9.1 Vectors

Vanessa Coffelt

9.1 Vectors

As we have seen numerous times in this book, Mathematics can be used to model and solve real-world problems. For many applications, real numbers suffice; that is, real numbers with the appropriate units attached can be used to answer questions like “How close is the nearest Sasquatch nest?”

There are other times though, when these kinds of quantities do not suffice. Perhaps it is important to know, for instance, how close the nearest Sasquatch nest is as well as the direction in which it lies. To answer questions like these which involve both a quantitative answer, or magnitude, along with a direction, we use the mathematical objects called vector.^[1]

A vector is represented geometrically as a directed line segment where the magnitude of the vector is taken to be the length of the line segment and the direction is made clear with the use of an arrow at one endpoint of the segment. When referring to vectors in this text, we shall adopt^[2] the `arrow’ notation, so the symbol $\vec{v}$ is read as `the vector $v$ ‘. Below is a typical vector $\vec{v}$ with endpoints $P\left(1, 2\right)$ and $Q\left(4, 6\right)$ .

The point $P$ is called the tail of $\vec{v}$ and the point $Q$ is called the terminal point or head of $\vec{v}$ . We can reconstruct $\vec{v}$ completely from $P$ and $Q$ , so we write $\vec{v} = \overrightarrow{PQ}$ , where the order of points $P$ (initial point) and $Q$ (terminal point) is important. (Think about this before moving on.)

A ray with endpoint P(1,2) and passing through the point Q(4,6). The arrow of the ray is at Q. The vector is named v = PQ — A Vector with initial point P and terminal point Q

While it is true that $P$ and $Q$ completely determine $\vec{v}$ , it is important to note that because vectors are defined in terms of their two characteristics, magnitude and direction, any directed line segment with the same length and direction as $\vec{v}$ is considered to be the same vector as $\vec{v}$ , regardless of its initial point.

In the case of our vector $\vec{v}$ above, any vector which moves three units to the right and four up^[3] from its initial point to arrive at its terminal point is considered the same vector as $\vec{v}$ . The notation we use to capture this idea is the component form of the vector, $\vec{v} = \left<3,4\right>$ , where the first number, $3$ , is called the $x$ –component of $\vec{v}$ and the second number, $4$ , is called the $y$ –component of $\vec{v}$ .

For example, if we wanted to reconstruct $\vec{v} = \left<3,4\right>$ with initial point $P'(-2,3)$ , then we would find the terminal point of $\vec{v}$ by adding $3$ to the $x$ -coordinate and adding $4$ to the $y$ -coordinate to obtain the terminal point $Q'(1,7)$ .

A ray with initial point point P'(-2,3) and terminal point Q'(1,7). There is a dashed horizontal line going from P' to the point under Q' and is labeled "over 3". Then a vertical dashed line from the end of the horizontal dashed line up to Q', labeled "up 4". The diagram is labeled vector v = <3,4> with initial point P'(-2,3). — Component Form of a Vector

The component form of a vector is what ties these very geometric objects back to Algebra and ultimately Trigonometry. We generalize our example in our definition below.

Definition 9.1

Suppose $\vec{v}$ is represented by a directed line segment with initial point $P\left(x_{0}, y_{0}\right)$ and terminal point $Q\left(x_{1}, y_{1}\right)$ . The component form of $\vec{v}$ is given by

$\vec{v} = \overrightarrow{PQ} = \left< x_{1} - x_{0}, y_{1} - y_{0} \right>$

Using the language of components, we have that two vectors are equal if and only if their corresponding components are equal. That is, $\left<v_{1}, v_{2}\right> = \left<v_{1}', v_{2}'\right>$ if and only if $v_{1} = v_{1}'$ and $v_{2} = v_{2}'$ . (Again, think about this before reading on.)

We now set about defining operations on vectors. Suppose we are given two vectors $\vec{v}$ and $\vec{w}$ . The sum, or resultant vector $\vec{v} + \vec{w}$ is obtained as follows. First, plot $\vec{v}$ . Next, plot $\vec{w}$ so that its initial point is the terminal point of $\vec{v}$ . To plot the vector $\vec{v} + \vec{w}$ we begin at the initial point of $\vec{v}$ and end at the terminal point of $\vec{w}$ . It is helpful to think of the vector $\vec{v} + \vec{w}$ as the `net result’ of moving along $\vec{v}$ then moving along $\vec{w}$ .

A triangle created by three vectors. Vector v starts at the lower left vertex of the triangle. The terminal point of the vector v is the initial point of the vector w. The third side of the triangle is has an initial point at the initial point of v and the terminal point is at the terminal point of w. The vector is label v+w. — Geometric Proof of the Sum of Two Vectors

Our next example makes good use of resultant vectors and reviews bearings and the Law of Cosines.^[4]

Example 9.1.1

A plane leaves an airport with an airspeed^[5] of 175 miles per hour at a bearing of N $40^{\circ}$ E. A 35 mile per hour wind is blowing at a bearing of S $60^{\circ}$ E. Find the true speed of the plane, rounded to the nearest mile per hour, and the true bearing of the plane, rounded to the nearest degree.

Solution:

For both the plane and the wind, we are given their speeds and their directions. Coupling speed (as a magnitude) with direction is the concept of velocity which we’ve seen a few times before.^[6]

We let $\vec{v}$ denote the plane’s velocity and $\vec{w}$ denote the wind’s velocity in the diagram below. The `true’ speed and bearing is found by analyzing the resultant vector, $\vec{v} + \vec{w}$ .

From the vector diagram, we get a triangle, the lengths of whose sides are the magnitude of $\vec{v}$ , which is 175, the magnitude of $\vec{w}$ , which is 35, and the magnitude of $\vec{v} + \vec{w}$ , which we’ll call $c$ .

From the given bearing information, we go through the usual geometry to determine that the angle between the sides of length 35 and 175 measures $100^{\circ}$ .

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From the Law of Cosines, we determine $c = \sqrt{31850 - 12250\cos(100^{\circ})} \approx 184$ , which means the true speed of the plane is (approximately) $184$ miles per hour.

To determine the true bearing of the plane, we need to determine the angle $\alpha$ . Using the Law of Cosines once more,^[7] we find $\cos(\alpha) = \frac{c^2+29400}{350c}$ so that $\alpha \approx 11^{\circ}$ .

Given the geometry of the situation, we add $\alpha$ to the given $40^{\circ}$ and find the true bearing of the plane to be (approximately) N $51^{\circ}$ E.

Our next step is to define addition of vectors component-wise to match the geometric action.^[8]

Definition 9.2

Suppose $\vec{v} = \left<v_{1},v_{2}\right>$ and $\vec{w} = \left<w_{1},w_{2}\right>$ . The vector sum, $\vec{v} + \vec{w}$ , is defined by

$\vec{v} + \vec{w} = \left< v_{1} + w_{1}}, v_{2} + w_{2} \right>$

Example 9.1.2

Let $\vec{v} = \left<3,4\right>$ and suppose $\vec{w} = \overrightarrow{PQ}$ where $P(-3,7)$ and $Q(-2,5)$ . Compute $\vec{v} + \vec{w}$ and interpret this sum geometrically.

Solution:

Before we can add the vectors using Definition 9.2, we need to write $\vec{w}$ in component form. Using Definition 9.1, we get $\vec{w} = \left<-2-(-3),5-7\right> = \left<1,-2\right>$ . Thus,

$\vec{v} + \vec{w} = \left<3,4\right> + \left<1,-2\right> = \left< 3 + 1, 4 + (-2) \right> = \left<4, 2\right>.$

To visualize this sum, we draw $\vec{v}$ with its initial point at $(0,0)$ (for convenience) so that its terminal point is $(3,4)$ . Next, we graph $\vec{w}$ with its initial point at $(3,4)$ . Moving one to the right and two down, we find the terminal point of $\vec{w}$ to be $(4,2)$ .

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We see the vector $\vec{v} + \vec{w}$ has initial point $(0,0)$ and terminal point $(4,2)$ so its component form is $\left<4,2\right>.$

In order for vector addition to enjoy the same kinds of properties as real number addition, it is necessary to extend our definition of vectors to include a `zero vector’, $\vec{0} = \left<0, 0\right>.$

Geometrically, $\vec{0}$ represents a point, which we can (very broadly) think of as a directed line segment with the same initial and terminal points. The reader may well object to the inclusion of $\vec{0}$ , because after all, vectors are supposed to have both a magnitude (length) and a direction.

