, and the vertical component is the \(y\) direction. For example, we can see in the graph in that the position vector \(〈2,3〉\) comes from adding the vectors v₁ and v₂. We have v₁ with initial point \((0,0)\) and terminal point \((2,0).\)

\[\begin{array}{l}{v}_{1}=〈2-0,0-0〉 \\ =〈2,0〉\end{array}\]

We also have v₂ with initial point \((0,0)\) and terminal point \((0,\ 3).\)

\[\begin{array}{l}{v}_{2}=〈0-0,3-0〉 \\ =〈0,3〉\end{array}\]

Therefore, the position vector is

\[\begin{array}{l}v=〈2+0,3+0〉 \\ =〈2,3〉\end{array}\]

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.

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

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

\[\begin{array}{l}\tan \ \theta =\frac{\left|{v}_{2}\right|}{\left|{v}_{1}\right|} \\ \tan \ \theta =\frac{3}{2} \\ \theta ={\tan }^{-1}(\frac{3}{2})=56.3^{\circ}\end{array}\]

Thus, the magnitude of \(v\) is \(\sqrt{13}\) and the direction is \({56.3}^{∘}\) off the horizontal.

Example

Try it.

Find the components of the vector \(v\) with initial point \((3,2)\) and terminal point \((7,4).\)

Solution

First find the standard position.

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

See the illustration in .

The horizontal component is \({v}_{1}\) \(=〈4,0〉\) and the vertical component is \({v}_{2}\) \(=〈0,2〉.\)

Finding the Unit Vector in the Direction of

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 \(i\) \(=〈1,0〉\) and is directed along the positive horizontal axis. The vertical unit vector is written as \(j\) \(=〈0,1〉\) and is directed along the positive vertical axis. See .

Example

Try it.

Find a unit vector in the same direction as \(v\) \(=〈-5,12〉.\)

Solution

First, we will find the magnitude.

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

Then we divide each component by \(|v|,\) which gives a unit vector in the same direction as v:

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

or, in component form

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

See .

Verify that the magnitude of the unit vector equals 1. The magnitude of \(-\frac{5}{13}i+\frac{12}{13}j\) is given as

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

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

Performing Operations with Vectors in Terms of

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.

Example

Try it.

Given a vector \(v\) with initial point \(P=(2,-6)\) and terminal point \(Q=(-6,6),\) write the vector in terms of \(i\) and \(j.\)

Solution

Begin by writing the general form of the vector. Then replace the coordinates with the given values.

\[\begin{array}{l}v=({x}_{2}-{x}_{1})i+({y}_{2}-{y}_{1})j \\ =(-6-2)i+(6-(-6))j \\ =-8i+12j\end{array}\]

Example

Try it.

Given initial point \({P}_{1}=(-1,3)\) and terminal point \({P}_{2}=(2,7),\) write the vector \(v\) in terms of \(i\) and \(j.\)

Solution

Begin by writing the general form of the vector. Then replace the coordinates with the given values.

\[\begin{array}{l}v=({x}_{2}-{x}_{1})i+({y}_{2}-{y}_{1})j \\ v=(2-(-1))i+(7-3)j \\ =3i+4j\end{array}\]

Performing Operations on Vectors in Terms of

When vectors are written in terms of \(i\) and \(j,\) we can carry out addition, subtraction, and scalar multiplication by performing operations on corresponding components.

Example

Try it.

Find the sum of \({v}_{1}=2i-3j\) and \({v}_{2}=4i+5j.\)

Solution

According to the formula, we have

\[\begin{array}{l}{v}_{1}+{v}_{2}=(2+4)i+(-3+5)j \\ =6i+2j\end{array}\]

Calculating the Component Form of a Vector: Direction

We have seen how to draw vectors according to their initial and terminal points and how to find the position vector. We have also examined notation for vectors drawn specifically in the Cartesian coordinate plane using \(i\ \text{and}\ j.\) For any of these vectors, we can calculate the magnitude. Now, we want to combine the key points, and look further at the ideas of magnitude and direction.

Calculating direction follows the same straightforward process we used for polar coordinates. We find the direction of the vector by finding the angle to the horizontal. We do this by using the basic trigonometric identities, but with \(|v|\) replacing \(r.\)

Example

Try it.

Given a vector with length 7 and an angle of 135°, write it in component form.

