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Vectors in the Plane

Describe a plane vector, using correct notation.

Vector Representation

A vector in a plane is represented by a directed line segment (an arrow). The endpoints of the segment are called the initial point and the terminal point of the vector. An arrow from the initial point to the terminal point indicates the direction of the vector. The length of the line segment represents its magnitude. We use the notation \(‖\text{v}‖\) to denote the magnitude of the vector \(\text{v}.\) A vector with an initial point and terminal point that are the same is called the zero vector, denoted \(0.\) The zero vector is the only vector without a direction, and by convention can be considered to have any direction convenient to the problem at hand.

Vectors with the same magnitude and direction are called equivalent vectors. We treat equivalent vectors as equal, even if they have different initial points. Thus, if \(\text{v}\) and \(\text{w}\) are equivalent, we write

\[\text{v}=\text{w}.\]

The arrows in (b) are equivalent. Each arrow has the same length and direction. A closely related concept is the idea of parallel vectors. Two vectors are said to be parallel if they have the same or opposite directions. We explore this idea in more detail later in the chapter. A vector is defined by its magnitude and direction, regardless of where its initial point is located.

The use of boldface, lowercase letters to name vectors is a common representation in print, but there are alternative notations. When writing the name of a vector by hand, for example, it is easier to sketch an arrow over the variable than to simulate boldface type: \(\overset{\to}{v}.\) When a vector has initial point \(P\) and terminal point \(Q,\) the notation \(\overset{\to}{PQ}\) is useful because it indicates the direction and location of the vector.

Example

Try it.

Sketch a vector in the plane from initial point \(P(1,1)\) to terminal point \(Q(8,5).\)

Solution

See . Because the vector goes from point \(P\) to point \(Q,\) we name it \(\overset{\to}{PQ}.\)

Combining Vectors

Vectors have many real-life applications, including situations involving force or velocity. For example, consider the forces acting on a boat crossing a river. The boat’s motor generates a force in one direction, and the current of the river generates a force in another direction. Both forces are vectors. We must take both the magnitude and direction of each force into account if we want to know where the boat will go.

A second example that involves vectors is a quarterback throwing a football. The quarterback does not throw the ball parallel to the ground; instead, he aims up into the air. The velocity of his throw can be represented by a vector. If we know how hard he throws the ball (magnitude—in this case, speed), and the angle (direction), we can tell how far the ball will travel down the field.

A real number is often called a scalar in mathematics and physics. Unlike vectors, scalars are generally considered to have a magnitude only, but no direction. Multiplying a vector by a scalar changes the vector’s magnitude. This is called scalar multiplication. Note that changing the magnitude of a vector does not indicate a change in its direction. For example, wind blowing from north to south might increase or decrease in speed while maintaining its direction from north to south.

As you might expect, if \(k=-1,\) we denote the product \(k\text{v}\) as

\[k\text{v}=(-1)\text{v}=\text{-}\text{v}.\]

Note that \(\text{-}\text{v}\) has the same magnitude as \(\text{v},\) but has the opposite direction ().

A second method for adding vectors is called the parallelogram method. With this method, we place the two vectors so they have the same initial point, and then we draw a parallelogram with the vectors as two adjacent sides, as in (b). The length of the diagonal of the parallelogram is the sum. Comparing (b) and (a), we can see that we get the same answer using either method. The vector \(v+w\) is called the vector sum.

\[‖\text{v}+\text{w}‖\le ‖\text{v}‖+‖\text{w}‖.\]

Condensed — the full section is in OpenStax Calculus Volume 3.

Vector Components

Working with vectors in a plane is easier when we are working in a coordinate system. When the initial points and terminal points of vectors are given in Cartesian coordinates, computations become straightforward.

Example

Try it.

Are \(\text{v}\) and \(\text{w}\) equivalent vectors?

\(\text{v}\) has initial point \((3,2)\) and terminal point \((7,2)\)
\(\text{w}\) has initial point \((1,-4)\) and terminal point \((1,0)\)
\(\text{v}\) has initial point \((0,0)\) and terminal point \((1,1)\)
\(\text{w}\) has initial point \((-2,2)\) and terminal point \((-1,3)\)

Solution

The vectors are each \(4\) units long, but they are oriented in different directions. So \(\text{v}\) and \(\text{w}\) are not equivalent ().
Based on , and using a bit of geometry, it is clear these vectors have the same length and the same direction, so \(\text{v}\) and \(\text{w}\) are equivalent.

