Scalars and Vectors Study Guide (page 3)
We start this lesson by briefly considering a reference frame and the coordinates of a point-like object relative to it. Next, we make a distinction between scalar and vector quantities, and present examples of both. At the end of the chapter, we take a quick look at the properties of vectors.
In order to be able to locate a point-like object on a line, on a surface, or in space, we need to choose a reference position called an origin and then give directions regarding how to get to the object starting from the origin.
A reference frame consists of an origin through which one or more axes specifying some given directions are passing. Coordinates are sets of numbers that uniquely specify the position of a point-like object with respect to this certain reference frame. These coordinates measure certain properties of the path from the origin to the object when we follow these axes according to some given instructions
The simplest case is the one-dimensional motion, along a straight line. In this case, the line is also the only reference axis we need. In order to completely specify the position of an object, we have to choose a fixed point as our origin, and then find the distance from the object to the origin using some convenient units. Because our object may be either to the left or to the right of the origin, we consider by convention that the distance is positive if the object lies to the right of the origin and negative if it lies to the left. This distance and the corresponding sign is what we call the coordinate of the object with respect to our frame of reference. The distance between two points in this case is just the difference between their coordinates. In Figure 2.1, the distance from P to Q is PQ; then PQ = [4.5 – (–2.5)] = 7 units.
In the two-dimensional case, a frame of reference is specified by an origin and a set of two axes passing through that origin. To get to the position of a given object at point P, we may say: Start from the origin and follow x-axis for a length of three units; then follow the direction of y-axis (that is, follow a parallel to the y-axis) for another four units. That is, our object has an x-coordinate of 3 and a y-coordinate of 4 (see Figure 2.2).
If the two axes are perpendicular to each other, we can find the distance from the origin to point P by applying Pythagoras' theorem:
d2 = x2 = y2
d2 = 32 = 42
d2 = 25
d = 5
The most widely known two-dimensional (2-D) system of coordinates or reference frames is the Cartesian one, when the two axes are perpendicular to each other. Given any point P, we draw parallel lines to the axes. The distance from where these lines intercept the axes to the origin are the two coordinates (x,y) of point P. This way of labeling points is known as the Cartesian system, in honor of the French mathematician and philosopher Renee Descartes (1596-1650), who was the first to think of this system.
A quick reference to Pythagoras's theorem gives the distance between two points with coordinates (x1, y1) and (x2, y2) as:
d2 = (x1 – x2)2 + (y1 – y2)2
Another widely used way of specifying a point's position on a plane surface is by using its so-called polar coordinates (r, θ) with r > 0.
One coordinate is the length r of the segment OP from the origin of the reference frame to the point, and the other is the angle θ that this segment OP makes with a reference axis Ox. This angle θ gives the direction to the point P, and r gives the distance to the origin. See Figure 2.3.
If the origin and the x-axis of the two representations are the same, then we can relate these sets of coordinates by the following relations:
x = r · cos θ
y = r · sin θ
r2 = x2 + y2
Depending on the symmetry of the problem we try to solve, it may be easier to use one or the other of these two representations.
The following Cartesian coordinates characterize a point: x = 7.0 and y = 2.0. Find the polar coordinates of this point.
Considering the previous explanation for polar coordinates, we can first find the distance to the point in polar coordinates, r, and then, applying trigonometric functions, we can determine the angle θ.
- x = 7.0
- y = –2.0
- r = ?
- θ = ?
- r2 = x2 + y2
- r2 = 72 + (–2)2
- r2 = 53
- r = 7.3
- sin θ = – 2.0/7.3
- θ = –16°
Both these two-dimensional systems of coordinates may be extended to three dimensions by adding a z-axis perpendicular to the plane, and we get the 3-D Cartesian and, respectively, the cylindrical coordinates systems.
For a 3-D Cartesian system of reference, the distance r of point P(x0, y0, z0) having coordinates x0, y0, and z0 to the origin is given by:
To go from this to the 3-D cylindrical system, we use the same relationship between (x, y) and (r, θ) in the horizontal plane, and the height z is same: (r,θ,z).