Solution

Using the conversion formulas \(x=|v|\cos \ \theta\) and \(y=|v|\sin \ \theta ,\) we find that

\[\begin{array}{l}x=7\cos (135^{\circ}) \\ =-\frac{7\sqrt{2}}{2} \\ y=7\sin (135^{\circ}) \\ =\frac{7\sqrt{2}}{2}\end{array}\]

This vector can be written as \(v=7\cos (135^{\circ})+7\sin (135^{\circ})\) or simplified as

\[v=-\frac{7\sqrt{2}}{2}i+\frac{7\sqrt{2}}{2}j\]

Finding the Dot Product of Two Vectors

As we discussed earlier in the section, scalar multiplication involves multiplying a vector by a scalar, and the result is a vector. As we have seen, multiplying a vector by a number is called scalar multiplication. If we multiply a vector by a vector, there are two possibilities: the dot product and the cross product. We will only examine the dot product here; you may encounter the cross product in more advanced mathematics courses.

The dot product of two vectors involves multiplying two vectors together, and the result is a scalar.

Example

Try it.

Find the dot product of \(v=〈5,12〉\) and \(u=〈-3,4〉.\)

Solution

Using the formula, we have

\[\begin{array}{l}v⋅u=〈5,12〉⋅〈-3,4〉 \\ =5⋅(-3)+12⋅4 \\ =-15+48 \\ =33\end{array}\]

Example

Try it.

Find the dot product of v₁ = 5i + 2j and v₂ = 3i + 7j. Then, find the angle between the two vectors.

Solution

Finding the dot product, we multiply corresponding components.

\[\begin{array}{l}{v}_{1}⋅{v}_{2}=〈5,2〉⋅〈3,7〉 \\ =5⋅3+2⋅7 \\ =15+14 \\ =29\end{array}\]

To find the angle between them, we use the formula \(\cos \ \theta =\frac{v}{|v|}⋅\frac{u}{|u|}.\)

\[\begin{array}{l}\frac{v}{|v|}⋅\frac{u}{|u|}=〈\frac{5}{\sqrt{29}},\frac{2}{\sqrt{29}}〉⋅〈\frac{3}{\sqrt{58}},\frac{7}{\sqrt{58}}〉 \\ =\frac{5}{\sqrt{29}}⋅\frac{3}{\sqrt{58}}+\frac{2}{\sqrt{29}}⋅\frac{7}{\sqrt{58}} \\ =\frac{15}{\sqrt{1682}}+\frac{14}{\sqrt{1682}}=\frac{29}{\sqrt{1682}} \\ \begin{array}{l}=0.707107 \\ {\cos }^{-1}(0.707107)=45^{\circ}\end{array}\end{array}\]

See .

Example

Try it.

Find the angle between \(u=〈-3,4〉\) and \(v=〈5,12〉.\)

Solution

Using the formula, we have

\[\begin{array}{l}\theta ={\cos }^{-1}(\frac{u}{|u|}⋅\frac{v}{|v|}) \\ (\frac{u}{|u|}⋅\frac{v}{|v|})=\frac{-3i+4j}{5}⋅\frac{5i+12j}{13} \\ =(-\frac{3}{5}⋅\frac{5}{13})+(\frac{4}{5}⋅\frac{12}{13}) \\ =-\frac{15}{65}+\frac{48}{65} \\ =\frac{33}{65} \\ \theta ={\cos }^{-1}(\frac{33}{65}) \\ ={59.5}^{∘}\end{array}\]

See .

Condensed — the full section is in OpenStax Algebra and Trigonometry 2e.

Key Concepts

The position vector has its initial point at the origin. See .
If the position vector is the same for two vectors, they are equal. See .
Vectors are defined by their magnitude and direction. See .
If two vectors have the same magnitude and direction, they are equal. See .
Vector addition and subtraction result in a new vector found by adding or subtracting corresponding elements. See .
Scalar multiplication is multiplying a vector by a constant. Only the magnitude changes; the direction stays the same. See and .
Vectors are comprised of two components: the horizontal component along the positive x-axis, and the vertical component along the positive y-axis. See .
The unit vector in the same direction of any nonzero vector is found by dividing the vector by its magnitude.
The magnitude of a vector in the rectangular coordinate system is \(|v|=\sqrt{{a}^{2}+{b}^{2}}.\) See .
In the rectangular coordinate system, unit vectors may be represented in terms of \(i\) and \(j\) where \(i\) represents the horizontal component and \(j\) represents the vertical component. Then, v = ai + bj is a scalar multiple of \(v\) by real numbers \(a\ \text{and}\ b.\) See and .
Adding and subtracting vectors in terms of i and j consists of adding or subtracting corresponding coefficients of i and corresponding coefficients of j. See .
A vector v = ai + bj is written in terms of magnitude and direction as \(v=|v|\cos \ \theta i+|v|\sin \ \theta j.\) See .
The dot product of two vectors is the product of the \(i\) terms plus the product of the \(j\) terms. See .
We can use the dot product to find the angle between two vectors. and .
Dot products are useful for many types of physics applications. See .