We have seen how to plot a vector when we are given an initial point and a terminal point. However, because a vector can be placed anywhere in a plane, it may be easier to perform calculations with a vector when its initial point coincides with the origin. We call a vector with its initial point at the origin a standard-position vector. Because the initial point of any vector in standard position is known to be \((0,0),\) we can describe the vector by looking at the coordinates of its terminal point. Thus, if vector v has its initial point at the origin and its terminal point at \((x,y),\) we write the vector in component form as

\[\text{v}=〈x,y〉.\]

When a vector is written in component form like this, the scalars x and y are called the components of \(\text{v}.\)

Recall that vectors are named with lowercase letters in bold type or by drawing an arrow over their name. We have also learned that we can name a vector by its component form, with the coordinates of its terminal point in angle brackets. However, when writing the component form of a vector, it is important to distinguish between \(〈x,y〉\) and \((x,y).\) The first ordered pair uses angle brackets to describe a vector, whereas the second uses parentheses to describe a point in a plane. The initial point of \(〈x,y〉\) is \((0,0);\) the terminal point of \(〈x,y〉\) is \((x,y).\)

\[‖\text{v}‖=\sqrt{{x}^{2}+{y}^{2}}.\]

Condensed — the full section is in OpenStax Calculus Volume 3.

Unit Vectors

A unit vector is a vector with magnitude \(1.\) For any nonzero vector \(\text{v},\) we can use scalar multiplication to find a unit vector \(\text{u}\) that has the same direction as \(\text{v}.\) To do this, we multiply the vector by the reciprocal of its magnitude:

\[\text{u}=\frac{1}{‖\text{v}‖}\text{v}.\]

Recall that when we defined scalar multiplication, we noted that \(‖k\text{v}‖=|k|\cdot ‖\text{v}‖.\) For \(\text{u}=\frac{1}{‖\text{v}‖}\text{v},\) it follows that \(‖\text{u}‖=\frac{1}{‖\text{v}‖}(‖\text{v}‖)=1.\) We say that \(\text{u}\) is the unit vector in the direction of\(\text{v}\) (). The process of using scalar multiplication to find a unit vector with a given direction is called normalization.

Example

Try it.

Let \(\text{v}=〈1,2〉.\)

Find a unit vector with the same direction as \(\text{v}.\)
Find a vector \(\text{w}\) with the same direction as \(\text{v}\) such that \(‖\text{w}‖=7.\)

Solution

First, find the magnitude of \(\text{v},\) then divide the components of \(\text{v}\) by the magnitude:
\[‖\text{v}‖=\sqrt{{1}^{2}+{2}^{2}}=\sqrt{1+4}=\sqrt{5}\]
\[\text{u}=\frac{1}{‖\text{v}‖}\text{v}=\frac{1}{\sqrt{5}}〈1,2〉=〈\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}〉.\]
The vector \(\text{u}\) is in the same direction as \(\text{v}\) and \(‖\text{u}‖=1.\) Use scalar multiplication to increase the length of \(\text{u}\) without changing direction:
\[\text{w}=7\text{u}=7〈\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}〉=〈\frac{7}{\sqrt{5}},\frac{14}{\sqrt{5}}〉.\]

We have seen how convenient it can be to write a vector in component form. Sometimes, though, it is more convenient to write a vector as a sum of a horizontal vector and a vertical vector. To make this easier, let’s look at standard unit vectors. The standard unit vectors are the vectors \(\text{i}=〈1,0〉\) and \(\text{j}=〈0,1〉\) ().

By applying the properties of vectors, it is possible to express any vector in terms of \(\text{i}\) and \(\text{j}\) in what we call a linear combination:

\[\text{v}=〈x,y〉=〈x,0〉+〈0,y〉=x〈1,0〉+y〈0,1〉=x\text{i}+y\text{j}.\]

Thus, \(\text{v}\) is the sum of a horizontal vector with magnitude \(x,\) and a vertical vector with magnitude \(y,\) as in the following figure.