While it seems clear that the magnitude of $\vec{0}$ should be $0$ , it is not clear what its direction is. As we shall see, the direction of $\vec{0}$ is in fact undefined, but this minor hiccup in the natural flow of things is worth the benefits we reap by including $\vec{0}$ in our discussions. We have the following theorem.

Theorem 9.1 Properties of Vector Addition

Commutative Property: For all vectors $\vec{v} \text{ and } \vec{w}$ , $\vec{v} + \vec{w} = \vec{w} + \vec{v}$
Associative Property: For all vectors $\vec{u}, \vec{v} \text{ and } \vec{w}$ , $\left(\vec{u} + \vec{v}\right) + \vec{w} = \vec{u} + \left(\vec{v} + \vec{w}\right)$
Identity Property: For all vectors $\vec{v}$ ,
$\vec{v} + \vec{0} = \vec{0} + \vec{v} = \vec{v}.$

The vector $\vec{0}$ acts as the additive identity for vector addition.
Inverse Property: For every vector $\vec{v} = \left< v_{1}, v_{2} \right>$ , the vector $\vec{w} = \left< - v_{1}, -v_{2} \right>$ satisfies
$\vec{v} + \vec{w} = \vec{w} + \vec{v} = \vec{0}.$

That is, the additive inverse of a vector is the vector of the additive inverses of its components.

The properties in Theorem 9.1 are easily verified using the definition of vector addition, and are a direct consequence of the definition of vector addition along with properties inherited from real number arithmetic.

For the commutative property, we note that if $\vec{v} = \left<v_{1},v_{2}\right>$ and $\vec{w} = \left<w_{1},w_{2}\right>$ then

$\begin{array}{rcl} \vec{v} + \vec{w} & = & \left< v_{1}, v_{2} \right> + \left< w_{1}, w_{2} \right> \\ & = & \left< v_{1} + w_{1}, v_{2} + w_{2} \right> \\ & = & \left< w_{1} + v_{1}, w_{2} + v_{2} \right> \\ & = & \vec{w} + \vec{v} \end{array}$

Geometrically, we can `see’ the commutative property by realizing that the sums $\vec{v}+\vec{w}$ and $\vec{w} + \vec{v}$ are the same directed diagonal determined by the parallelogram below.

Five vectors look like a parallelogram. The lower left vertex of the parallelogram. The vector v starts at this vertex and then vector w starts at the terminal point of vector v. On the other side of the parallelogram, vector w starts at the lower left vertex and then vector v starts at the terminal point of w. The terminal point of v is the same point as the terminal point of w on the other side of the parallelogram. The diagonal of the parallelogram represents the sum of the two vectors on each side of the parallelogram. — Demonstrating the Commutative Property of Vector Addition

The proofs of the associative and identity properties proceed similarly, and the reader is encouraged to verify them and provide accompanying diagrams.

The additive identity property is likewise verified algebraically using a calculation. If $\vec{v} = \left<v_{1},v_{2}\right>$ , then

$\begin{array}{rcl} \vec{v} + \vec{0} &=& \left<v_{1},v_{2}\right> + \left<0, 0\right>\\ &=& \left< v_{1} + 0, v_{2} + 0 \right>\\ &=& \left<v_{1},v_{2}\right>\\ &=& \vec{v} \end{array}$

From the commutative property of vector addition, we get that $\vec{0} + \vec{v} = \vec{v}$ as well. Again, the reader is encouraged to visualize what this means geometrically.^[9]

Regarding additive inverses, we can verify by direct computation that if $\vec{v} = \left< v_{1}, v_{2} \right>$ and $\vec{w} = \left< - v_{1}, -v_{2} \right>$ ,

$\begin{array}{rcl} \vec{v} + \vec{w} &=& \left< v_{1}, v_{2} \right> + \left< - v_{1}, -v_{2} \right> \\ &=& \left< v_{1} + ( - v_{1}), v_{2} + ( - v_{2}) \right> \\ &=& \left< 0,0 \right> \\ &=& \vec{0} \end{array}$

Once again, the commutative property of vector addition assures us that, likewise, $\vec{w} + \vec{v} = \vec{0}.$

Moreover, additive inverses of vectors are unique. That is, given a vector $\vec{v} = \left<v_{1}, v_{2}\right>$ , there is precisely only one vector $\vec{w}$ so that $\vec{v} + \vec{w} = \vec{0}.$

To see this, suppose a vector $\vec{w} = \left<w_{1},w_{2}\right>$ satisfies $\vec{v} + \vec{w} = \vec{0}$ . By the definition of vector addition, we have $\left<v_{1} + w_{1}, v_{2} + w_{2}\right> = \left<0,0\right>$ . Hence, $v_{1} + w_{1} = 0$ and $v_{2} + w_{2} = 0$ . We get $w_{1} = -v_{1}$ and $w_{2} = -v_{2}$ so that $\vec{w} = \left<-v_{1} , -v_{2} \right>$ as prescribed in Theorem 9.1.

Hence, every vector $\vec{v}$ has one, and only one, additive inverse. In general, we denote the additive inverse of a vector $\vec{v}$ with the (highly suggestive) notation $- \vec{v}.$

Geometrically, the vectors $\vec{v} = \left<v_{1}, v_{2}\right>$ and $-\vec{v} = \left<-v_{1}, -v_{2}\right>$ have the same length, but opposite directions. As a result, when adding the vectors geometrically, the sum $\vec{v} + (-\vec{v})$ results in starting at the initial point of $\vec{v}$ and ending back at the initial point of $\vec{v}$ . That is, the net result of moving $\vec{v}$ then $-\vec{v}$ is not moving at all.

Two vectors side by side. The first vector increases left two right and is labeled v. The second vector is the same length, but the terminal point and initial point are at the opposite ends from vector v. The second vector is labeled -v. — Graphical Representation of Additive Inverse of a Vector

Using the additive inverse of a vector, we can define the difference of two vectors: $\vec{v} - \vec{w} = \vec{v} + (-\vec{w})$ . Looking at this at the level of components, we see if $\vec{v} = \left<v_{1},v_{2}\right>$ and $\vec{w} = \left<w_{1},w_{2}\right>$ then

$\begin{array}{rcl} \vec{v} - \vec{w} & = & \vec{v} + (-\vec{w}) \\ & = & \left<v_{1},v_{2}\right> + \left<-w_{1},-w_{2}\right> \\ & = & \left<v_{1} + \left(-w_{1}\right),v_{2} + \left(-w_{2}\right) \right>\\ & = & \left<v_{1} -w_{1},v_{2} - w_{2}\right> \\ \end{array}$

In other words, like vector addition, vector subtraction works component-wise.

To interpret the vector $\vec{v} - \vec{w}$ geometrically, we note

$\begin{array}{rcll} \vec{w} + \left(\vec{v} - \vec{w}\right) & = & \vec{w} + \left(\vec{v} +(-\vec{w})\right) & \text{Definition of Vector Subtraction} \\ & = & \vec{w} + \left((-\vec{w})+\vec{v}\right) & \text{Commutativity of Vector Addition} \\ & = & (\vec{w} + (-\vec{w})) + \vec{v} & \text{Associativity of Vector Addition} \\ & = & \vec{0} + \vec{v} & \text{Definition of Additive Inverse}\\ & = & \vec{v} & \text{Definition of Additive Identity} \\ \end{array}$

This means that the `net result’ of moving along $\vec{w}$ then moving along $\vec{v} - \vec{w}$ is just $\vec{v}$ itself.

From the diagram below on the left, we see that $\vec{v}-\vec{w}$ may be interpreted as the vector whose initial point is the terminal point of $\vec{w}$ and whose terminal point is the terminal point of $\vec{v}.$

Two diagrams side by side. The first has vectors v and w with the same initial points and different terminal points. The terminal point of w is the the initial point of the vector representing the difference between vector v and vector w. The terminal point of vector v is the terminal point of the difference vector. The second diagram has the first diagram along with vector v starting at the original vector w and the vector w starting at the terminal point of vector v. The terminal point of the new versions of v and w have the same point. The second diagram is a parallelogram. — Geometric Interpretation of Vector Difference

It is also worth mentioning that in the parallelogram determined by the vectors $\vec{v}$ and $\vec{w}$ above on the right, the vector $\vec{v}-\vec{w}$ is one of the diagonals — the other being $\vec{v} + \vec{w}$ .

Next, we discuss scalar multiplication — that is, taking a real number times a vector.

