Many quantities we think about daily can be described by a single number: temperature, speed, cost, weight and height. There are also many other concepts we encounter daily that cannot be described with just one number. For instance, a weather forecaster often describes wind with its speed and its direction (“\(\ldots\) with winds from the southeast gusting up to 30 mph \(\ldots\)”). When applying a force, we are concerned with both the magnitude and direction of that force. In both of these examples, direction is important. Because of this, we study vectors, mathematical objects that convey both magnitude and direction information.
One “bare-bones” definition of a vector is based on what we wrote above: “a vector is a mathematical object with magnitude and direction parameters.” This definition leaves much to be desired, as it gives no indication as to how such an object is to be used. Several other definitions exist; we choose here a definition rooted in a geometric visualization of vectors. It is very simplistic but readily permits further investigation.
Given points \(P\) and \(Q\) (either in the plane or in space), we denote with \(\overrightarrow{PQ}\) the vector from \(P\) to \(Q\text{.}\) The point \(P\) is said to be the initial point of the vector, and the point \(Q\) is the terminal point.
The magnitude, length or norm of \(\overrightarrow{PQ}\) is the length of the line segment \(\overline{PQ}\text{:}\)\(\norm{\overrightarrow{PQ}} = \norm{\overline{PQ}}\text{.}\)
Figure 12.2.3 shows multiple instances of the same vector. Each directed line segment has the same direction and length (magnitude), hence each is the same vector.
The \(x\) and \(y\) axes are drawn from \(-4\) to \(4\text{.}\) There are four vectors drawn. In the first quadrant is the first vector drawn from point \((0,0)\) to \((3,1)\text{.}\) The second vector is drawn in the second quadrant from point \((-4,1)\) to point \((-1,2)\text{.}\) In the fourth quadrant the vector is drawn from \((2, -4)\) to \((5, -3)\text{.}\) The last vector is drawn from \((-2, -3)\) to \((1, -2)\) and it crosses the third .
We use \(\mathbb{R}^2\) (pronounced “r two”) to represent all the vectors in the plane, and use \(\mathbb{R}^3\) (pronounced “r three”) to represent all the vectors in space.
The \(x\) and \(y\) axes are drawn from \(-4\) to \(4\text{.}\) There are two vectors \(PQ\) and \(RS\text{.}\) The vector \(PQ\) is drawn in the first quadrant from point \(P=(1,0)\) and \(Q=(3,1)\text{.}\) The second vector is in the second quadrant from point \(R=(-3, 1)\) to \(S=(-1,2)\text{.}\)
Consider the vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{RS}\) as shown in Figure 12.2.4. The vectors look to be equal; that is, they seem to have the same length and direction. Indeed, they are. Both vectors move 2 units to the right and 1 unit up from the initial point to reach the terminal point. One can analyze this movement to measure the magnitude of the vector, and the movement itself gives direction information (one could also measure the slope of the line passing through \(P\) and \(Q\) or \(R\) and \(S\)). Since they have the same length and direction, these two vectors are equal.
This demonstrates that inherently all we care about is displacement; that is, how far in the \(x\text{,}\)\(y\) and possibly \(z\) directions the terminal point is from the initial point. Both the vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{RS}\) in Figure 12.2.4 have an \(x\)-displacement of 2 and a \(y\)-displacement of 1. This suggests a standard way of describing vectors in the plane. A vector whose \(x\)-displacement is \(a\) and whose \(y\)-displacement is \(b\) will have terminal point \((a,b)\) when the initial point is the origin, \((0,0)\text{.}\) This leads us to a definition of a standard and concise way of referring to vectors.