A Geometric View of Vectors

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

\[\begin{array}{l}\overset{\to}{AB}=〈{x}_{2}-{x}_{1},{y}_{2}-{y}_{1}〉 \\ =〈a,b〉\end{array}\]

In , we see the original vector \(\overset{\to}{CD}\) and the position vector \(\overset{\to}{AB}.\)

Example

Try it.

Consider the vector whose initial point is \(P(2,3)\) and terminal point is \(Q(6,4).\) Find the position vector.

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

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

The position vector begins at \((0,0)\) and terminates at \((4,1).\) The graphs of both vectors are shown in .

We see that the position vector is \(〈4,1〉.\)

Condensed — the full section is in OpenStax Precalculus 2e.

Finding Magnitude and Direction

Example

Try it.

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

Solution

First, find the position vector.

\[\begin{array}{l}u=〈-2,-(-8),-5-1〉 \\ =〈6,-6〉\end{array}\]

We use the Pythagorean Theorem to find the magnitude.

\[\begin{array}{l}|u|=\sqrt{{(6)}^{2}+{(-6)}^{2}} \\ =\sqrt{72} \\ =6\sqrt{2}\end{array}\]

The direction is given as

\[\begin{array}{l}\tan \ \theta =\frac{-6}{6}=-1⇒\theta ={\tan }^{-1}(-1) \\ =-45^{\circ}\end{array}\]

However, the angle terminates in the fourth quadrant, so we add 360° to obtain a positive angle. Thus, \(-45^{\circ}+360^{\circ}=315^{\circ}.\) See .

Condensed — the full section is in OpenStax Precalculus 2e.

Performing Vector Addition and Scalar Multiplication

Example

Try it.

Given \(u\) \(=〈3,-2〉\) and \(v\) \(=〈-1,4〉,\) find two new vectors u + v, and u − v.

Solution

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

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

See (a).

To find the difference of two vectors, add the negative components of \(v\) to \(u.\) Thus,

\[\begin{array}{l}u+(-v)=〈3,-2〉+〈1,-4〉 \\ =〈3+1,-2+(-4)〉 \\ =〈4,-6〉\end{array}\]

See (b).

Multiplying By a Scalar

Example

Try it.

Given vector \(v\) \(=〈3,1〉,\) find 3v, \(\frac{1}{2}\) \(v,\) and −v.

Solution

See for a geometric interpretation. If \(v\) \(=〈3,1〉,\) then

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

Example

Try it.

Given \(u\) \(=〈3,-2〉\) and \(v\) \(=〈-1,4〉,\) find a new vector w = 3u + 2v.

Solution

First, we must multiply each vector by the scalar.

\[\begin{array}{l}3u=3〈3,-2〉 \\ =〈9,-6〉 \\ 2v=2〈-1,4〉 \\ =〈-2,8〉\end{array}\]

Then, add the two together.

\[\begin{array}{l}w=3u+2v \\ =〈9,-6〉+〈-2,8〉 \\ =〈9-2,-6+8〉 \\ =〈7,2〉\end{array}\]

So, \(w\) \(=〈7,2〉.\)

Finding Component Form

\[\begin{array}{l}{v}_{1}=〈2-0,0-0〉 \\ =〈2,0〉\end{array}\]

We also have v₂ with initial point \((0,0)\) and terminal point \((0,\ 3).\)

\[\begin{array}{l}{v}_{2}=〈0-0,3-0〉 \\ =〈0,3〉\end{array}\]

Therefore, the position vector is

\[\begin{array}{l}v=〈2+0,3+0〉 \\ =〈2,3〉\end{array}\]

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.

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

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

\[\begin{array}{l}\tan \ \theta =\frac{\left|{v}_{2}\right|}{\left|{v}_{1}\right|} \\ \tan \ \theta =\frac{3}{2} \\ \theta ={\tan }^{-1}(\frac{3}{2})=56.3^{\circ}\end{array}\]

Thus, the magnitude of \(v\) is \(\sqrt{13}\) and the direction is \({56.3}^{∘}\) off the horizontal.

Example

Try it.

Find the components of the vector \(v\) with initial point \((3,2)\) and terminal point \((7,4).\)

Solution

First find the standard position.

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

See the illustration in .

The horizontal component is \({v}_{1}\) \(=〈4,0〉\) and the vertical component is \({v}_{2}\) \(=〈0,2〉.\)

Finding the Unit Vector in the Direction of

Example

Try it.

Find a unit vector in the same direction as \(v\) \(=〈-5,12〉.\)

Solution

First, we will find the magnitude.