Definition 9.3

If $k$ is a real number and $\vec{v} = \left<v_{1},v_{2}\right>$ , we define $k\vec{v}$ by

$k\vec{v} = k\left<v_{1},v_{2}\right> =\left<k v_{1},k v_{2}\right>$

Scalar multiplication by $k$ in vectors can be understood geometrically as scaling the vector (if $k > 0$ ) or scaling the vector and reversing its direction (if $k < 0$ ) as demonstrated below.

For vectors side by side. The first is vector v with the initial point on a dashed line. The second vector is in the same direction as v, but twice as long. The third vector is in the same direction as v, but half as long. The last vector is the same length as 2v, but in the opposite direction. All four vectors start at the horizontal dashed line. — Examples of Scalar Multiplication of Vectors

Note by Definition 9.3, $(-1)\vec{v} = (-1)\left<v_{1},v_{2}\right> = \left<(-1)v_{1}, (-1)v_{2}\right> = \left<-v_{1},-v_{2}\right> = -\vec{v}$ , which is what we would expect. This and other properties of scalar multiplication are summarized in the next theorem.

Theorem 9.2 Properties of Scalar Multiplication

Associative Property: For every vector $\vec{v}$ and scalars $k$ and $r$ , $(kr)\vec{v} = k(r\vec{v})$
Identity Property: For all vectors $\vec{v}$ , $1\vec{v} = \vec{v}$
Additive Inverse Property: For all vectors $\vec{v}$ , $-\vec{v} = (-1)\vec{v}$ .
Distributive Property of Scalar Multiplication over Scalar Addition: For every vector $\vec{v}$ and scalars $k$ and $r$ ,
$(k+r)\vec{v} = k\vec{v} + r\vec{v}$
Distributive Property of Scalar Multiplication over Vector Addition: For all vectors $\vec{v}$ and $\vec{w}$ and scalars $k$ ,
$k(\vec{v}+\vec{w}) = k\vec{v} + k\vec{w}$
Zero Product Property: If $\vec{v}$ is vector and $k$ is a scalar, then
$k\vec{v} = \vec{0} \quad \text{if and only if} \quad k=0 \quad \text{or} \quad \vec{v} =\vec{0}$

The proof of Theorem 9.2, like the proof of Theorem 9.1, ultimately boils down to the definition of scalar multiplication and properties of real numbers.

For example, to prove the associative property, we let $\vec{v} = \left<v_{1},v_{2}\right>$ . If $k$ and $r$ are scalars then

$\begin{array}{rcll} (kr) \vec{v} & = & (kr) \left<v_{1},v_{2}\right> & \\ [3pt] & = & \left<(kr) v_{1}, (kr) v_{2}\right> & \text{Definition of Scalar Multiplication} \\ [3pt] & = &\left<k (r v_{1}), k (r v_{2})\right> & \text{Associative Property of Real Number Multiplication} \\ [3pt] & = & k\left<r v_{1}, r v_{2}\right> & \text{Definition of Scalar Multiplication} \\ [3pt] & = & k\left(r\left<v_{1}, v_{2}\right>\right) & \text{Definition of Scalar Multiplication} \\ [3pt] & = & k(r\vec{v}) & \\ \end{array}$

The reader is invited to think about what this property means geometrically. The remaining properties are proved similarly and are left as exercises.

Our next example demonstrates how Theorem 9.2 allows us to do the same kind of algebraic manipulations with vectors as we do with variables — multiplication and division of vectors notwithstanding. If the pedantry seems familiar, it should.

Example 9.1.3

Solve $5\vec{v} - 2\left(\vec{v} + \left<1,-2\right>\right) = \vec{0}$ for $\vec{v}.$

Solution:

$\begin{array}{rcl} 5\vec{v} - 2\left(\vec{v} +\ \left<1,-2\right>\right) & = & \vec{0} \\ 5\vec{v} + (-1)\left[ 2\left(\vec{v} +\left<1,-2\right>\right)\right] & = & \vec{0}\\ 5\vec{v} + [(-1)(2)]\left(\vec{v} + \left<1,-2\right>\right) & = & \vec{0}\\ 5\vec{v} + (-2)\left(\vec{v} + \left<1,-2\right>\right) & = & \vec{0}\\ 5\vec{v} + \left[(-2)\vec{v} + (-2)\left<1,-2\right> \right] & = & \vec{0}\\ 5\vec{v} + \left[(-2)\vec{v} + \left<(-2)(1),(-2)(-2)\right> \right] & = & \vec{0}\\ \left[5\vec{v} + (-2)\vec{v}\right] + \left<-2,4\right> & = & \vec{0}\\ (5 + (-2)) \vec{v} + \left<-2,4\right> & = & \vec{0}\\ 3\vec{v} + \left<-2,4\right> & = & \vec{0}\\ \left(3\vec{v} + \left<-2,4\right>\right) + \left(- \left<-2,4\right>\right) & = & \vec{0} + \left(- \left<-2,4\right>\right)\\ 3\vec{v} + \left[\left<-2,4\right> + \left(- \left<-2,4\right>\right)\right] & = & \vec{0} + (-1) \left<-2,4\right>\\ 3\vec{v} + \vec{0} & = & \vec{0} + \left<(-1)(-2),(-1)(4)\right>\\ 3\vec{v} & = & \left<2,-4\right>\\[3pt] \frac{1}{3} \left(3\vec{v} \right) & = & \frac{1}{3} \left(\left<2,-4\right>\right)\\[3pt] \left[\left(\frac{1}{3}\right) (3) \right]\vec{v} & = & \left< \left(\frac{1}{3}\right)(2), \left(\frac{1}{3}\right)(-4)\right>\\[3pt] 1 \vec{v} & = & \left< \frac{2}{3}, -\frac{4}{3} \right> \\[3pt] \vec{v} & =& \left< \frac{2}{3}, -\frac{4}{3} \right> \\ \end{array}$

The reader is invited to check our solution in the original equation.

A vector whose initial point is $(0,0)$ is said to be in standard position. If $\vec{v} = \left<v_{1},v_{2}\right>$ is plotted in standard position, then its terminal point is necessarily $\left(v_{1},v_{2}\right)$ . (Once more, think about this before reading on.)

A vector in the first quadrant of the coordinate plane. The vector starts at the origin and a terminal point of (v sub 1, v sub 2). The magnitude of the vector is the distance of between the two points. The angle from the x-axis and the vector is labeled as theta. The graph is titled "v = <v_1, v_2> in standard position." — A Vector in Standard Position

Plotting a vector in standard position enables us to more easily quantify the concepts of magnitude and direction of the vector.

Recall the magnitude of vector $\vec{v}$ is the length of the directed line segment representing $\vec{v}$ . When plotted in standard position, the length of this line segment is none other than the distance from the origin $(0,0)$ to the point $\left(v_{1},v_{2}\right)$ . Hence, the magnitude of $\vec{v}$ , which we denote $\| \vec{v} \|$ , is given by $\| \vec{v} \| = \sqrt{v_{1}^2 + v_{2}^2}.$

Turning to the notion of direction, we note that the point $\left(v_{1},v_{2}\right)$ is on the terminal side of the angle $\theta$ depicted in the diagram above. From Theorem 7,4, we have $v_{1} = \| \vec{v} \| \cos(\theta)$ and $v_{2} = \| \vec{v} \| \sin(\theta)$ . From the definition of scalar multiplication and vector equality, we get

$\begin{array}{rcl} \vec{v} & = & \left< v_{1} , v_{2} \right> \\ [3pt] & = & \left< \| \vec{v} \| \cos(\theta), \| \vec{v} \| \sin(\theta) \right> \\ [3pt] & = & \| \vec{v} \| \left< \cos(\theta),\sin(\theta) \right> \\ \end{array}$

This motivates the following definition.

Definition 9.4

Suppose $\vec{v}$ is a vector with component form $\vec{v} =\left< v_{1} , v_{2} \right>$ .

The magnitude of $\vec{v}$ , denoted $\| \vec{v} \|$ , is given by $\| \vec{v} \| = \sqrt{v_{1}^2 + v_{2}^2}$
The direction angle of $\vec{v}$ , denoted $\theta$ , is given by $\displaystyle{\theta = \arctan \left( \frac{v_{2}}{v_{1}} \right)}$
Note: $\theta$ is an angle in standard position whose terminal side contains the point $\left(v_{1},v_{2}\right)$ .
If $\vec{v} \neq \vec{0}$ , directional vector of $\vec{v}$ , denoted $\bm\hat{v}$ is given by $\bm\hat{v} = \left< \cos(\theta ), \sin(\theta ) \right>$

Taken together, we get $\vec{v} = \left< \| \vec{v} \| \cos(\theta), \| \vec{v} \| \sin(\theta) \right>$ .

