The position vector retains all of the regular properties of a vector that I discuss in Chapter 4 in that it still has a magnitude (length), sense (direction), and point of application (location in space).

Figure 5-3 shows the vector from the example in the preceding section and helps demonstrate the representation used to define a position vector.

Figure 5.3. Position vector for as a crow flies example.

By establishing the locations of Point A and Point B, you can define the Cartesian coordinates for these points. The order of the start point (vector tail) and stop point (vector head) is extremely important when you're creating a position vector. The first step is to establish a coordinate system if one isn't already declared. In the Figure 5-2 example, I assume that a positive Cartesian x-direction is to the east, and a positive Cartesian y-direction is to the north. A positive elevation is vertical upwards from the xy plane.

The following formulas show you how you can easily find the relative distance traveled along a line between the start and end points.

To find the relative distance traveled from end to start, simply reverse the x_END and x_START terms in the previous calculations. In the event that your destination is at a different elevation (such as being on a hill or in a valley) than your starting point, you need to include one extra dimension, the Cartesian z-direction — you just substitute the z values: Δz = (z_END − z_START)

In this direction, a positive change in the Cartesian z-direction indicates that the end point is above the start point. Likewise, a negative value indicates that the end point of the vector is below the start point. These generic relationships easily take into account positive and negative distance values.

In the first part of Figure 5-3 earlier in the chapter, you can see that Point B is located northeast of Point A. You can define the Cartesian coordinates for Point A and Point B from the dimensions and units indicated:

where (x,y,z) are the coordinates of their respective points. The distance that needs to be traveled to the right (or east, or positive) is given by the distance (x_B − x_A), and the distance north would be given by the distance (y_B − y_A). More generally:

If you travel in the opposite direction from Point B to Point A (refer to the second part of Figure 5-3), you get the same numeric values, but the signs in front are now different (−1,200 feet, −700 feet, and −200 feet).

In practice, you encounter the magnitude of a random vector F denoted as

Pythagoras of Samos was a Greek philosopher and mathematician (circa 570"B.C.) who is credited with discovering the Pythagorean theorem, which proves the relationship between the sides of a right (or 90-degree) triangle (see Figure 5-4). To calculate the magnitude of a vector, you need to use your ability to locate right triangles and apply some basic geometry.

Figure 5.4. The Pythagorean theorem.

Going vertical: The Pythagorean theorem in three dimensions

For a three dimensional problem, you need to do a bit more calculation but you can still employ the Pythagorean theorem by simply constructing two right triangles inside a box of known dimensions A, B, and C as shown in Figure 5-5.

Figure 5.5. Threedimensional Pythagorean theorem.

The distance between two points at different elevations is equal to the square root of the sum of the squares of the sides of a right-angle box that fully contains the start and end points on opposite corners.

For example, in Figure 5-5, D is the hypotenuse of the first right triangle, ABD. The Pythagorean theorem tells you that for triangle ABD, D² = A² + B²

Similarly, you can also create a second, vertically-oriented right triangle ABC with a hypotenuse of E and sides C (the height of the box) and D (the previously calculated hypotenuse of the first triangle). Using the Pythagorean theorem yet again, you can state that E² = C² + D². You can then substitute the equation for triangle ABD (which represents the value of D²) into this equation to get E² = A² + B² + C² or

Putting Pythagoras to work

In the case of the navigation example in Figure 5-2 earlier in the chapter, which shows a position vector between Points A and B, the box that contains these two points has sides of distances Δ_x, Δ_y, and Δ_z. Substituting these values into the final equation in the preceding section allows you to compute the magnitude of the position vector between Point A and Point B:

Fortunately, the magnitude of a position vector has a specific meaning that you can actually see and measure in that it precisely represents the direct distance between those two points.

Figure 5-6 illustrates the three-dimensional box required to compute the magnitude of the position vector between Point A and Point B.

The distance calculated is the magnitude of the position vector r_AB starting at Point A and ending at Point B.

Figure 5.6. Position vector.