The component form of a vector \(\vec{v}\) in \(\mathbb{R}^2\text{,}\) whose terminal point is \((a,b)\) when its initial point is \((0,0)\text{,}\) is \(\la a,b\ra\text{.}\)
The component form of a vector \(\vec{v}\) in \(\mathbb{R}^3\text{,}\) whose terminal point is \((a,b,c)\) when its initial point is \((0,0,0)\text{,}\) is \(\la a,b,c\ra\text{.}\)
Using \(P\) as the initial point, we move 2 units in the positive \(x\)-direction and \(-1\) units in the positive \(y\)-direction to arrive at the terminal point \(P\,'=(5,1)\text{,}\) as drawn in Figure 12.2.7.(a). The magnitude of \(\vec v\) is determined directly from the component form:
The \(x\) and \(y\) axes are drawn from \(-4\) to \(4\text{.}\) Two vectors are drawn. The vector \(PP’\) is in the first quadrant it is downward facing, it starts from \(P=(3, 2)\) and ends at \(P'=(5, 1)\text{.}\) The second vector \(RS\) from \((-3, -2)\) to \((-1, 2)\text{.}\) This vector lies midway between the second and the third quadrant.
One can readily see from Figure 12.2.7.(a) that the \(x\)- and \(y\)-displacement of \(\overrightarrow{RS}\) is 2 and 4, respectively, as the component form suggests.
Using \(Q\) as the initial point, we move 2 units in the positive \(x\)-direction, \(-1\) unit in the positive \(y\)-direction, and 3 units in the positive \(z\)-direction to arrive at the terminal point \(Q' = (3,0,4)\text{,}\) illustrated in Figure 12.2.7.(b). The magnitude of \(\vec u\) is:
Now that we have defined vectors, and have created a nice notation by which to describe them, we start considering how vectors interact with each other. That is, we define an algebra on vectors.
The \(x\) and y axes are drawn from \(0\) to \(4\text{.}\) Two vectors \(\vec u\) and \(\vec v\) are shown along with the vector addition of the two. The \(\vec u\) vector is drawn from point \((0,0)\) to \((1,3)\text{.}\) The \(\vec v\) vector is drawn from origin to \((2,1)\text{.}\) The vector \(\vec u\) is longer than the \(\vec v\) vector. The \(\vec u + \vec v\) vector is drawn from the origin to \((3, 4)\text{.}\) The \(\vec u + \vec v\) vector is the longest and is in between the two vectors.
As vectors convey magnitude and direction information, the sum of vectors also convey length and magnitude information. Adding \(\vec u+\vec v\) suggests the following idea:
This idea is sketched in Figure 12.2.12, where the initial point of \(\vec v\) is the terminal point of \(\vec u\text{.}\) This is known as the “Head to Tail Rule” of adding vectors. Vector addition is very important. For instance, if the vectors \(\vec u\) and \(\vec v\) represent forces acting on a body, the sum \(\vec u+\vec v\) gives the resulting force. Because of various physical applications of vector addition, the sum \(\vec u+\vec v\) is often referred to as the resultant vector, or just the “resultant.”
The \(x\) and \(y\) axes are drawn from \(0\) to \(4\text{.}\) Two vectors \(\vec u\) and \(\vec v\) are shown along with the vector addition of the two. The \(\vec u\) vector is drawn from point \((0,0)\) to \((1,3)\text{.}\) The \(\vec v\) vector is drawn from origin to \((2,1)\text{.}\)
The \(\vec v\) vector is translated to start from the point \((1, 3)\) to \((3, 4)\) and it forms a triangle with \(\vec u\) and \(u+v\text{.}\) The \(\vec u\) vector is translated to start from point \((2, 1)\) to point \((3, 4)\text{.}\) The vector \(\vec u\) is longer than the \(\vec v\) vector and it forms a triangle with \(\vec u\) and \(u+v\text{.}\) The \(\vec u + \vec v\) vector is drawn from the origin to \((3, 4)\text{.}\) The \(\vec u + \vec v\) vector is the longest and is in between the two vectors.
Analytically, it is easy to see that \(\vec u+\vec v = \vec v+\vec u\text{.}\)Figure 12.2.12 also gives a graphical representation of this, using gray vectors. Note that the vectors \(\vec u\) and \(\vec v\text{,}\) when arranged as in the figure, form a parallelogram. Because of this, the Head to Tail Rule is also known as the Parallelogram Law: the vector \(\vec u+\vec v\) is defined by forming the parallelogram defined by the vectors \(\vec u\) and \(\vec v\text{;}\) the initial point of \(\vec u+\vec v\) is the common initial point of parallelogram, and the terminal point of the sum is the common terminal point of the parallelogram.