Condensed — the full section is in OpenStax Calculus Volume 3.

Applications of Vectors

Because vectors have both direction and magnitude, they are valuable tools for solving problems involving such applications as motion and force. Recall the boat example and the quarterback example we described earlier. Here we look at two other examples in detail.

Example

Try it.

Jane’s car is stuck in the mud. Lisa and Jed come along in a truck to help pull her out. They attach one end of a tow strap to the front of the car and the other end to the truck’s trailer hitch, and the truck starts to pull. Meanwhile, Jane and Jed get behind the car and push. The truck generates a horizontal force of \(300\) lb on the car. Jane and Jed are pushing at a slight upward angle and generate a force of \(150\) lb on the car. These forces can be represented by vectors, as shown in . The angle between these vectors is \(15\text{^{\circ}}.\) Find the resultant force (the vector sum) and give its magnitude to the nearest tenth of a pound and its direction angle from the positive x-axis.

Solution

To find the effect of combining the two forces, add their representative vectors. First, express each vector in component form or in terms of the standard unit vectors. For this purpose, it is easiest if we align one of the vectors with the positive x-axis. The horizontal vector, then, has initial point \((0,0)\) and terminal point \((300,0).\) It can be expressed as \(〈300,0〉\) or \(300\text{i}.\)

The second vector has magnitude \(150\) and makes an angle of \(15\text{^{\circ}}\) with the first, so we can express it as \(〈150\ \text{cos}(15\text{^{\circ}}),150\ \text{sin}(15\text{^{\circ}})〉,\) or \(150\ \text{cos}(15\text{^{\circ}})\text{i}+150\ \text{sin}(15\text{^{\circ}})\text{j}.\) Then, the sum of the vectors, or resultant vector, is \(\text{r}=〈300,0〉+〈150\ \text{cos}(15\text{^{\circ}}),150\ \text{sin}(15\text{^{\circ}})〉,\) and we have

\[\begin{array}{ll}‖\text{r}‖ & =\sqrt{{(300+150\ \text{cos}(15\text{^{\circ}}))}^{2}+{(150\ \text{sin}(15\text{^{\circ}}))}^{2}} \\ & \approx 446.6.\end{array}\]

The angle \(\theta\) made by \(\text{r}\) and the positive x-axis has \(\text{tan}\ \theta =\frac{150\ \text{sin}\ 15\text{^{\circ}}}{(300+150\ \text{cos}\ 15\text{^{\circ}})}\approx 0.09,\) so \(\theta \approx ta{n}^{-1}(0.09)\approx 5\text{^{\circ}},\) which means the resultant force \(\text{r}\) has an angle of \(5\text{^{\circ}}\) above the horizontal axis.

Condensed — the full section is in OpenStax Calculus Volume 3.

Key Concepts

Vectors are used to represent quantities that have both magnitude and direction.
We can add vectors by using the parallelogram method or the triangle method to find the sum. We can multiply a vector by a scalar to change its length or give it the opposite direction.
Subtraction of vectors is defined in terms of adding the negative of the vector.
A vector is written in component form as \(\text{v}=〈x,y〉.\)
The magnitude of a vector is a scalar: \(‖\text{v}‖=\sqrt{{x}^{2}+{y}^{2}}.\)
A unit vector \(\text{u}\) has magnitude \(1\) and can be found by dividing a vector by its magnitude: \(\text{u}=\frac{1}{‖\text{v}‖}\text{v}.\) The standard unit vectors are \(\text{i}=〈1,0〉\ \text{and}\ \text{j}=〈0,1〉.\) A vector \(\text{v}=〈x,y〉\) can be expressed in terms of the standard unit vectors as \(\text{v}=x\text{i}+y\text{j}.\)
Vectors are often used in physics and engineering to represent forces and velocities, among other quantities.

Vectors in the Plane

For the following exercises, consider points \(P(-1,3),\) \(Q(1,5),\) and \(R(-3,7).\) Determine the requested vectors and express each of them a. in component form and b. by using the standard unit vectors.