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

Then we divide each component by \(|v|,\) which gives a unit vector in the same direction as v:

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

or, in component form

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

See .

Verify that the magnitude of the unit vector equals 1. The magnitude of \(-\frac{5}{13}i+\frac{12}{13}j\) is given as

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

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

Performing Operations with Vectors in Terms of

Example

Try it.

Given a vector \(v\) with initial point \(P=(2,-6)\) and terminal point \(Q=(-6,6),\) write the vector in terms of \(i\) and \(j.\)

Solution

Begin by writing the general form of the vector. Then replace the coordinates with the given values.

\[\begin{array}{l}v=({x}_{2}-{x}_{1})i+({y}_{2}-{y}_{1})j \\ =(-6-2)i+(6-(-6))j \\ =-8i+12j\end{array}\]

Example

Try it.

Given initial point \({P}_{1}=(-1,3)\) and terminal point \({P}_{2}=(2,7),\) write the vector \(v\) in terms of \(i\) and \(j.\)

Solution

Begin by writing the general form of the vector. Then replace the coordinates with the given values.

\[\begin{array}{l}v=({x}_{2}-{x}_{1})i+({y}_{2}-{y}_{1})j \\ v=(2-(-1))i+(7-3)j \\ =3i+4j\end{array}\]

Performing Operations on Vectors in Terms of

When vectors are written in terms of \(i\) and \(j,\) we can carry out addition, subtraction, and scalar multiplication by performing operations on corresponding components.

Example

Try it.

Find the sum of \({v}_{1}=2i-3j\) and \({v}_{2}=4i+5j.\)

Solution

According to the formula, we have

\[\begin{array}{l}{v}_{1}+{v}_{2}=(2+4)i+(-3+5)j \\ =6i+2j\end{array}\]

Calculating the Component Form of a Vector: Direction

Example

Try it.

Given a vector with length 7 and an angle of 135°, write it in component form.

Solution

Using the conversion formulas \(x=|v|\cos \ \theta\) and \(y=|v|\sin \ \theta ,\) we find that

\[\begin{array}{l}x=7\cos (135^{\circ}) \\ =-\frac{7\sqrt{2}}{2} \\ y=7\sin (135^{\circ}) \\ =\frac{7\sqrt{2}}{2}\end{array}\]

This vector can be written as \(v=7\cos (135^{\circ})+7\sin (135^{\circ})\) or simplified as

\[v=-\frac{7\sqrt{2}}{2}i+\frac{7\sqrt{2}}{2}j\]

Finding the Dot Product of Two Vectors

The dot product of two vectors involves multiplying two vectors together, and the result is a scalar.

Example

Try it.

Find the dot product of \(v=〈5,12〉\) and \(u=〈-3,4〉.\)

Solution

Using the formula, we have

\[\begin{array}{l}v⋅u=〈5,12〉⋅〈-3,4〉 \\ =5⋅(-3)+12⋅4 \\ =-15+48 \\ =33\end{array}\]

Example

Try it.

Find the dot product of v₁ = 5i + 2j and v₂ = 3i + 7j. Then, find the angle between the two vectors.

Solution

Finding the dot product, we multiply corresponding components.

\[\begin{array}{l}{v}_{1}⋅{v}_{2}=〈5,2〉⋅〈3,7〉 \\ =5⋅3+2⋅7 \\ =15+14 \\ =29\end{array}\]

To find the angle between them, we use the formula \(\cos \ \theta =\frac{v}{|v|}⋅\frac{u}{|u|}.\)

See .

Example

Try it.

Find the angle between \(u=〈-3,4〉\) and \(v=〈5,12〉.\)

Solution

Using the formula, we have

See .

Condensed — the full section is in OpenStax Precalculus 2e.

Key Concepts

The position vector has its initial point at the origin. See .
If the position vector is the same for two vectors, they are equal. See .
Vectors are defined by their magnitude and direction. See .
If two vectors have the same magnitude and direction, they are equal. See .
Vector addition and subtraction result in a new vector found by adding or subtracting corresponding elements. See .
Scalar multiplication is multiplying a vector by a constant. Only the magnitude changes; the direction stays the same. See and .
Vectors are comprised of two components: the horizontal component along the positive x-axis, and the vertical component along the positive y-axis. See .
The unit vector in the same direction of any nonzero vector is found by dividing the vector by its magnitude.
The magnitude of a vector in the rectangular coordinate system is \(|v|=\sqrt{{a}^{2}+{b}^{2}}.\) See .
In the rectangular coordinate system, unit vectors may be represented in terms of \(i\) and \(j\) where \(i\) represents the horizontal component and \(j\) represents the vertical component. Then, v = ai + bj is a scalar multiple of \(v\) by real numbers \(a\ \text{and}\ b.\) See and .
Adding and subtracting vectors in terms of i and j consists of adding or subtracting corresponding coefficients of i and corresponding coefficients of j. See .
A vector v = ai + bj is written in terms of magnitude and direction as \(v=|v|\cos \ \theta i+|v|\sin \ \theta j.\) See .
The dot product of two vectors is the product of the \(i\) terms plus the product of the \(j\) terms. See .
We can use the dot product to find the angle between two vectors. and .
Dot products are useful for many types of physics applications. See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Consider the vector whose initial point is \(P(2,3)\) and terminal point is \(Q(6,4).\) Find the position vector.