The computation of \(\vec u-\vec v\) is straightforward, and we show all steps below. Usually the formal step of multiplying by \((-1)\) is omitted and we “just subtract.”
The \(x\) axis is drawn from \(0\) to \(4\) and the \(y\) axis is drawn from \(-1\) to \(3\text{.}\) The \(\vec u\) vector is drawn from origin to point \((1, 2)\text{.}\) The \(\vec v\) vector is drawn from the origin to \((2,1)\text{.}\) The vector \(\vec u - \vec v\) is drawn from origin to \((2, -1)\) and it lies in the fourth quadrant.
The \(\vec u - \vec v\) vector is translated to start from point \((1, 2)\) and ends at point \((3, 1)\text{.}\) The \(\vec v\) vector is also translated but in the opposite direction and it starts from \((3, 1)\) and ends at point \((2, -1)\text{.}\)
Figure 12.2.14 illustrates, using the Head to Tail Rule, how the subtraction can be viewed as the sum \(\vec u + (-\vec v)\text{.}\) The figure also illustrates how \(\vec u-\vec v\) can be obtained by looking only at the terminal points of \(\vec u\) and \(\vec v\) (when their initial points are the same).
The \(x\) axis is drawn from \(0\) to \(4\) and the \(y\) axis is drawn from \(-1\) to \(3\text{.}\) The \(\vec v\) vector is drawn from origin to point \((2,1)\text{,}\) another vector \(2\vec v\) is also drawn from origin to point \((4, 2)\text{.}\) The two vectors have the same direction.
Both \(\vec v\) and \(2\vec v\) are sketched in Figure 12.2.16. Make note that \(2\vec v\) does not start at the terminal point of \(\vec v\text{;}\) rather, its initial point is also the origin.
The zero vector is the vector whose initial point is also its terminal point. It is denoted by \(\vec 0\text{.}\) Its component form, in \(\mathbb{R}^2\text{,}\) is \(\la 0,0\ra\text{;}\) in \(\mathbb{R}^3\text{,}\) it is \(\la 0,0,0\ra\text{.}\) Usually the context makes is clear whether \(\vec 0\) is referring to a vector in the plane or in space.
Our examples have illustrated key principles in vector algebra: how to add and subtract vectors and how to multiply vectors by a scalar. The following theorem states formally the properties of these operations.
The following are true for all scalars \(c\) and \(d\text{,}\) and for all vectors \(\vec u\text{,}\)\(\vec v\) and \(\vec w\text{,}\) where \(\vec u\text{,}\)\(\vec v\) and \(\vec w\) are all in \(\mathbb{R}^2\) or where \(\vec u\text{,}\)\(\vec v\) and \(\vec w\) are all in \(\mathbb{R}^3\text{:}\)
As stated before, each nonzero vector \(\vec v\) conveys magnitude and direction information. We have a method of extracting the magnitude, which we write as \(\norm{\vec v}\text{.}\)Unit vectors are a way of extracting just the direction information from a vector.
Consider this scenario: you are given a vector \(\vec v\) and are told to create a vector of length 10 in the direction of \(\vec v\text{.}\) How does one do that? If we knew that \(\vec u\) was the unit vector in the direction of \(\vec v\text{,}\) the answer would be easy: \(10\vec u\text{.}\) So how do we find \(\vec u\text{?}\)
Property 8 of Theorem 12.2.17 holds the key. If we divide \(\vec v\) by its magnitude, it becomes a vector of length 1. Consider:
\begin{align*}
\snorm{\frac{1}{\norm{\vec v}}\vec v} \amp = \frac{1}{\norm{\vec v}}\norm{\vec v} \amp \text{ (we can pull out \(\ds \frac{1}{\norm{\vec v}}\) as it is a positive scalar)}\\
\amp = 1\text{.}
\end{align*}
To create a vector with magnitude 5 in the direction of \(\vec v\text{,}\) we multiply the unit vector \(\vec u\) by 5. Thus \(5\vec u = \la 15/\sqrt{10},5/\sqrt{10}\ra\) is the vector we seek. This is sketched in Figure 12.2.22.