For the following exercises, use the given vectors \(\text{a}\) and \(\text{b}.\)

Determine the vector sum \(\text{a}+\text{b}\) and express it in both the component form and by using the standard unit vectors.
Find the vector difference \(\text{a}-\text{b}\) and express it in both the component form and by using the standard unit vectors.
Verify that the vectors \(\text{a},\) \(\text{b},\) and \(\text{a}+\text{b},\) and, respectively, \(\text{a},\) \(\text{b},\) and \(\text{a}-\text{b}\) satisfy the triangle inequality.
Determine the vectors \(2\text{a},\) \(\text{-}\text{b},\) and \(2\text{a}-\text{b}.\) Express the vectors in both the component form and by using standard unit vectors.

For the following exercises, find vector \(\text{v}\) with the given magnitude and in the same direction as vector \(\text{u}.\)

For the following exercises, find the component form of vector \(\text{u},\) given its magnitude and the angle the vector makes with the positive x-axis. Give exact answers when possible.

For the following exercises, vector \(\text{u}\) is given. Find the angle \(\theta \in [0,2\pi )\) that vector \(\text{u}\) makes with the positive direction of the x-axis, in a counter-clockwise direction.

Practice (40)

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

Sketch a vector in the plane from initial point \(P(1,1)\) to terminal point \(Q(8,5).\)

Vis svaret

See . Because the vector goes from point \(P\) to point \(Q,\) we name it \(\overset{\to}{PQ}.\)
Sketch the vector \(\overset{\to}{ST}\) where \(S\) is point \((3,-1)\) and \(T\) is point \((-2,3).\)

Vis svaret
Given the vectors \(\text{v}\) and \(\text{w}\) shown in , sketch the vectors
1. \(3\text{w}\)
2. \(\text{v}+\text{w}\)
3. \(2\text{v}-\text{w}\)
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1. The vector \(3\text{w}\) has the same direction as \(\text{w};\) it is three times as long as \(\text{w}.\)
  
  Vector \(3\text{w}\) has the same direction as \(\text{w}\) and is three times as long.
2. Use either addition method to find \(\text{v}+\text{w}.\)
3. To find \(2\text{v}-\text{w},\) we can first rewrite the expression as \(2\text{v}+(\text{-}\text{w}).\) Then we can draw the vector \(\text{-}\text{w},\) then add it to the vector \(2\text{v}.\)
Using vectors \(\text{v}\) and \(\text{w}\) from , sketch the vector \(2\text{w}-\text{v}.\)

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Are \(\text{v}\) and \(\text{w}\) equivalent vectors?
1. \(\text{v}\) has initial point \((3,2)\) and terminal point \((7,2)\)
  \(\text{w}\) has initial point \((1,-4)\) and terminal point \((1,0)\)
2. \(\text{v}\) has initial point \((0,0)\) and terminal point \((1,1)\)
  \(\text{w}\) has initial point \((-2,2)\) and terminal point \((-1,3)\)
Vis svaret
1. The vectors are each \(4\) units long, but they are oriented in different directions. So \(\text{v}\) and \(\text{w}\) are not equivalent ().
2. Based on , and using a bit of geometry, it is clear these vectors have the same length and the same direction, so \(\text{v}\) and \(\text{w}\) are equivalent.
Which of the following vectors are equivalent?

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Vectors \(\text{a},\) \(\text{b},\) and \(\text{e}\) are equivalent.
Express vector \(\text{v}\) with initial point \((-3,4)\) and terminal point \((1,2)\) in component form.
Vis svaret
1. Geometric
  Sketch the vector in the coordinate plane ().
  The terminal point is 4 units to the right and 2 units down from the initial point.
  Find the point that is 4 units to the right and 2 units down from the origin.
  In standard position, this vector has initial point \((0,0)\) and terminal point \((4,-2)\text{:}\)
  \[\text{v}=〈4,-2〉.\]
2. Algebraic
  In the first solution, we used a sketch of the vector to see that the terminal point lies 4 units to the right. We can accomplish this algebraically by finding the difference of the x-coordinates:
  \[{x}_{t}-{x}_{i}=1-(-3)=4.\]
  Similarly, the difference of the y-coordinates shows the vertical length of the vector.
  \[{y}_{t}-{y}_{i}=2-4=-2.\]
  So, in component form,
  \[\begin{array}{ll}\text{v} & =〈{x}_{t}-{x}_{i},{y}_{t}-{y}_{i}〉 \\ & =〈1-(-3),2-4〉 \\ & =〈4,-2〉.\end{array}\]
Vector \(\text{w}\) has initial point \((-4,-5)\) and terminal point \((-1,2).\) Express \(\text{w}\) in component form.