A few remarks are in order. First, we note that if $\vec{v} \neq 0$ then there are infinitely many angles $\theta$ which satisfy Definition 9.4. However, the fact that all of them must contain the same point $\left(v_{1},v_{2}\right)$ on their terminal sides means they are all coterminal.

Hence, if $\theta$ and $\theta'$ both satisfy the conditions of Definition 9.4, then $\cos(\theta) = \cos(\theta')$ and $\sin(\theta) = \sin(\theta')$ , and as such, $\left< \cos(\theta), \sin(\theta) \right> = \left< \cos(\theta'), \sin(\theta') \right>$ making $\bm\hat{v}$ is well-defined.

For $\vec{0} = \left< 0, 0 \right>$ , note that $\| \vec{0} \| = \sqrt{0^2 + 0^2} = 0$ . Hence, $\| \vec{0} \| \left< \cos(\theta), \sin(\theta) \right> = 0 \left< \cos(\theta), \sin(\theta) \right> = <0,0>$ for every angle $\theta$ . In other words, every angle $\theta$ satisfies the equation $\vec{v} = \left< \| \vec{v} \| \cos(\theta), \| \vec{v} \| \sin(\theta) \right>$ in Definition 9.4, so for this reason, $\bm\hat{0}$ is undefined.

The following theorem summarizes the important facts about the magnitude and direction of a vector.

Theorem 9.3 Properties of Magnitude and Direction

Suppose $\vec{v}$ is a vector.

$\| \vec{v} \| \geq 0$ and $\| \vec{v} \| = 0$ if and only if $\vec{v} = \vec{0}$
For all scalars $k$ , $\| k \, \vec{v} \| = |k| \| \vec{v} \|$
If $\vec{v} \neq \vec{0}$ then $\vec{v} = \| \vec{v} \| \bm\hat{v}$ , so that $\bm\hat{v} = \left(\dfrac{1}{\|\vec{v}\|}\right) \vec{v}$ . ^[10]

The proof of the first property in Theorem 9.3 is a direct consequence of the definition of $\| \vec{v} \|$ . Given $\vec{v} = \left< v_{1} ,v_{2}\right>$ , then $\| \vec{v} \| = \sqrt{v_{1}^2 + v_{2}^2}$ which is by definition greater than or equal to $0$ . Moreover, $\sqrt{v_{1}^2 + v_{2}^2} = 0$ if and only of $v_{1}^2 + v_{2}^2 = 0$ if and only if $v_{1} = v_{2} = 0$ . Hence, $\| \vec{v} \| = 0$ if and only if $\vec{v} = \left<0,0\right> = \vec{0}$ , as required.

The second property is a result of the definition of magnitude and scalar multiplication along with a property of radicals. If $\vec{v} = \left< v_{1} ,v_{2}\right>$ and $k$ is a scalar then

$\begin{array}{rcll} \| k \, \vec{v} \| & = & \| k \left< v_{1}, v_{2}\right> \| & \\ [3pt]& = & \| \left<kv_{1},kv_{2}\right>\| & \text{Definition of scalar multiplication} \\ [3pt]& = & \sqrt{\left(kv_{1}\right)^2 + \left(kv_{2}\right)^2} & \text{Definition of magnitude} \\ [3pt]& = & \sqrt{k^2v_{1}^2 + k^2v_{2}^2} & \\[3pt]& = & \sqrt{k^2(v_{1}^2+v_{2}^2)} & \\ [3pt] & = & \sqrt{k^2} \sqrt{v_{1}^2+v_{2}^2} & \text{Product Rule for Radicals} \\ [3pt]& = & |k| \sqrt{v_{1}^2+v_{2}^2} & \text{$\sqrt{k^2} = |k|$} \\& = & |k| \| \vec{v} \| & \\ \end{array}$

The equation $\vec{v} = \| \vec{v} \| \bm\hat{v}$ in Theorem 9.3 is a consequence of the definitions of $\| \vec{v} \|$ and $\bm\hat{v}$ and was worked out in the discussion just prior to Definition 9.4. In words, the equation $\vec{v} = \| \vec{v} \| \bm\hat{v}$ says that any given vector is the product of its magnitude and its direction — an important concept to keep in mind when studying and using vectors.

The formula for $\bm\hat{v}$ stated in Theorem 9.3 is a consequence of solving $\vec{v} = \| \vec{v} \| \bm\hat{v}$ for $\bm\hat{v}$ by multiplying^[11] both sides of the equation by $\frac{1}{\| \vec{v} \|}$ and using the properties of Theorem 9.2. We leave these details to the reader. We are overdue for an example.

Example 9.1.4

Example 9.1.4.1

Write the component form of the vector $\vec{v}$ with $\|\vec{v}\| = 5$ so that when $\vec{v}$ is plotted in standard position, it lies in Quadrant II and makes a $60^{\circ}$ angle^[12] with the negative $x$ -axis.

Solution:

Write the component form of the vector $\vec{v}$ with $\|\vec{v}\| = 5$ so that when $\vec{v}$ is plotted in standard position, it lies in Quadrant II and makes a $60^{\circ}$ angle with the negative $x$ -axis.

We are told that $\| \vec{v} \| = 5$ and are given information about its direction, so we can use the formula $\vec{v} = \| \vec{v} \| \bm\hat{v}$ to get the component form of $\vec{v}.$

To determine $\bm\hat{v}$ , we appeal to Definition 9.4. Because $\vec{v}$ lies in Quadrant II and makes a $60^{\circ}$ angle with the negative $x$ -axis, one angle $\theta$ satisfying the criteria of Definition 9.4 is $\theta = 120^{\circ}.$

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Hence,

$\begin{array}{rcl} \bm\hat{v} &=& \left< \cos\left(120^{\circ}\right), \sin\left(120^{\circ}\right) \right>\\[10pt] &=& \left< - \frac{1}{2} , \frac{\sqrt{3}}{2} \right> \end{array}$

so the result is

$\begin{array}{rcl} \vec{v} &=& \| \vec{v} \| \bm\hat{v}\\ &=& 5 \left< - \frac{1}{2} , \frac{\sqrt{3}}{2} \right>\\[10pt] &=& \left< - \frac{5}{2} , \frac{5\sqrt{3}}{2} \right> \end{array}$

Example 9.1.4.2

For $\vec{v} = \left<3, -3\sqrt{3}\right>$ , compute $\|\vec{v}\|$ and $\theta$ , $0 \leq \theta < 2\pi$ so that $\vec{v} = \| \vec{v} \| \left<\cos(\theta), \sin(\theta)\right>$ .

Solution:

For $\vec{v} = \left<3, -3\sqrt{3}\right>$ , compute $\|\vec{v}\|$ and $\theta$ , $0 \leq \theta < 2\pi$ so that $\vec{v} = \| \vec{v} \| \left<\cos(\theta), \sin(\theta)\right>.$

For $\vec{v} = \left<3, -3\sqrt{3}\right>$ , we get $\| \vec{v} \| = \sqrt{(3)^2+(-3\sqrt{3})^2} = 6$ .

In light of Definition 9.4, we can find the $\theta$ we’re after by finding a Quadrant IV angle whose terminal side contains the point $(3, -3\sqrt{3})$ .

We compute $\theta$ using the definition of the direction angle of $\vec{v}$ , $\displaystyle{\theta = \arctan \left( \frac{v_{2}}{v_{1}} \right).}$

$\begin{array}{lcl} \theta & = & \arctan \left( \dfrac{v_{2}}{v_{1}} \right) \\[4pt] & = & \arctan \left( \dfrac{-3\sqrt{3}}{3} \right) \\[4pt] & = & \arctan \left( - \sqrt{3} \right) \\[4pt] & = & - \frac{\pi}{3} + \pi k, \text{ where } k \text{ is any integer} \\ \end{array}$

As the terminal side of $\vec{v}$ is in Quadrant IV, we let $k = 2$ resulting in $\theta = -\frac{ \pi}{3} + 2\pi = \frac{5\pi}{3}.$

We may check our answer by verifying $6\left<\cos\left(\frac{5\pi}{3}\right), \sin\left(\frac{5\pi}{3}\right) \right> = \left<3, -3\sqrt{3}\right> = \vec{v}$ .

Example 9.1.4.3a

For the vectors $\vec{v} = \left<3,4\right>$ and $\vec{w} = \left<1, -2\right>$ , determine the following.