The \(x\) axis is drawn from \(0\) to \(5\) and the \(y\) axis is drawn from \(0\) to \(3\text{.}\) The \(\vec u\) vector is drawn from origin to point \((1, 0.4)\text{,}\) another vector \(\vec v\) is also drawn from origin to point \((3, 1)\text{.}\) The third vector \(5\vec u\) is also drawn from the origin to point \((5, 1.5)\text{.}\) The three vectors have the same direction.
This equation illustrates the fact that a nonzero vector has both magnitude and direction, where we view a unit vector as supplying only direction information. Identifying unit vectors with direction allows us to define parallel vectors.
It is equivalent to say that vectors \(\vec v_1\) and \(\vec v_2\) are parallel if there is a scalar \(c\neq 0\) such that \(\vec v_1 = c\vec v_2\) (see marginal note).
If one graphed all unit vectors in \(\mathbb{R}^2\) with the initial point at the origin, then the terminal points would all lie on the unit circle. Based on what we know from trigonometry, we can then say that the component form of all unit vectors in \(\mathbb{R}^2\) is \(\la \cos(\theta) ,\sin(\theta) \ra\) for some angle \(\theta\text{.}\)
A similar construction in \(\mathbb{R}^3\) shows that the terminal points all lie on the unit sphere. These vectors also have a particular component form, but its derivation is not as straightforward as the one for unit vectors in \(\mathbb{R}^2\text{.}\) Important concepts about unit vectors are given in the following Key Idea.
A vector \(\vec u\) in \(\mathbb{R}^2\) is a unit vector if, and only if, its component form is \(\la \cos\theta,\sin\theta\ra\) for some angle \(\theta\text{.}\)
A vector \(\vec u\) in \(\mathbb{R}^3\) is a unit vector if, and only if, its component form is \(\la \sin(\theta) \cos(\varphi) ,\sin(\theta) \sin(\varphi) ,\cos(\theta) \ra\) for some angles \(\theta\) and \(\varphi\text{.}\)
Consider a weight of 50lb hanging from two chains, as shown in Figure 12.2.27. One chain makes an angle of \(30^\circ\) with the vertical, and the other an angle of \(45^\circ\text{.}\) Find the force applied to each chain.
Image is of a weight of \(50\) pounds suspended by two chains. The chain on the left forms an angle of \(30\) degrees with the vertical and the chain on the left forms a degree of \(45\) with the vertical.
We can view each chain as “pulling” the weight up, preventing it from falling. We can represent the force from each chain with a vector. Let \(\vec F_1\) represent the force from the chain making an angle of \(30^\circ\) with the vertical, and let \(\vec F_2\) represent the force form the other chain. Convert all angles to be measured from the horizontal (as shown in Figure 12.2.28), and apply Key Idea 12.2.25. As we do not yet know the magnitudes of these vectors, (that is the problem at hand), we use \(m_1\) and \(m_2\) to represent them.
Image shows the force vectors from the exercise. The vector \(\vec F_2\) is at an angle \(45\) from the horizontal and the vector \(\vec F_1\) forms an angle of \(120\) from the horizontal. A third vector representing the downward pull by gravity marked as \(\vec F\text{.}\)
The sum of the entries in the first component is 0, and the sum of the entries in the second component is also 0. This leads us to the following two equations:
It might seem odd that the sum of the forces applied to the chains is more than 50lb. We leave it to a physics class to discuss the full details, but offer this short explanation. Our equations were established so that the vertical components of each force sums to 50lb, thus supporting the weight. Since the chains are at an angle, they also pull against each other, creating an “additional” horizontal force while holding the weight in place.
Unit vectors were very important in the previous calculation; they allowed us to define a vector in the proper direction but with an unknown magnitude. Our computations were then computed component-wise. Because such calculations are often necessary, the standard unit vectors can be useful.
These two examples demonstrate that converting between component form and the standard unit vectors is rather straightforward. Many mathematicians prefer component form, and it is the preferred notation in this text. Many engineers prefer using the standard unit vectors, and many engineering text use that notation.
A weight of 25lb is suspended from a chain of length 2ft while a wind pushes the weight to the right with constant force of 5lb as shown in Figure 12.2.32. What angle will the chain make with the vertical as a result of the wind’s pushing? How much higher will the weight be?