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\(〈3,7〉\)
Let \(\text{v}\) be the vector with initial point \((2,5)\) and terminal point \((8,13),\) and let \(\text{w}=〈-2,4〉.\)
1. Express \(\text{v}\) in component form and find \(‖\text{v}‖.\) Then, using algebra, find
2. \(\text{v}+\text{w},\)
3. \(3\text{v},\) and
4. \(\text{v}-2\text{w}.\)
Vis svaret
1. To place the initial point of \(\text{v}\) at the origin, we must translate the vector \(2\) units to the left and \(5\) units down (). Using the algebraic method, we can express \(\text{v}\) as \(\text{v}=〈8-2,13-5〉=〈6,8〉\text{:}\)
  \[‖\text{v}‖=\sqrt{{6}^{2}+{8}^{2}}=\sqrt{36+64}=\sqrt{100}=10.\]
2. To find \(\text{v}+\text{w},\) add the x-components and the y-components separately:
  \[\text{v}+\text{w}=〈6,8〉+〈-2,4〉=〈4,12〉.\]
3. To find \(3\text{v},\) multiply \(\text{v}\) by the scalar \(k=3\text{:}\)
  \[3\text{v}=3\cdot 〈6,8〉=〈3\cdot 6,3\cdot 8〉=〈18,24〉.\]
4. To find \(\text{v}-2\text{w},\) find \(-2\text{w}\) and add it to \(\text{v}\text{:}\)
  \[\text{v}-2\text{w}=〈6,8〉-2\cdot 〈-2,4〉=〈6,8〉+〈4,-8〉=〈10,0〉.\]
Let \(\text{a}=〈7,1〉\) and let \(\text{b}\) be the vector with initial point \((3,2)\) and terminal point \((-1,-1).\)
1. Find \(‖\text{a}‖.\)
2. Express \(\text{b}\) in component form.
3. Find \(3\text{a}-4\text{b}.\)
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a. \(‖\text{a}‖=5\sqrt{2},\) b. \(\text{b}=〈-4,-3〉,\) c. \(3\text{a}-4\text{b}=〈37,15〉\)
Prove the additive inverse property.
Find the component form of a vector with magnitude 4 that forms an angle of \(-45\text{^{\circ}}\) with the x-axis.

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Let \(x\) and \(y\) represent the components of the vector (). Then \(x=4\ \text{cos}(-45\text{^{\circ}})=2\sqrt{2}\) and \(y=4\ \text{sin}(-45\text{^{\circ}})=-2\sqrt{2}.\) The component form of the vector is \(〈2\sqrt{2},-2\sqrt{2}〉.\)
Find the component form of vector \(\text{v}\) with magnitude \(10\) that forms an angle of \(120\text{^{\circ}}\) with the positive x-axis.

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\(\text{v}=〈-5,5\sqrt{3}〉\)
Let \(\text{v}=〈1,2〉.\)
1. Find a unit vector with the same direction as \(\text{v}.\)
2. Find a vector \(\text{w}\) with the same direction as \(\text{v}\) such that \(‖\text{w}‖=7.\)
Vis svaret
1. First, find the magnitude of \(\text{v},\) then divide the components of \(\text{v}\) by the magnitude:
  \[‖\text{v}‖=\sqrt{{1}^{2}+{2}^{2}}=\sqrt{1+4}=\sqrt{5}\]
  \[\text{u}=\frac{1}{‖\text{v}‖}\text{v}=\frac{1}{\sqrt{5}}〈1,2〉=〈\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}〉.\]
2. The vector \(\text{u}\) is in the same direction as \(\text{v}\) and \(‖\text{u}‖=1.\) Use scalar multiplication to increase the length of \(\text{u}\) without changing direction:
  \[\text{w}=7\text{u}=7〈\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}〉=〈\frac{7}{\sqrt{5}},\frac{14}{\sqrt{5}}〉.\]
Let \(\text{v}=〈9,2〉.\) Find a vector with magnitude \(5\) in the opposite direction as \(\text{v}.\)