$\bm\hat{v}$

Solution:

For the vectors $\vec{v} = \left<3,4\right>$ and $\vec{w} = \left<1, -2\right>$ , determine $\bm\hat{v}.$

We are given the component form of $\vec{v}$ , sowe’ll use the formula $\bm\hat{v} = \left(\frac{1}{\|\vec{v}\|}\right) \vec{v}.$

For $\vec{v} = \left<3,4\right>$ , we have $\| \vec{v} \| = \sqrt{3^2+4^2} = \sqrt{25} = 5$ .

Hence,

$\begin{array}{rcl} \bm\hat{v} &=& \frac{1}{5} \left< 3, 4 \right>\\[4pt] &=& \left<\frac{3}{5}, \frac{4}{5}\right> \end{array}$

Example 9.1.4.3b

For the vectors $\vec{v} = \left<3,4\right>$ and $\vec{w} = \left<1, -2\right>$ , determine the following.

$\| \vec{v} \| -2 \|\vec{w}\|$

Solution:

For the vectors $\vec{v} = \left<3,4\right>$ and $\vec{w} = \left<1, -2\right>$ , determine $\| \vec{v} \| -2 \|\vec{w}\|.$

We know from our work above that $\| \vec{v} \| = 5$ , so to find $\| \vec{v} \| -2 \|\vec{w}\|$ , we need only find $\| \vec{w} \|$ .

Given $\vec{w} = \left<1, -2\right>$ , we get $\| \vec{w} \| = \sqrt{1^2+(-2)^2} = \sqrt{5}$ .

Hence, $\| \vec{v} \| -2 \|\vec{w}\| = 5 - 2\sqrt{5}$ .

Example 9.1.4.3c

For the vectors $\vec{v} = \left<3,4\right>$ and $\vec{w} = \left<1, -2\right>$ , determine the following.

$\| \vec{v} -2\vec{w}\|$

Solution:

For the vectors $\vec{v} = \left<3,4\right>$ and $\vec{w} = \left<1, -2\right>$ , determine $\| \vec{v} -2\vec{w}\|.$

In the expression $\| \vec{v} -2\vec{w}\|$ , notice that the arithmetic on the vectors comes first, then the magnitude.

Hence, our first step is to find the component form of the vector $\vec{v} - 2\vec{w}$ . We get $\vec{v} - 2 \vec{w} = \left<3,4\right> - 2\left<1,-2\right> = \left<1, 8\right>$ .

Hence,

$\begin{array}{rcl} \| \vec{v} -2\vec{w}\| &=& \| \left<1, 8\right>\| \\ &=& \sqrt{1^2+8^2} \\ &=& \sqrt{65} \end{array}$

Example 9.1.4.3d

For the vectors $\vec{v} = \left<3,4\right>$ and $\vec{w} = \left<1, -2\right>$ , determine the following.

$\| \bm\hat{w} \|$

Solution:

For the vectors $\vec{v} = \left<3,4\right>$ and $\vec{w} = \left<1, -2\right>$ , determine $\| \bm\hat{w} \|.$

One approach to find $\| \bm\hat{w} \|$ , is to first find $\bm\hat{w}$ and then take the magnitude.

Using the formula $\bm\hat{w} = \left(\frac{1}{\| \vec{w} \|}\right) \vec{w}$ along with $\| \vec{w} \| = \sqrt{5}$ , which we found the in the previous problem, we get $\bm\hat{w} = \frac{1}{\sqrt{5}} \left<1, -2\right> = \left< \frac{1}{\sqrt{5}}, -\frac{2}{\sqrt{5}}\right> = \left< \frac{\sqrt{5}}{5}, -\frac{2\sqrt{5}}{5}\right>.$

Hence,

$\begin{array}{rcl} \| \bm\hat{w} \| &=& \sqrt{\left( \frac{\sqrt{5}}{5}\right)^2 + \left(-\frac{2\sqrt{5}}{5}\right)^2} \\[10pt] &=& \sqrt{\frac{5}{25} + \frac{20}{25}} \\[10pt] &=& \sqrt{1} \\ &=& 1 \end{array}$

Alternatively, we can use Theorem 9.3. Given $\bm\hat{w} = \left(\frac{1}{\| \vec{w} \|} \right) \vec{w}$ , where $\frac{1}{\| \vec{w} \|}>0$ is a scalar,

$\begin{array}{rcl} \| \bm\hat{w} \| &=& \left\| \left(\frac{1}{\| \vec{w} \|} \right) \vec{w} \right\| \\[10pt] &=& \frac{1}{\| \vec{w} \|} \| \vec{w} \| \\ &=& \frac{\| \vec{w} \|}{\| \vec{w} \|} \\ &=& 1 \end{array}$

For a third way to show $\| \bm\hat{w} \| = 1$ , we can appeal to Definition 9.4. Because $\bm\hat{w} = \left< \cos(\theta), \sin(\theta) \right>$ for some angle $\theta,$ $\| \bm\hat{w} \| = \sqrt{\cos^{2}(\theta) + \sin^{2}(\theta)} = \sqrt{1} = 1$ , where we have used the Pythagorean Identity, $\cos^{2}(\theta) + \sin^{2}(\theta) = 1$ . No matter how we approach the problem, $\| \bm\hat{w} \| = 1.$

Note that the second and third solutions to number 3d in Example 9.1.4 above work for any nonzero vector, $\vec{w}$ . We will have more to say about this shortly.

The process exemplified by number 1 in Example 9.1.4 above by which we take information about the magnitude and direction of a vector and find the component form of a vector is called resolving a vector into its components. As an application of this process, we revisit Example 9.1.1 below.

Example 9.1.5

A plane leaves an airport with an airspeed of 175 miles per hour with bearing N $40^{\circ}$ E. A 35 mile per hour wind is blowing at a bearing of S $60^{\circ}$ E. Find the true speed of the plane, rounded to the nearest mile per hour, and the true bearing of the plane, rounded to the nearest degree.

Solution:

We proceed as we did in Example 9.1.1 and let $\vec{v}$ denote the plane’s velocity and $\vec{w}$ denote the wind’s velocity, and set about determining $\vec{v} + \vec{w}.$

If we regard the airport as being at the origin, the positive $y$ -axis acting as due north and the positive $x$ -axis acting as due east, we see that the vectors $\vec{v}$ and $\vec{w}$ are in standard position and their directions correspond to the angles $50^{\circ}$ and $-30^{\circ}$ , respectively.

Hence, the component form of

$\begin{array}{rcl} \vec{v} &=& 175\left<\cos(50^{\circ}), \sin(50^{\circ})\right>\\ &=& \left<175\cos(50^{\circ}), 175\sin(50^{\circ})\right> \end{array}$

and the component form of

$\vec{w} = \left<35\cos(-30^{\circ}), 35\sin(-30^{\circ}) \right>$

We have no convenient way to express the exact values of cosine and sine of $50^{\circ}$ , so we leave both vectors in terms of cosines and sines.^[13] Adding corresponding components, we find the resultant vector

$\vec{v} + \vec{w} = \left< 175\cos(50^{\circ}) + 35\cos(-30^{\circ}), 175\sin(50^{\circ}) + 35\sin(-30^{\circ})\right>$

To find the `true’ speed of the plane, we compute the magnitude of this resultant vector

$\| \vec{v} + \vec{w}\| = \sqrt{ (175\cos(50^{\circ}) + 35\cos(-30^{\circ}))^2 + (175\sin(50^{\circ}) + 35\sin(-30^{\circ}))^2} \approx 184$

Hence, the `true’ speed of the plane is approximately 184 miles per hour.

To find the true bearing, we need to find the angle $\theta$ whose terminal side when graphed in standard position contains

$(x,y) = (175\cos(50^{\circ}) + 35\cos(-30^{\circ}), 175\sin(50^{\circ}) + 35\sin(-30^{\circ}))$

As both of these coordinates are positive,^[14] we know $\theta$ is a Quadrant I angle, as depicted below. Furthermore,

$\begin{array}{rcl} \tan(\theta) &=& \frac{y}{x} \\[10pt] &=& \frac{175\sin(50^{\circ}) + 35\sin(-30^{\circ})}{175\cos(50^{\circ}) + 35\cos(-30^{\circ})} \end{array}$

so using the arctangent function,^[15] we get $\theta \approx 39^{\circ}.$ Because, for the purposes of bearing, we need the angle between $\vec{v} + \vec{w}$ and the positive $y$ -axis, we take the complement of $\theta$ and find the `true’ bearing of the plane to be approximately N $51^{\circ}$ E.