Image shows a weight of \(25\) pounds being suspended by a chain. The chain forms an angle of \(\theta\) with the vertical, and an angle \(\varphi\) with the horizontal. The wind is pushing the weight to the right with force \(\vec F_w\text{.}\) The chain is of length \(2\) feet.
The force of the wind is represented by the vector \(\vec F_w = 5\veci\text{.}\) The force of gravity on the weight is represented by \(\vec F_g = -25\vecj\text{.}\) The direction and magnitude of the vector representing the force on the chain are both unknown. We represent this force with
for some magnitude \(m\) and some angle with the horizontal \(\varphi\text{.}\) (Note: \(\theta\) is the angle the chain makes with the vertical; \(\varphi\) is the angle with the horizontal.)
This is enough to determine \(\vec F_c\) already, as we know \(m\cos(\varphi) = -5\) and \(m\sin(\varphi) =25\text{.}\) Thus \(F_c = \la -5,25\ra\text{.}\) We can use this to find the magnitude \(m\text{:}\)
\begin{equation*}
m = \sqrt{(-5)^2+25^2} = 5\sqrt{26}\approx 25.5\text{ lb }\text{.}
\end{equation*}
We can then use either equality from Equation (12.2.1) to solve for \(\varphi\text{.}\) We choose the first equality as using arccosine will return an angle in the \(2\)nd quadrant:
We can now use trigonometry to find out how high the weight is lifted. The diagram shows that a right triangle is formed with the 2ft chain as the hypotenuse with an interior angle of \(11.31^\circ\text{.}\) The length of the adjacent side (in the diagram, the dashed vertical line) is \(2\cos(11.31^\circ) \approx 1.96\)ft. Thus the weight is lifted by about \(0.04\)ft, almost 1/2in.
The algebra we have applied to vectors is already demonstrating itself to be very useful. There are two more fundamental operations we can perform with vectors, the dot product and the cross product. The next two sections explore each in turn.
In the following exercises, points \(P\) and \(Q\) are given. Write the vector \(\overrightarrow{PQ}\) in component form and using the standard unit vectors.
The \(x\) and \(y\) axes are uncalibrated. Two vectors \(\vec u\) and \(\vec v\) are shown, both starting at the origin and facing away. The vector \(\vec u\) is in the first quadrant, and is bent close to the positive \(x\) axis, the \(\vec v\) vector is in the third quadrant and is bent close to the negative \(y\) axis. The vector \(\vec v\) appears to be slightly longer than \(\vec u\text{.}\)
The \(x\) and \(y\) axes are uncalibrated. Two vectors \(\vec u\) and \(\vec v\) are shown, both start at the origin and are facing away from each other. The \(\vec u\) vector is in the first quadrant while the \(\vec v\) is in the third quadrant, the \(\vec v\) vector appears to be \(1 / 4\) th \(\vec u\text{.}\)
The \(x\text{,}\)\(y\) and \(z\) axes are uncalibrated. Two vectors \(\vec u\) and \(\vec v\) are shown, both start at the origin and face away from each other. The vector \(\vec u\) is longer and appears to be in the \(zy\) plane, and the vector \(\vec v\) is shorter and appears to be in the \(xz\) plane.
The \(x\text{,}\)\(y\) and \(z\) axes are uncalibrated. Two vectors \(\vec u\) and \(\vec v\) are shown, both start at the origin. The \(\vec u\) vector is along the positive \(z\) axis and the \(\vec v\) vector is along the positive \(y\) axis.
Image is of a weight of \(100\) pounds suspended by two chains. The chain on the left forms an angle of \(\varphi\) degrees with the vertical and the chain on the right forms a degree of \(\theta\) with the vertical.
A weight of \(p\)lb is suspended from a chain of length \(\ell\) while a constant force of \(\vec F_w\) pushes the weight to the right, making an angle of \(\theta\) with the vertical, as shown in the figure below.
Image shows a weight of \(p\) pounds being suspended by a chain. The chain forms an angle of \(\theta\) with the vertical. The wind is pushing the weight to the right with force \(\vec F_w\text{.}\) The chain is of length \(\ell\) feet.
In the following exercises, a force \(\vec F_w\) and length \(\ell\) are given. Find the angle \(\theta\) and the height the weight is lifted as it moves to the right.