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\(〈-\frac{45}{\sqrt{85}},-\frac{10}{\sqrt{85}}〉\)
1. Express the vector \(\text{w}=〈3,-4〉\) in terms of standard unit vectors.
2. Vector \(\text{u}\) is a unit vector that forms an angle of \(60\text{^{\circ}}\) with the positive x-axis. Use standard unit vectors to describe \(\text{u}.\)
Vis svaret
1. Resolve vector \(\text{w}\) into a vector with a zero y-component and a vector with a zero x-component:
  \[\text{w}=〈3,-4〉=3\text{i}-4\text{j}.\]
2. Because \(\text{u}\) is a unit vector, the terminal point lies on the unit circle when the vector is placed in standard position ().
  \[\begin{array}{ll}u & =〈\text{cos}\ 60\text{^{\circ}},\text{sin}\ 60\text{^{\circ}}〉 \\ & =〈\frac{1}{2},\frac{\sqrt{3}}{2}〉 \\ & =\frac{1}{2}\text{i}+\frac{\sqrt{3}}{2}\text{j}.\end{array}\]
Let \(\text{a}=〈16,-11〉\) and let \(\text{b}\) be a unit vector that forms an angle of \(225\text{^{\circ}}\) with the positive x-axis. Express \(\text{a}\) and \(\text{b}\) in terms of the standard unit vectors.

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\(\text{a}=16\text{i}-11\text{j},\) \(\text{b}=-\frac{\sqrt{2}}{2}\text{i}-\frac{\sqrt{2}}{2}\text{j}\)
Jane’s car is stuck in the mud. Lisa and Jed come along in a truck to help pull her out. They attach one end of a tow strap to the front of the car and the other end to the truck’s trailer hitch, and the truck starts to pull. Meanwhile, Jane and Jed get behind the car and push. The truck generates a horizontal force of \(300\) lb on the car. Jane and Jed are pushing at a slight upward angle and generate a force of \(150\) lb on the car. These forces can be represented by vectors, as shown in . The angle between these vectors is \(15\text{^{\circ}}.\) Find the resultant force (the vector sum) and give its magnitude to the nearest tenth of a pound and its direction angle from the positive x-axis.

Vis svaret

To find the effect of combining the two forces, add their representative vectors. First, express each vector in component form or in terms of the standard unit vectors. For this purpose, it is easiest if we align one of the vectors with the positive x-axis. The horizontal vector, then, has initial point \((0,0)\) and terminal point \((300,0).\) It can be expressed as \(〈300,0〉\) or \(300\text{i}.\)
The second vector has magnitude \(150\) and makes an angle of \(15\text{^{\circ}}\) with the first, so we can express it as \(〈150\ \text{cos}(15\text{^{\circ}}),150\ \text{sin}(15\text{^{\circ}})〉,\) or \(150\ \text{cos}(15\text{^{\circ}})\text{i}+150\ \text{sin}(15\text{^{\circ}})\text{j}.\) Then, the sum of the vectors, or resultant vector, is \(\text{r}=〈300,0〉+〈150\ \text{cos}(15\text{^{\circ}}),150\ \text{sin}(15\text{^{\circ}})〉,\) and we have
\[\begin{array}{ll}‖\text{r}‖ & =\sqrt{{(300+150\ \text{cos}(15\text{^{\circ}}))}^{2}+{(150\ \text{sin}(15\text{^{\circ}}))}^{2}} \\ & \approx 446.6.\end{array}\]
The angle \(\theta\) made by \(\text{r}\) and the positive x-axis has \(\text{tan}\ \theta =\frac{150\ \text{sin}\ 15\text{^{\circ}}}{(300+150\ \text{cos}\ 15\text{^{\circ}})}\approx 0.09,\) so \(\theta \approx ta{n}^{-1}(0.09)\approx 5\text{^{\circ}},\) which means the resultant force \(\text{r}\) has an angle of \(5\text{^{\circ}}\) above the horizontal axis.
An airplane flies due west at an airspeed of \(425\) mph. The wind is blowing from the northeast at \(40\) mph. What is the ground speed of the airplane? What is the bearing of the airplane?