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In part 3d of Example 9.1.4, we saw that the length of the direction vector, $\bm\hat{w}$ , $\| \bm\hat{w} \| = 1$ . Vectors of length $1$ play such an important role that they are given a special name.

Definition 9.5

Let $\vec{v}$ be a vector. If $\| \vec{v} \| = 1$ , then we say that $\vec{v}$ is a unit vector.

Note that if $\vec{v}$ is a unit vector, then necessarily, $\vec{v} = \| \vec{v} \| \bm\hat{v} = 1 \cdot \bm\hat{v} = \bm\hat{v}$ . Conversely, in the solution of part 3d of Example 9.1.4, two different arguments show for any nonzero vector $\vec{v}$ , $\| \bm\hat{v} \| = 1$ , so $\bm\hat{v}$ is a unit vector.

In other words, unit vectors are direction vectors and vice-versa. Indeed, the vector $\bm\hat{v}$ which we have defined as `the directional vector of $\vec{v}$ ‘ is often described as `the unit vector in the direction of $\vec{v}$ .’

In practice, if $\vec{v}$ is a unit vector we write it as $\bm\hat{v}$ as opposed to $\vec{v}$ because we have reserved the ` $\bm\hat{~}$ ‘ notation for unit vectors. The process of multiplying a nonzero vector by the factor $\frac{1}{\| \vec{v} \|}$ to produce a unit vector is called `normalizing the vector.’

The terminal points of unit vectors, when plotted in standard position, lie on the Unit Circle. (You should take the time to show this.) As a result, we visualize normalizing a nonzero vector $\vec{v}$ as shrinking^[16] its terminal point, when plotted in standard position, back to the Unit Circle.

A unit circle on the cartesian plane. Vector v is in the first quadrant with the initial point at the origin. Along vector v is the unit vector for v with a terminal point at the unit circle. The diagram is labeled "Visualizing vector normalization" and included the formula — Graphical Representation of Vector Normalization

Of all of the unit vectors, two deserve special mention.

Definition 9.6 The Principal Unit Vectors

The vector $\bm\hat{\text{i}}$ is defined by $\bm\hat{\text{i}} = \left<1,0\right>$
The vector $\bm\hat{\text{j}}$ is defined by $\bm\hat{\text{j}} = \left<0,1\right>$

Geometrically, in the $xy$ -plane, the vector $\bm\hat{\text{i}}$ as represents the positive $x$ -direction, whereas the vector $\bm\hat{\text{j}}$ represents the positive $y$ -direction. We have the following `decomposition’ theorem.^[17]

Theorem 9.4 Principal Vector Decomposition Theorem

Let $\vec{v}$ be a vector with component form $\vec{v} = \left< v_{1} ,v_{2}\right>$ . Then $\vec{v} = v_{1} \bm\hat{\text{i}} + v_{2} \bm\hat{\text{j}}$ .

The proof of Theorem 9.4 is straightforward. Given $\bm\hat{\text{i}} = \left<1,0\right>$ and $\bm\hat{\text{j}} = \left< 0,1\right>$ , we have from the definition of scalar multiplication and vector addition that

$\begin{array}{rcl} v_{1} \bm\hat{\text{i}} + v_{2} \bm\hat{\text{j}} &=& v_{1}\left<1,0\right> + v_{2}\left<0,1\right>\\ &=& \left<v_{1},0\right> + \left<0,v_{2}\right> \\ &=& \left<v_{1},v_{2}\right> \\ &=& \vec{v} \end{array}$

Geometrically, the situation looks like this:

A unit circle and the first quadrant of the cartesian plane. The vector v has an initial point at the origin. The principal unit vectors i and j are on the x- and y-axis respectively. The multiple of each principal vector is also drawn on the respective axes. — Geometric Principal Vector Decomposition of a Vector

We conclude this section with a classic example which demonstrates how vectors are used in physics to study forces. A `force’ is defined as a `push’ or a `pull.’ The intensity of the push or pull is the magnitude of the force, and is measured in Newtons (N) in the SI system or pounds (lbs.) $\!$ in the English system.^[18]

The following example uses all of the concepts in this section, and should be studied in great detail.

Example 9.1.6

A $50$ pound speaker is suspended from the ceiling by two support braces. If one of them makes a $60^{\circ}$ angle with the ceiling and the other makes a $30^{\circ}$ angle with the ceiling, what are the tensions on each of the supports?

Solution:

We represent the problem schematically below along with the corresponding vector diagram.

Two diagrams side by side. The first is of a box representing a rectangle and a box. The rectangle represents the ceiling beam in the example and the box represents the speaker and is labeled 50 lbs. There are two lines extending from opposite ends of the bottom of the rectangle to a center point on the top of the box. The angle between the line on the left and the rectangle is labeled 30 degrees, while the line on the right and the rectangle create a 60 degree angle. The second diagram removes the box and rectangle and adds dashed and dotted lines in their place, respectively. The left line is drawn as a vector T sub 2 and the right line is drawn as a vector T sub 1. An additional vector w is included going straight down from the common initial point of the other two vectors. — Graphical Representation of Example 9.1.6

We have three forces acting on the speaker: the weight of the speaker, which we’ll call $\vec{w}$ , pulling the speaker directly downward, and the forces on the support rods, which we’ll call $\vec{T_{1}}$ and $\vec{T_{2}}$ (for `tensions’) acting upward at angles $60^{\circ}$ and $30^{\circ}$ , respectively.

We are looking for the tensions on the support, which are the magnitudes $\| \vec{T_{1}} \|$ and $\| \vec{T_{2}} \|$ . In order for the speaker to remain stationary,^[19] we require $\vec{w} + \vec{T_{1}} + \vec{T_{2}} = \vec{0}$ .

Viewing the common initial point of these vectors as the origin and the dashed line as the $x$ -axis, we use Theorem 9.3 to get component representations for the three vectors involved. We can model the weight of the speaker as a vector pointing directly downwards with a magnitude of 50 pounds. That is, $\| \vec{w} \| = 50$ and $\bm\hat{w} = -\bm\hat{\text{j}} = \left<0,-1\right>$ .

Hence,

$\begin{array}{rcl} \vec{w} &=& 50\left<0,-1\right>\\ &=& \left<0,-50\right> \end{array}$

For the force in the first support, we get

$\begin{array}{rcl} \vec{T_{1}} & = & \| \vec{T_{1}} \|\left<\cos\left(60^{\circ}\right), \sin\left(60^{\circ}\right)\right> \\ [8pt] & = & \left< \dfrac{\| \vec{T_{1}} \|}{2} , \dfrac{\| \vec{T_{1}} \|\sqrt{3}}{2}\right> \\ \end{array}$

For the second support, we note that the angle $30^{\circ}$ is measured from the negative $x$ -axis, so the angle needed to write $\vec{T_{2}}$ in component form is $150^{\circ}$ . Hence

$\begin{array}{rcl} \vec{T_{2}} & = & \| \vec{T_{2}} \|\left<\cos\left(150^{\circ}\right), \sin\left(150^{\circ}\right)\right>\\ [8pt] & = & \left<-\dfrac{\| \vec{T_{2}} \|\sqrt{3}}{2}, \dfrac{\| \vec{T_{2}} \|}{2} \right> \\ \end{array}$

The requirement $\vec{w} + \vec{T_{1}} + \vec{T_{2}} = \vec{0}$ gives us the vector equation:

$\begin{array}{rcl} \vec{w} + \vec{T_{1}} + \vec{T_{2}} & = & \vec{0} \\ [8pt] \left<0,-50\right> + \left< \dfrac{\| \vec{T_{1}} \|}{2} , \dfrac{\| \vec{T_{1}} \|\sqrt{3}}{2}\right> + \left<-\dfrac{\| \vec{T_{2}} \|\sqrt{3}}{2}, \dfrac{\| \vec{T_{2}} \|}{2} \right> & = & \left<0, 0\right> \\ [8pt] \left< \dfrac{\| \vec{T_{1}} \|}{2} -\dfrac{\| \vec{T_{2}} \|\sqrt{3}}{2}, \dfrac{\| \vec{T_{1}} \|\sqrt{3}}{2} + \dfrac{\| \vec{T_{2}} \|}{2} -50 \right> & = & \left<0, 0\right> \\ \end{array}$

Equating the corresponding components of the vectors on each side, we get a system of linear equations in the variables $\| \vec{T_{1}} \|$ and $\| \vec{T_{2}} \|.$

$\left\{ \begin{array}{lrcl} (E1) & \dfrac{\| \vec{T_{1}} \|}{2} -\dfrac{\| \vec{T_{2}} \|\sqrt{3}}{2} & = & 0 \\ [8pt] (E2) & \dfrac{\| \vec{T_{1}} \|\sqrt{3}}{2} + \dfrac{\| \vec{T_{2}} \|}{2} -50 & = & 0 \\ \end{array} \right.$

From $(E1)$ , we get $\| \vec{T_{1}} \| = \| \vec{T_{2}} \| \sqrt{3}$ .