Revelar la respuesta

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
\[\begin{array}{l}v=〈6-2,4-3〉 \\ =〈4,1〉\end{array}\]
The position vector begins at \((0,0)\) and terminates at \((4,1).\) The graphs of both vectors are shown in .
We see that the position vector is \(〈4,1〉.\)
Find the position vector given that vector \(v\) has an initial point at \((-3,2)\) and a terminal point at \((4,5),\) then graph both vectors in the same plane.

Revelar la respuesta

The position vector is found using the following calculation:
\[\begin{array}{l}v=〈4-(-3),5-2〉 \\ =〈7,3〉\end{array}\]
Thus, the position vector begins at \((0,0)\) and terminates at \((7,3).\) See .
Draw a vector \(v\) that connects from the origin to the point \((3,5).\)
Find the magnitude and direction of the vector with initial point \(P(-8,1)\) and terminal point \(Q(-2,-5).\) Draw the vector.

Revelar la respuesta

First, find the position vector.
\[\begin{array}{l}u=〈-2,-(-8),-5-1〉 \\ =〈6,-6〉\end{array}\]
We use the Pythagorean Theorem to find the magnitude.
\[\begin{array}{l}|u|=\sqrt{{(6)}^{2}+{(-6)}^{2}} \\ =\sqrt{72} \\ =6\sqrt{2}\end{array}\]
The direction is given as
\[\begin{array}{l}\tan \ \theta =\frac{-6}{6}=-1⇒\theta ={\tan }^{-1}(-1) \\ =-45^{\circ}\end{array}\]
However, the angle terminates in the fourth quadrant, so we add 360° to obtain a positive angle. Thus, \(-45^{\circ}+360^{\circ}=315^{\circ}.\) See .
Show that vector v with initial point at \((5,-3)\) and terminal point at \((-1,2)\) is equal to vector u with initial point at \((-1,-3)\) and terminal point at \((-7,2).\) Draw the position vector on the same grid as v and u. Next, find the magnitude and direction of each vector.

Revelar la respuesta

As shown in , draw the vector \(v\) starting at initial \((5,-3)\) and terminal point \((-1,2).\) Draw the vector \(u\) with initial point \((-1,-3)\) and terminal point \((-7,2).\) Find the standard position for each.
Next, find and sketch the position vector for v and u. We have
\[\begin{array}{l}v=〈-1-5,2-(-3)〉 \\ =〈-6,5〉 \\ \\ u=〈-7-(-1),2-(-3)〉 \\ =〈-6,5〉\end{array}\]
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.
\[\begin{array}{l}|v|=\sqrt{{(-1-5)}^{2}+{(2-(-3))}^{2}} \\ =\sqrt{{(-6)}^{2}+{(5)}^{2}} \\ =\sqrt{36+25} \\ =\sqrt{61} \\ |u|=\sqrt{{(-7-(-1))}^{2}+{(2-(-3))}^{2}} \\ =\sqrt{{(-6)}^{2}+{(5)}^{2}} \\ =\sqrt{36+25} \\ =\sqrt{61}\end{array}\]
As the magnitudes are equal, we now need to verify the direction. Using the tangent function with the position vector gives
\[\begin{array}{l}\tan \ \theta =-\frac{5}{6}⇒\theta ={\tan }^{-1}(-\frac{5}{6}) \\ =-39.8^{\circ}\end{array}\]
However, we can see that the position vector terminates in the second quadrant, so we add \(180^{\circ}.\) Thus, the direction is \(-39.8^{\circ}+180^{\circ}=140.2^{\circ}.\)
Given \(u\) \(=〈3,-2〉\) and \(v\) \(=〈-1,4〉,\) find two new vectors u + v, and u − v.