Vis svaret

Let’s start by sketching the situation described ().
Set up a sketch so that the initial points of the vectors lie at the origin. Then, the plane’s velocity vector is \(\text{p}=-425\text{i}.\) The vector describing the wind makes an angle of \(225\text{^{\circ}}\) with the positive x-axis:
\[\text{w}=〈40\ \text{cos}(225\text{^{\circ}}),40\ \text{sin}(225\text{^{\circ}})〉=〈-\frac{40}{\sqrt{2}},-\frac{40}{\sqrt{2}}〉=-\frac{40}{\sqrt{2}}\text{i}-\frac{40}{\sqrt{2}}\text{j}.\]
When the airspeed and the wind act together on the plane, we can add their vectors to find the resultant force:
\[p+w=-425\text{i}+(-\frac{40}{\sqrt{2}}\text{i}-\frac{40}{\sqrt{2}}\text{j})=(-425-\frac{40}{\sqrt{2}})\text{i}-\frac{40}{\sqrt{2}}\text{j}.\]
The magnitude of the resultant vector shows the effect of the wind on the ground speed of the airplane:
\[‖p+w‖=\sqrt{{(-425-\frac{40}{\sqrt{2}})}^{2}+{(-\frac{40}{\sqrt{2}})}^{2}}\approx 454.17\ \text{mph}\]
As a result of the wind, the plane is traveling at approximately \(454\) mph relative to the ground.
To determine the bearing of the airplane, we want to find the direction of the vector \(p+w\text{:}\)
\[\begin{array}{lll}\text{tan}\ \theta & = & \frac{-\frac{40}{\sqrt{2}}}{(-425-\frac{40}{\sqrt{2}})}\approx 0.06 \\ \theta & \approx & 3.57\text{^{\circ}}.\end{array}\]
The overall direction of the plane is \(3.57\text{^{\circ}}\) south of west.
An airplane flies due north at an airspeed of \(550\) mph. The wind is blowing from the northwest at \(50\) mph. What is the ground speed of the airplane?

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Approximately \(516\) mph
\(\overset{\to}{PQ}\)

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a. \(\overset{\to}{PQ}=〈2,2〉;\) b. \(\overset{\to}{PQ}=2\text{i}+2\text{j}\)
\(\overset{\to}{PR}\)
\(\overset{\to}{QP}\)

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a. \(\overset{\to}{QP}=〈-2,-2〉;\) b. \(\overset{\to}{QP}=-2\text{i}-2\text{j}\)
\(\overset{\to}{RP}\)
\(\overset{\to}{PQ}+\overset{\to}{PR}\)

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a. \(\overset{\to}{PQ}+\overset{\to}{PR}=〈0,6〉;\) b. \(\overset{\to}{PQ}+\overset{\to}{PR}=6\text{j}\)
\(\overset{\to}{PQ}-\overset{\to}{PR}\)
\(2\overset{\to}{PQ}-2\overset{\to}{PR}\)

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a. \(2\overset{\to}{PQ}-2\overset{\to}{PR}=〈8,-4〉;\) b. \(2\overset{\to}{PQ}-2\overset{\to}{PR}=8\text{i}-4\text{j}\)
\(2\overset{\to}{PQ}+\frac{1}{2}\overset{\to}{PR}\)
The unit vector in the direction of \(\overset{\to}{PQ}\)

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a. \(〈\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}〉;\) b. \(\frac{1}{\sqrt{2}}\text{i}+\frac{1}{\sqrt{2}}\text{j}\)
The unit vector in the direction of \(\overset{\to}{PR}\)
A vector \(\text{v}\) has initial point \((-1,-3)\) and terminal point \((2,1).\) Find the unit vector in the direction of \(\text{v}.\) Express the answer in component form.