Substituting that into $(E2)$ gives $\frac{(\| \vec{T_{2}} \| \sqrt{3})\sqrt{3}}{2} + \frac{\| \vec{T_{2}} \|}{2} - 50 = 0.$

Solving, we get $2\| \vec{T_{2}} \| - 50 =0$ , so $\| \vec{T_{2}} \| = 25$ pounds.

Hence, $\| \vec{T_{1}} \| = \| \vec{T_{2}} \| \sqrt{3} = 25 \sqrt{3}$ pounds.

Note that the sum of the tensions on the wires in Example 9.1.6 exceed the $50$ pounds of the speaker. Explaining why this happens is a good exercise and gets at the heart of the concept of vectors and resolution of forces. Speaking of exercises $\ldots$

9.1.1 Section Exercises

In Exercises 1 – 10, use the given pair of vectors $\vec{v}$ and $\vec{w}$ to compute the following quantities. State whether the result is a vector or a scalar.

$\vec{v} + \vec{w}$
$\vec{w} - 2\vec{v}$
$\| \vec{v} + \vec{w} \|$
$\| \vec{v} \| + \| \vec{w} \|$
$\| \vec{v} \| \vec{w} - \| \vec{w} \| \vec{v}$
$\|\vec{w}\| \hat{v}$

Finally, verify that the vectors satisfy the Parallelogram Law

$\|\vec{v}\|^2 + \|\vec{w}\|^2 = \dfrac{1}{2}\left[ \| \vec{v} + \vec{w}\|^2 + \|\vec{v} - \vec{w}\|^2\right]$

$\vec{v} = \left<12, -5\right>$ , $\vec{w} = \left<3, 4\right>$
$\vec{v} = \left<-7, 24 \right>$ , $\vec{w} = \left<-5, -12\right>$
$\vec{v} = \left<2, -1 \right>$ , $\vec{w} = \left<-2, 4 \right>$
$\vec{v} = \left<10, 4 \right>$ , $\vec{w} = \left<-2, 5 \right>$
$\vec{v} = \left<-\sqrt{3}, 1\right>$ , $\vec{w} = \left<2\sqrt{3}, 2\right>$
$\vec{v} = \left<\frac{3}{5}, \frac{4}{5}\right>$ , $\vec{w} = \left<-\frac{4}{5}, \frac{3}{5}\right>$
$\vec{v} = \left<\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right>$ , $\vec{w} = \left<-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right>$
$\vec{v} = \left<\frac{1}{2}, \frac{\sqrt{3}}{2} \right>$ , $\vec{w} = \left< -1, -\sqrt{3} \right>$
$\vec{v} = 3\bm\hat{\text{i}} + 4\bm\hat{\text{j}}$ , $\vec{w} = -2\bm\hat{\text{j}}$
$\vec{v} =\frac{1}{2} \left(\bm\hat{\text{i}} + \bm\hat{\text{j}}\right)$ , $\vec{w} = \frac{1}{2} \left(\bm\hat{\text{i}} - \bm\hat{\text{j}}\right)$

In Exercises 11 – 25, write the component form of the vector $\vec{v}$ using the information given about its magnitude and direction. Give exact values.

$\|\vec{v}\| = 6$ ; when drawn in standard position $\vec{v}$ lies in Quadrant I and makes a $60^{\circ}$ angle with the positive $x$ -axis
$\|\vec{v}\| = 3$ ; when drawn in standard position $\vec{v}$ lies in Quadrant I and makes a $45^{\circ}$ angle with the positive $x$ -axis
$\|\vec{v}\| = \frac{2}{3}$ ; when drawn in standard position $\vec{v}$ lies in Quadrant I and makes a $60^{\circ}$ angle with the positive $y$ -axis
$\|\vec{v}\| = 12$ ; when drawn in standard position $\vec{v}$ lies along the positive $y$ -axis
$\|\vec{v}\| = 4$ ; when drawn in standard position $\vec{v}$ lies in Quadrant II and makes a $30^{\circ}$ angle with the negative $x$ -axis
$\|\vec{v}\| = 2\sqrt{3}$ ; when drawn in standard position $\vec{v}$ lies in Quadrant II and makes a $30^{\circ}$ angle with the positive $y$ -axis
$\|\vec{v}\| = \frac{7}{2}$ ; when drawn in standard position $\vec{v}$ lies along the negative $x$ -axis
$\|\vec{v}\| = 5\sqrt{6}$ ; when drawn in standard position $\vec{v}$ lies in Quadrant III and makes a $45^{\circ}$ angle with the negative $x$ -axis
$\|\vec{v}\| = 6.25$ ; when drawn in standard position $\vec{v}$ lies along the negative $y$ -axis
$\|\vec{v}\| = 4\sqrt{3}$ ; when drawn in standard position $\vec{v}$ lies in Quadrant IV and makes a $30^{\circ}$ angle with the positive $x$ -axis
$\|\vec{v}\| = 5\sqrt{2}$ ; when drawn in standard position $\vec{v}$ lies in Quadrant IV and makes a $45^{\circ}$ angle with the negative $y$ -axis
$\| \vec{v}\| = 2\sqrt{5}$ ; when drawn in standard position $\vec{v}$ lies in Quadrant I and makes an angle measuring $\arctan(2)$ with the positive $x$ -axis
$\| \vec{v}\| = \sqrt{10}$ ; when drawn in standard position $\vec{v}$ lies in Quadrant II and makes an angle measuring $\arctan(3)$ with the negative $x$ -axis
$\| \vec{v}\| = 5$ ; when drawn in standard position $\vec{v}$ lies in Quadrant III and makes an angle measuring $\arctan\left(\frac{4}{3}\right)$ with the negative $x$ -axis
$\| \vec{v}\| = 26$ ; when drawn in standard position $\vec{v}$ lies in Quadrant IV and makes an angle measuring $\arctan\left(\frac{5}{12}\right)$ with the positive $x$ -axis

In Exercises 26 – 31, approximate the component form of the vector $\vec{v}$ using the information given about its magnitude and direction. Round your approximations to two decimal places.

$\|\vec{v}\| = 392$ ; when drawn in standard position $\vec{v}$ makes a $117^{\circ}$ angle with the positive $x$ -axis
$\|\vec{v}\| = 63.92$ ; when drawn in standard position $\vec{v}$ makes a $78.3^{\circ}$ angle with the positive $x$ -axis
$\|\vec{v}\| = 5280$ ; when drawn in standard position $\vec{v}$ makes a $12^{\circ}$ angle with the positive $x$ -axis
$\|\vec{v}\| = 450$ ; when drawn in standard position $\vec{v}$ makes a $210.75^{\circ}$ angle with the positive $x$ -axis
$\|\vec{v}\| = 168.7$ ; when drawn in standard position $\vec{v}$ makes a $252^{\circ}$ angle with the positive $x$ -axis
$\| \vec{v}\| = 26$ ; when drawn in standard position $\vec{v}$ makes a $304.5^{\circ}$ angle with the positive $x$ -axis

In Exercises 32- 52, for the given vector $\vec{v}$ , compute the magnitude $\|\vec{v}\|$ and the direction angle $\theta$ with $0 \leq \theta < 360^{\circ}$ so that $\vec{v} = \|\vec{v}\| \left<\cos(\theta), \sin(\theta) \right>$ (See Definition 9.4.) Round approximations to two decimal places.