Revelar la respuesta

To find the sum of two vectors, we add the components. Thus,
\[\begin{array}{l}u+v=〈3,-2〉+〈-1,4〉 \\ =〈3+(-1),-2+4〉 \\ =〈2,2〉\end{array}\]
See (a).
To find the difference of two vectors, add the negative components of \(v\) to \(u.\) Thus,
\[\begin{array}{l}u+(-v)=〈3,-2〉+〈1,-4〉 \\ =〈3+1,-2+(-4)〉 \\ =〈4,-6〉\end{array}\]
See (b).
Given vector \(v\) \(=〈3,1〉,\) find 3v, \(\frac{1}{2}\) \(v,\) and −v.

Revelar la respuesta

See for a geometric interpretation. If \(v\) \(=〈3,1〉,\) then
\[\begin{array}{l}3v=〈3⋅3,3⋅1〉 \\ =〈9,3〉 \\ \frac{1}{2}v=〈\frac{1}{2}⋅3,\frac{1}{2}⋅1〉 \\ =〈\frac{3}{2},\frac{1}{2}〉 \\ -v=〈-3,-1〉\end{array}\]
Find the scalar multiple 3 \(u\) given \(u\) \(=〈5,4〉.\)

Revelar la respuesta

\(3u=〈15,12〉\)
Given \(u\) \(=〈3,-2〉\) and \(v\) \(=〈-1,4〉,\) find a new vector w = 3u + 2v.

Revelar la respuesta

First, we must multiply each vector by the scalar.
\[\begin{array}{l}3u=3〈3,-2〉 \\ =〈9,-6〉 \\ 2v=2〈-1,4〉 \\ =〈-2,8〉\end{array}\]
Then, add the two together.
\[\begin{array}{l}w=3u+2v \\ =〈9,-6〉+〈-2,8〉 \\ =〈9-2,-6+8〉 \\ =〈7,2〉\end{array}\]
So, \(w\) \(=〈7,2〉.\)
Find the components of the vector \(v\) with initial point \((3,2)\) and terminal point \((7,4).\)

Revelar la respuesta

First find the standard position.
\[\begin{array}{l}v=〈7-3,4-2〉 \\ =〈4,2〉\end{array}\]
See the illustration in .
The horizontal component is \({v}_{1}\) \(=〈4,0〉\) and the vertical component is \({v}_{2}\) \(=〈0,2〉.\)
Find a unit vector in the same direction as \(v\) \(=〈-5,12〉.\)

Revelar la respuesta

First, we will find the magnitude.
\[\begin{array}{l}|v|=\sqrt{{(-5)}^{2}+{(12)}^{2}} \\ =\sqrt{25+144} \\ =\sqrt{169} \\ =13\end{array}\]
Then we divide each component by \(|v|,\) which gives a unit vector in the same direction as v:
\[\frac{v}{|v|}=-\frac{5}{13}i+\frac{12}{13}j\]
or, in component form
\[\frac{v}{|v|}=〈-\frac{5}{13},\frac{12}{13}〉\]
See .
Verify that the magnitude of the unit vector equals 1. The magnitude of \(-\frac{5}{13}i+\frac{12}{13}j\) is given as
\[\begin{array}{l}\sqrt{{(-\frac{5}{13})}^{2}+{(\frac{12}{13})}^{2}}=\sqrt{\frac{25}{169}+\frac{144}{169}} \\ =\sqrt{\frac{169}{169}}=1\end{array}\]
The vector u \(=\frac{5}{13}\) i \(+\frac{12}{13}\) j is the unit vector in the same direction as v \(=〈-5,12〉.\)
Given a vector \(v\) with initial point \(P=(2,-6)\) and terminal point \(Q=(-6,6),\) write the vector in terms of \(i\) and \(j.\)

Revelar la respuesta

Begin by writing the general form of the vector. Then replace the coordinates with the given values.
\[\begin{array}{l}v=({x}_{2}-{x}_{1})i+({y}_{2}-{y}_{1})j \\ =(-6-2)i+(6-(-6))j \\ =-8i+12j\end{array}\]
Given initial point \({P}_{1}=(-1,3)\) and terminal point \({P}_{2}=(2,7),\) write the vector \(v\) in terms of \(i\) and \(j.\)

Revelar la respuesta

Begin by writing the general form of the vector. Then replace the coordinates with the given values.
\[\begin{array}{l}v=({x}_{2}-{x}_{1})i+({y}_{2}-{y}_{1})j \\ v=(2-(-1))i+(7-3)j \\ =3i+4j\end{array}\]
Write the vector \(u\) with initial point \(P=(-1,6)\) and terminal point \(Q=(7,-5)\) in terms of \(i\) and \(j.\)

Revelar la respuesta

\(u=8i-11j\)
Find the sum of \({v}_{1}=2i-3j\) and \({v}_{2}=4i+5j.\)

Revelar la respuesta

According to the formula, we have
\[\begin{array}{l}{v}_{1}+{v}_{2}=(2+4)i+(-3+5)j \\ =6i+2j\end{array}\]
Given a vector with length 7 and an angle of 135°, write it in component form.