Find the cylindrical system coordinates of point Q of (Cartesian) coordinates (3,4,6) units.
r2 = x2 + y2
r2 = 32 + 42 = 25
r = 5
sin θ = component on y/r
sin θ = 4/5 = 0.8
θ = 53°
Point Q coordinates in the cylindrical system are (5,53°,6).
Scalars and Vectors
Each of the physical quantities we may encounter in physics can be categorized as either scalars or vectors. A quantity that is completely specified only by its magnitude using a certain unit is called a scalar quantity. When you say, "I bought 3 lbs of tomatoes: " you have pretty much specified all the relevant information about your purchase. Therefore, mass is a scalar quantity. Other examples of scalar quantities include temperature, volume, pressure, density, and time. When you say that the temperature outside is 55° F, you do not need to specify a direction. Scalar quantities can be used in regular arithmetic computations with the only precaution being to express them using the same units.
At other times, in order to fully characterize a physical quantity, we need to specify both a magnitude and a direction. For example, in order to get from Lansing, Michigan, to Detroit, Michigan, you have to travel approximately 70 miles due east. Going in the wrong direction will take you to Grand Rapids or maybe Jackson, instead of Detroit.
A vector quantity therefore has both a magnitude and a direction. In the previous example, we say that displacement is a vector. We usually draw a vector as an arrow. The direction of the arrow gives the direction of the vector, and its length is usually scaled proportionally to the magnitude of the vector, using appropriate units. The magnitude of a vector is always a positive number. A vector symbol is a bold letter, for instance, A.
For a 3-D Cartesian system, we define a unit vector along each one of the axes; we do so by making a vector along the axis, pointing it in the positive direction, and making it have a unit length. The usual notation is i for the unit vector along the x-axis, j for the unit vector along the y-axis, and k for the unit vector along the z-axis.
| i | = | j | = | k | = 1
Unit vectors are used to specify the directions of the reference axes. We also define a vector's components Ax, Ay, and Az as the difference of its starting point and end tip coordinates (see Figure 2.4).
Then, we may represent the vector as:
A = i Ax + j Ay + k Az
Assuming the starting point of vector A is P(xs, ys, zs) and the end tip is at point Q(xt, yt, zt) then:
Ax = xt – xs Ay = yt – ys Az = zt – zs
The magnitude of a vector is:
A2 = Ax2 + Ay2 + Az2
Vectors enjoy certain properties that can make their handling easier. For example, the components of any vector can be used in place of the vector itself in any calculation where it is convenient to do so.
Two vectors are equal if they have the same magnitude and the same direction. As the direction is specified by a whole bundle of parallel lines to each other, this allows us most of the time to move a vector parallel to itself without affecting the outcome of the problem.
All three vectors A, B, and C in Figure 2.5 are equal to one another. Their directions are parallel, and their magnitudes are equal.
We define the sum of two vectors as a new vector, also called a resultant vector.
One method to obtain the resultant vector is by moving one of the vectors parallel to itself until its starting point coincides with the end of the first one. The resultant vector starts at the tail of the first vector and ends at the end tip of the second vector.
We write the resultant vector as:
R = A + B
Without any loss of generality, we can limit ourselves to the two-dimensional case of vectors in the same plane.
Show graphically the addition of two vectors A and B.
As defined previously, we move the second vector parallel to itself so that its tail intersects the tip of the first
vector (see Figure 2.6). If the two vectors being added have components (Ax, Ay, Az) and (Bx, By, Bz), the sum or resultant can be easily seen to have the components (Rx, Ry, Rz) given by:
Rx = Ax + Bx
Ry = Ay + By
Rz = Az + Bz
and the magnitude of the resultant is:
R2 = Rx2 + Ry2 + Rz2
While the resultant vector can be written as:
R = i · Rx + j · Ry + k · Rz
The opposite of a vector A is a vector – A, having the same magnitude, but oriented in the opposite sense relative to the initial vector (see Figure 2.7).
This allows us to define the difference of two vectors by means of adding one of the vectors with the opposite of the other.