$\vec{v} = \left<1,\sqrt{3}\right>$
$\vec{v} = \left<5,5\right>$
$\vec{v} = \left<-2\sqrt{3}, 2 \right>$
$\vec{v} = \left<-\sqrt{2}, \sqrt{2} \right>$
$\vec{v} = \left<-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right>$
$\vec{v} = \left<-\frac{1}{2}, -\frac{\sqrt{3}}{2} \right>$
$\vec{v} = \left<6, 0\right>$
$\vec{v} = \left<-2.5, 0\right>$
$\vec{v} = \left<0, \sqrt{7} \right>$
$\vec{v} = -10 \bm\hat{\text{j}}$
$\vec{v} = \left< 3,4\right>$
$\vec{v} = \left<12, 5\right>$
$\vec{v} = \left<-4, 3 \right>$
$\vec{v} = \left<-7, 24\right>$
$\vec{v} = \left<-2, -1 \right>$
$\vec{v} = \left<-2, -6\right>$
$\vec{v} = \bm\hat{\text{i}} + \bm\hat{\text{j}}$
$\vec{v} = \bm\hat{\text{i}} - 4\bm\hat{\text{j}}$
$\vec{v} = \left<123.4, -77.05\right>$
$\vec{v} = \left<965.15, 831.6\right>$
$\vec{v} = \left<-114.1, 42.3\right>$
A small boat leaves the dock at Camp DuNuthin and heads across the Nessie River at 17 miles per hour (that is, with respect to the water) at a bearing of S $68^{\circ}$ W. The river is flowing due east at 8 miles per hour. What is the boat’s true speed and heading? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree.
The HMS Sasquatch leaves port with bearing S $20^{\circ}$ E maintaining a speed of 42 miles per hour (that is, with respect to the water). If the ocean current is 5 miles per hour with a bearing of N $60^{\circ}$ E, find the HMS Sasquatch’s true speed and bearing. Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree.
If the captain of the HMS Sasquatch in Exercise 54 wishes to reach Chupacabra Cove, an island 100 miles away at a bearing of S $20^{\circ}$ E from port, in three hours, what speed and heading should she set to take into account the ocean current? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree.
HINT: If $\vec{v}$ denotes the velocity of the HMS Sasquatch and $\vec{w}$ denotes the velocity of the current, what does $\vec{v} + \vec{w}$ need to be to reach Chupacabra Cove in three hours?
In calm air, a plane flying from the Pedimaxus International Airport can reach Cliffs of Insanity Point in two hours by following a bearing of N $8.2^{\circ}$ E at 96 miles an hour. (The distance between the airport and the cliffs is 192 miles.) If the wind is blowing from the southeast at 25 miles per hour, what speed and bearing should the pilot take so that she makes the trip in two hours along the original heading? Round the speed to the nearest hundredth of a mile per hour and your angle to the nearest tenth of a degree.
The SS Bigfoot leaves Yeti Bay on a course of N $37^{\circ}$ W at a speed of 50 miles per hour. After traveling half an hour, the captain determines he is 30 miles from the bay and his bearing back to the bay is S $40^{\circ}$ E. What is the speed and bearing of the ocean current? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree.
A $600$ pound Sasquatch statue is suspended by two cables from a gymnasium ceiling. If each cable makes a $60^{\circ}$ angle with the ceiling, find the tension on each cable. Round your answer to the nearest pound.
Two cables are to support an object hanging from a ceiling. If the cables are each to make a $42^{\circ}$ angle with the ceiling, and each cable is rated to withstand a maximum tension of $100$ pounds, what is the heaviest object that can be supported? Round your answer down to the nearest pound.
A $300$ pound metal star is hanging on two cables which are attached to the ceiling. The left hand cable makes a $72^{\circ}$ angle with the ceiling while the right hand cable makes a $18^{\circ}$ angle with the ceiling. What is the tension on each of the cables? Round your answers to three decimal places.
Two drunken college students have filled an empty beer keg with rocks and tied ropes to it in order to drag it down the street in the middle of the night. The stronger of the two students pulls with a force of 100 pounds at a heading of N $77^{\circ}$ E and the other pulls at a heading of S $68^{\circ}$ E. What force should the weaker student apply to his rope so that the keg of rocks heads due east? What resultant force is applied to the keg? Round your answer to the nearest pound.
Emboldened by the success of their late night keg pull in Exercise 61 above, our intrepid young scholars have decided to pay homage to the chariot race scene from the movie `Ben-Hur’ by tying three ropes to a couch, loading the couch with all but one of their friends and pulling it due west down the street. The first rope points N $80^{\circ}$ W, the second points due west and the third points S $80^{\circ}$ W. The force applied to the first rope is 100 pounds, the force applied to the second rope is 40 pounds and the force applied (by the non-riding friend) to the third rope is 160 pounds. They need the resultant force to be at least 300 pounds otherwise the couch won’t move. Does it move? If so, is it heading due west?
Let $\vec{v} = \langle v_{1}, v_{2} \rangle$ be any non-zero vector. Show that $\dfrac{1}{\|\vec{v}\|} \vec{v}$ has length 1.
We say that two non-zero vectors $\vec{v}$ and $\vec{w}$ are parallel if they have same or opposite directions. That is, $\vec{v} \neq \vec{0}$ and $\vec{w} \neq \vec{0}$ are parallel if either $\hat{v} = \hat{w}$ or $\hat{v} = -\hat{w}$ . Show that this means $\vec{v} = k\vec{w}$ for some non-zero scalar $k$ and that $k > 0$ if the vectors have the same direction and $k < 0$ if they point in opposite directions.
The goal of this exercise is to use vectors to describe non-vertical lines in the plane. To that end, consider the line $y = 2x - 4$ . Let $\vec{v}_{\text{\0} = \langle 0, -4 \rangle$ and let $\vec{s} = \langle 1, 2 \rangle$ . Let $t$ be any real number. Show that the vector defined by $\vec{v} = \vec{v}_{0} + t\vec{s}$ , when drawn in standard position, has its terminal point on the line $y = 2x - 4$ . (Hint: Show that $\vec{v}_{0} + t\vec{s} = \langle t, 2t - 4 \rangle$ for any real number $t$ .) Now consider the non-vertical line $y = mx + b$ . Repeat the previous analysis with $\vec{v}_{0} = \langle 0, b \rangle$ and let $\vec{s} = \langle 1, m \rangle$ . Thus any non-vertical line can be thought of as a collection of terminal points of the vector sum of $\langle 0, b \rangle$ (the position vector of the $y$ -intercept) and a scalar multiple of the slope vector $\vec{s} = \langle 1, m \rangle$ .
Prove the associative and identity properties of vector addition in Theorem 9.1.
Prove the properties of scalar multiplication in Theorem 9.2.

Section 9.1 Exercise Answers can be found in the Appendix … Coming soon

The word `vector' comes from the Latin vehere meaning `to convey' or `to carry.' ↵
Other textbook authors use bold vectors such as $\pmb{v}$ . We find that writing in bold font on the chalkboard is inconvenient at best, so we have chosen the `arrow' notation. ↵
If this idea of `over' and `up' seems familiar, it should. The slope of the line segment containing $\vec{v}$ is $\frac{4}{3}$ . ↵
If necessary, review Sections 8.4.1 and 8.5. ↵
That is, the speed of the plane relative to the air around it. If there were no wind, plane's airspeed would be the same as its speed as observed from the ground. How does wind affect this? Keep reading! ↵
See Section 7.1.3, for instance. ↵
Or, as our given angle, $100^{\circ}$ , is obtuse, we could use the Law of Sines without any ambiguity here. ↵
In more advanced courses. chief among them Linear Algebra, vectors are actually defined as $1\times n$ or $n \times 1$ matrices, depending on the situation. ↵
Recall, $\vec{0}$ is represented geometrically as a point $\ldots$ ↵
We will see in Definition 9.5 that we also call $\bm \hat{v}$ the unit vector in the direction of $\vec{v}$ . ↵
Of course, to go from $\vec{v} = \| \vec{v} \| \bm\hat{v}$ to $\bm\hat{v} = \left( \frac{1}{\|\vec{v}\|}\right) \vec{v}$ , we are essentially `dividing both sides' of the equation by the scalar $\| \vec{v} \|$ . The authors encourage the reader, however, to work out the details carefully to gain an appreciation of the properties in play. ↵
Due to the utility of vectors in `real-world' applications, we will usually use degree measure for the angle when giving the vector's direction. There are examples and exercises in which radians are used as well. ↵
Keeping things `calculator' friendly, for once! ↵
Yes, a calculator approximation is the quickest way to see this, but you can also use good old-fashioned inequalities and the fact that $45^{\circ} \leq 50^{\circ} \leq 60^{\circ}$ . ↵
We could just have easily used arcsine or arccosine here $\ldots$ ↵
$\ldots$ if $\| \vec{v} \| > 1 \; \ldots$ ↵
We will see a generalization of Theorem 9.4 in Section 9.2. Stay tuned! ↵
See also Section 8.3.3. ↵
This is the criteria for `static equilbrium'. ↵