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\(〈\frac{3}{5},\frac{4}{5}〉\)
A vector \(\text{v}\) has initial point \((-2,5)\) and terminal point \((3,-1).\) Find the unit vector in the direction of \(\text{v}.\) Express the answer in component form.
The vector \(\text{v}\) has initial point \(P(1,0)\) and terminal point \(Q\) that is on the y-axis and above the initial point. Find the coordinates of terminal point \(Q\) such that the magnitude of the vector \(\text{v}\) is \(\sqrt{5}.\)

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\(Q(0,2)\)
The vector \(\text{v}\) has initial point \(P(1,1)\) and terminal point \(Q\) that is on the x-axis and left of the initial point. Find the coordinates of terminal point \(Q\) such that the magnitude of the vector \(\text{v}\) is \(\sqrt{10}.\)
\(\text{a}=2\text{i}+\text{j},\) \(\text{b}=\text{i}+3\text{j}\)

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a. \(\text{a}+\text{b}=3\text{i}+4\text{j},\) \(\text{a}+\text{b}=〈3,4〉;\) b. \(\text{a}-\text{b}=\text{i}-2\text{j},\) \(\text{a}-\text{b}=〈1,-2〉;\) c. Answers will vary; d. \(2\text{a}=4\text{i}+2\text{j},\) \(2\text{a}=〈4,2〉,\) \(\text{-}\text{b}=\text{-}\text{i}-3\text{j},\) \(\text{-}\text{b}=〈-1,-3〉,\) \(2\text{a}-\text{b}=3\text{i}-\text{j},\) \(2\text{a}-\text{b}=〈3,-1〉\)
\(\text{a}=2\text{i},\) \(\text{b}=-2\text{i}+2\text{j}\)
Let \(\text{a}\) be a standard-position vector with terminal point \((-2,-4).\) Let \(\text{b}\) be a vector with initial point \((1,2)\) and terminal point \((-1,4).\) Find the magnitude of vector \(-3\text{a}+\text{b}-4\text{i}+\text{j}.\)

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\(15\)
Let \(\text{a}\) be a standard-position vector with terminal point at \((2,5).\) Let \(\text{b}\) be a vector with initial point \((-1,3)\) and terminal point \((1,0).\) Find the magnitude of vector \(\text{a}-3\text{b}+14\text{i}-14\text{j}.\)
Let \(\text{u}\) and \(\text{v}\) be two nonzero vectors that are nonequivalent. Consider the vectors \(\text{a}=4\text{u}+5\text{v}\) and \(\text{b}=\text{u}+2\text{v}\) defined in terms of \(\text{u}\) and \(\text{v}.\) Find the scalar \(\lambda\) such that vectors \(\text{a}+\lambda \text{b}\) and \(\text{u}-\text{v}\) are equivalent.

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\(\lambda =-3\)
Let \(\text{u}\) and \(\text{v}\) be two nonzero vectors that are nonequivalent. Consider the vectors \(\text{a}=2\text{u}-4\text{v}\) and \(\text{b}=3\text{u}-7\text{v}\) defined in terms of \(\text{u}\) and \(\text{v}.\) Find the scalars \(\alpha\) and \(\beta\) such that vectors \(\alpha \text{a}+\beta \text{b}\) and \(\text{u}-\text{v}\) are equivalent.

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.

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

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

imaginary unit
i² = −1.

^\circ

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

\approx

approximately equal
Equal to the precision shown, not exactly.

\leq,\ \geq

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

\frac{\partial f}{\partial x}

partial derivative
Derivative with respect to x, holding the other variables fixed.

\nabla f

gradient (nabla, del)
Vector of partial derivatives; points uphill.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\iint,\ \oint

double / contour integral
Integral over a region of the plane; integral around a closed curve.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

\mathbf{u} \cdot \mathbf{v},\ \|\mathbf{v}\|

dot product, norm
Σ u_i v_i; the length of v, √(v·v).

How to: Vectors in the Plane

Describe a plane vector, using correct notation.
Perform basic vector operations (scalar multiplication, addition, subtraction).
Express a vector in component form.
Explain the formula for the magnitude of a vector.
Express a vector in terms of unit vectors.
Give two examples of vector quantities.
The vector
Use either addition method to find

Questions people ask

What is a partial derivative?

The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.

What does the gradient point at?

Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.

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Parts of this page are adapted from OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Mer i Multivariable Calculus

Functions of several variables Partial derivatives and the gradient Optimisation in several variables Double and triple integrals Vector fields, line integrals and the big theorems