Revelar la respuesta

Using the conversion formulas \(x=|v|\cos \ \theta\) and \(y=|v|\sin \ \theta ,\) we find that
\[\begin{array}{l}x=7\cos (135^{\circ}) \\ =-\frac{7\sqrt{2}}{2} \\ y=7\sin (135^{\circ}) \\ =\frac{7\sqrt{2}}{2}\end{array}\]
This vector can be written as \(v=7\cos (135^{\circ})+7\sin (135^{\circ})\) or simplified as
\[v=-\frac{7\sqrt{2}}{2}i+\frac{7\sqrt{2}}{2}j\]
A vector travels from the origin to the point \((3,5).\) Write the vector in terms of magnitude and direction.

Revelar la respuesta

\(v=\sqrt{34}\cos (59^{\circ})i+\sqrt{34}\sin (59^{\circ})j\)
Magnitude = \(\sqrt{34}\)
\(\theta ={\tan }^{-1}(\frac{5}{3})=59.04^{\circ}\)
Find the dot product of \(v=〈5,12〉\) and \(u=〈-3,4〉.\)

Revelar la respuesta

Using the formula, we have
\[\begin{array}{l}v⋅u=〈5,12〉⋅〈-3,4〉 \\ =5⋅(-3)+12⋅4 \\ =-15+48 \\ =33\end{array}\]
Find the dot product of v₁ = 5i + 2j and v₂ = 3i + 7j. Then, find the angle between the two vectors.

Revelar la respuesta

Finding the dot product, we multiply corresponding components.
\[\begin{array}{l}{v}_{1}⋅{v}_{2}=〈5,2〉⋅〈3,7〉 \\ =5⋅3+2⋅7 \\ =15+14 \\ =29\end{array}\]
To find the angle between them, we use the formula \(\cos \ \theta =\frac{v}{|v|}⋅\frac{u}{|u|}.\)
\[\begin{array}{l}\frac{v}{|v|}⋅\frac{u}{|u|}=〈\frac{5}{\sqrt{29}},\frac{2}{\sqrt{29}}〉⋅〈\frac{3}{\sqrt{58}},\frac{7}{\sqrt{58}}〉 \\ =\frac{5}{\sqrt{29}}⋅\frac{3}{\sqrt{58}}+\frac{2}{\sqrt{29}}⋅\frac{7}{\sqrt{58}} \\ =\frac{15}{\sqrt{1682}}+\frac{14}{\sqrt{1682}}=\frac{29}{\sqrt{1682}} \\ \begin{array}{l}=0.707107 \\ {\cos }^{-1}(0.707107)=45^{\circ}\end{array}\end{array}\]
See .
Find the angle between \(u=〈-3,4〉\) and \(v=〈5,12〉.\)

Revelar la respuesta

Using the formula, we have
\[\begin{array}{l}\theta ={\cos }^{-1}(\frac{u}{|u|}⋅\frac{v}{|v|}) \\ (\frac{u}{|u|}⋅\frac{v}{|v|})=\frac{-3i+4j}{5}⋅\frac{5i+12j}{13} \\ =(-\frac{3}{5}⋅\frac{5}{13})+(\frac{4}{5}⋅\frac{12}{13}) \\ =-\frac{15}{65}+\frac{48}{65} \\ =\frac{33}{65} \\ \theta ={\cos }^{-1}(\frac{33}{65}) \\ ={59.5}^{∘}\end{array}\]
See .
We now have the tools to solve the problem we introduced in the opening of the section.
An airplane is flying at an airspeed of 200 miles per hour headed on a SE bearing of 140°. A north wind (from north to south) is blowing at 16.2 miles per hour. What are the ground speed and actual bearing of the plane? See .

Revelar la respuesta

The ground speed is represented by \(x\) in the diagram, and we need to find the angle \(\alpha\) in order to calculate the adjusted bearing, which will be \(140^{\circ}+\alpha .\)
Notice in , that angle \(BCO\) must be equal to angle \(AOC\) by the rule of alternating interior angles, so angle \(BCO\) is 140°. We can find \(x\) by the Law of Cosines:
\[\begin{array}{l}{x}^{2}={(16.2)}^{2}+{(200)}^{2}-2(16.2)(200)\cos (140^{\circ}) \\ {x}^{2}=45,226.41 \\ x=\sqrt{45,226.41} \\ x=212.7\end{array}\]
The ground speed is approximately 213 miles per hour. Now we can calculate the bearing using the Law of Sines.
\[\begin{array}{l}\frac{\sin \ \alpha }{16.2}=\frac{\sin (140^{\circ})}{212.7} \\ \sin \ \alpha =\frac{16.2\sin (140^{\circ})}{212.7} \\ =0.04896 \\ {\sin }^{-1}(0.04896)=2.8^{\circ}\end{array}\]
Therefore, the plane has a SE bearing of 140°+2.8°=142.8°. The ground speed is 212.7 miles per hour.
What are the characteristics of the letters that are commonly used to represent vectors?