A – B = A + (–B)
In Figure 2.8, the resultant vector of A – A = A + (–A) = 0 is demonstrated. The two vectors have intentionally not been perfectly superimposed as an aid to the eye.
Find the magnitude of the resultant vector of the following vectors: (3,–5,–2) m and (–2,0,8) m, and of the difference of the same vectors.
The components of the resultant are:
Rx = Ax + Bx = 3 – 2 = 1 m
Ry = Ay + By = –5 + 0 = –5 m
Rz = Az + Bz = – 2 + 8 = 6 m
R2 = 1 + 25 + 36 = 62 m2
and the magnitude of the resultant vector is:
R = 7.9 m
For the difference:
A – B = A + (–B)
and the components of the A + (– B) are:
Rx = Ax + (–Bx) = 3 + [– (–2)] = 5 m
Ry = Ay + (–By) = –5 + 0 = –5 m
Rz = Az + (–Bz) = –2 + (–8) = –10 m
R2 = 25 + 25 + 100 = 150 m2
and the magnitude of the resultant vector is:
R = 12 m
We call the scalar product (or dot product) of two vectors A and B a scalar quantity. A · B is obtained by multiplying the magnitudes of the two vectors and the cosine of the angle made by their directions (see Figure 2.9).
A · B = A · B · cos θ
What is the angle between two vectors that have a scalar product equal to zero?
When the scalar product of two vectors of nonzero magnitude is zero:
A · B = 0
It follows that the cosine of the angle made by their directions should be zero, and therefore, the two vectors must be perpendicular to each other.
cos θ = 0
θ = π/2
For the scalar product, the order of multiplication is not important: A · B = B · A (scalar product is commutative). We apply the previous definition to both sides of the equality:
A · B = A · B · cos θ
B · A = B · A · cos (–θ )
But A and B are numbers, so A · B will equal B. A, and the cosine function is symmetric: cos θ = cos (–θ). Hence, both sides yield the same answer.
We call the vector product (or cross product) of two vectors A and B a vector quantity: A × B with a magnitude obtained by multiplying the magnitudes of the two vectors and the sine of the angle made by their directions. The direction of the resultant vector is perpendicular on the plane defined by these two vectors (see Figure 2.10).
C = A × B
C = A · B · sin θ
The direction of the resultant vector is given by the right-hand rule (also known as the corkscrew rule). If you position the right hand around the direction of the angle between the two vectors such that the fingers point in the direction of moving the first vector on top of the second vector, and you extend your thumb, the direction of the resultant vector is given by the thumb. Also, for the corkcrew rule: If we rotate the corkscrew such that the first vector advances on top of the second vector, the direction of advance of the screw is the direction of the resultant vector. The two rules are identical, so whichever you find less weird and are able to apply to your problem should bring you to the same result.
For the vector product, the order of multiplication is important because the result is a vector. Vector product is not commutative:
A × B = –B × A
We apply the above definition to both sides of the equality:
|A × B| = A · B · sin θ
|B × A| = B · A · sin (–θ)
But A and B are numbers, so A · B = B · A, and the sine function is antisymmetric: sin θ = –sin (–θ). Hence, although the magnitude is the same, the direction is opposite:
|A × B| = A · B · sin θ
|B × A| = –B · A · sin θ
The vector product is distributive. This means that when having three vectors, the following equality holds true:
(A + B) × C = A ×C + B × C
We should also note the following remarkable identities between the unit vectors i, j, and k:
i · i = j · j = k · k = 1
i · j = j · k = k · i = 0
i × j = k, j × k = i, and k × i = j
Using these identities and the other properties of vectors, we can calculate the scalar and vector product of any two vectors when we know their x, y, and z components.
Show that i × j = k.
First, we will show the magnitudes of the left-hand side and right-hand side to be equal:
|i × j| = 1 · 1 · sin 90° = 1
|k| = 1
For the orientation, we can show through a diagram the orientation of the i × j vector. In Figure 2.11, we see the screw rotating from i unit vector toward the j unit vector in the direction of the angel, and the screw advances upward in the direction of k.
Practice problems of this concept can be found at: Scalars and Vectors Practice Questions
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