Revelar la respuesta

lowercase, bold letter, usually \(u,v,w\)
How is a vector more specific than a line segment?
What are \(i\) and \(j,\) and what do they represent?

Revelar la respuesta

They are unit vectors. They are used to represent the horizontal and vertical components of a vector. They each have a magnitude of 1.
What is component form?
When a unit vector is expressed as \(〈a,b〉,\) which letter is the coefficient of the \(i\) and which the \(j?\)

Revelar la respuesta

The first number always represents the coefficient of the \(i,\) and the second represents the \(j.\)
Given a vector with initial point \((5,2)\) and terminal point \((-1,-3),\) find an equivalent vector whose initial point is \((0,0).\) Write the vector in component form \(〈a,b〉.\)
Given a vector with initial point \((-4,2)\) and terminal point \((3,-3),\) find an equivalent vector whose initial point is \((0,0).\) Write the vector in component form \(〈a,b〉.\)

Revelar la respuesta

\(〈7,-5〉\)
Given a vector with initial point \((7,-1)\) and terminal point \((-1,-7),\) find an equivalent vector whose initial point is \((0,0).\) Write the vector in component form \(〈a,b〉.\)
\({P}_{1}=(5,1),{P}_{2}=(3,-2),{P}_{3}=(-1,3),\) and \({P}_{4}=(9,-4)\)

Revelar la respuesta

not equal
\({P}_{1}=(2,-3),{P}_{2}=(5,1),{P}_{3}=(6,-1),\) and \({P}_{4}=(9,3)\)
\({P}_{1}=(-1,-1),{P}_{2}=(-4,5),{P}_{3}=(-10,6),\) and \({P}_{4}=(-13,12)\)

Revelar la respuesta

equal
\({P}_{1}=(3,7),{P}_{2}=(2,1),{P}_{3}=(1,2),\) and \({P}_{4}=(-1,-4)\)
\({P}_{1}=(8,3),{P}_{2}=(6,5),{P}_{3}=(11,8),\) and \({P}_{4}=(9,10)\)

Revelar la respuesta

equal
Given initial point \({P}_{1}=(-3,1)\) and terminal point \({P}_{2}=(5,2),\) write the vector \(v\) in terms of \(i\) and \(j.\)
Given initial point \({P}_{1}=(6,0)\) and terminal point \({P}_{2}=(-1,-3),\) write the vector \(v\) in terms of \(i\) and \(j.\)

Revelar la respuesta

\(-7i-3j\)
Find u + (v − w)
Find 4v + 2u

Revelar la respuesta

\(-6i-2j\)
\(u=〈2,-3〉,v=〈1,5〉\)
\(u=〈-3,4〉,v=〈-2,1〉\)

Revelar la respuesta

\(u+v=〈-5,5〉,u-v=〈-1,3〉,2u-3v=〈0,5〉\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\theta

theta
The usual name for an angle.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

^\circ

degrees
1/360 of a full turn. 180° = π radians.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\arcsin,\ \sin^{-1}

inverse sine
The angle whose sine is the given value (and likewise arccos, arctan).

How to: Vectors

View vectors geometrically.
Find magnitude and direction.
Perform vector addition and scalar multiplication.
Find the component form of a vector.
Find the unit vector in the direction of
Perform operations with vectors in terms of
Find the dot product of two vectors.
Lower case, boldfaced type, with or without an arrow on top such as

Questions people ask

Why radians instead of degrees?

A radian is the angle whose arc equals the radius, so it is a pure ratio rather than an arbitrary 1/360 of a turn. With radians the derivative of sin x is exactly cos x; with degrees a factor of π/180 appears everywhere.

Why does sin x = 1/2 have infinitely many solutions?

Sine repeats every full turn, and within one turn it reaches 1/2 twice (at π/6 and 5π/6). Add any whole number of turns to either and it is still true.

How do I remember the exact values?

Two triangles: the 45-45-90 with sides 1, 1, √2 and the 30-60-90 with sides 1, √3, 2. Every value for 30°, 45° and 60° is a ratio of those sides; symmetry gives the rest of the circle.

Prueba tu propio

Parts of this page are adapted from OpenStax Algebra and Trigonometry 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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