Note

One final note on position vectors: Although the distance between Point A and Point B is a specific value, the vector r_AB that defines the path from Point A to Point B isn't the same as the position vector r_BA from Point B to Point A. That is, the magnitudes are the same (

In this book, I denote a unit vector by the label u and attach a label as a subscript to help describe the direction. Just as before, I denote the magnitude of a unit vector as

Note

Unlike the magnitude of the position vector, the magnitude of a unit vector is always exactly one unit long. That is,

Whenever you use the Cartesian coordinate system (see "Taking Direction from the Cartesian Coordinate System" earlier in the chapter), you can make use of three very special unit vectors. As shown in Figure 5-7, you can define a vector of magnitude 1.0 in the positive direction for each of the principle Cartesian axes.

Figure 5.7. A unit vector.

To write a vector in Cartesian vector notation, you take full advantage of those three unit vectors (i, j, and k) that I talk about in the preceding section. For example, if you have a vector V with magnitude

You can also have a vector going in a combination of directions. Suppose that vector V has a piece in the x-direction of magnitude

Because there's no y-direction component, the coefficient in front of the j (y-direction) is zero. You also notice that the coefficient in front of the k (z-direction) is a negative value. This negative value simply indicates that the magnitude of that piece of the vector is acting in the opposite direction from the assumed positive z-direction.

Position vectors can be Cartesian too!

In terms of the generic coordinates (x_A, y_A, z_A) for Point A and coordinates (x_B, y_B, z_B) for Point B, you can write the position vector from Point A to Point B as

or even more generically as

As long as you choose the points for the start and the end of the vector correctly, and you can correctly determine the Cartesian coordinates of each of those points, the signs of the coefficients in front of each of the unit vectors i, j, and k take care of themselves. In fact, the signs of the scalar values (dealing only with magnitude and not with sense) are what help you determine the sense of the vector — a negative scalar coefficient tells you that piece of the vector is acting in the negative direction of the unit vector noted immediately after that scalar value.

For example, try to write the actual position vector in Cartesian coordinates from Point A to Point B for a navigation example where Point A has coordinates of (+400,+300,0) and Point B has coordinates of (+1600,+1000,+200). (For a visual, check out Figure 5-2 earlier in the chapter.) The notation for the label of this vector is r_AB.

In this example, the change in the x-dimension (Δ_x) is 1,200 feet, the change in the y-direction (Δ_y) is 700 feet, and the change in the z-direction (Δ_z) is 200 feet. (To see the calculations that produce these figures, check out "Moving from Point A to Point B and back again" earlier in the chapter.)

You can now write the position vector from Point A to Point B as

Relationship between a vector, its magnitude, and its direction

In the example in the preceding section, you create a simple position vector V = 10i + 0j + 0k meters by using a known distance (the magnitude) and its direction. You can simplify this vector even further by omitting the terms that have zero as their coefficients. This step leaves a new form of the vector V:

The original vector V was created from the given information that the magnitude of the vector is 10 meters parallel to the positive Cartesian x-axis. If you examine this setup in general terms, you may conclude the following relationship:

vector=magnitude.direction

For the navigation example, you can substitute the following terms into the equation:

• vector:

  

• magnitude:

  

• direction: Unit vector in the positive Cartesian x-direction

  :

  

Thus, the expression above is more accurately written as:

Notice that this expression directly relates the vector itself as the multiplication of its scalar magnitude and a unit vector in the direction of that magnitude. Imagine that! Creating a new vector from a unit vector!

In the section "Relationship between a vector, its magnitude, and its direction" earlier in the chapter, I develop the expression for a vector based on its magnitude and direction. If you do a little rearranging of the last equation of that section, you get the following expression:

This expression illustrates that you can establish the direction of a vector's line of action by taking the vector representation over the magnitude of that same vector. Suppose you have a position vector from Point A to Point B given as

with a magnitude of

(You may recognize these numbers from the example in "Unit Vectors Tell Direction, Too!" earlier in the chapter.) You can easily create a unit vector to describe the orientation of the line of action between Point A and Point B (the line that connects those two points):

Notice that a unit vector has no actual units because the vector terms have units of feet in the numerator of each term, and the magnitude (or distance) gives units of feet in the denominator, so they cancel out. To verify that this vector meets the criteria of being a unit vector, you simply need to calculate its magnitude:

Realize that because the coefficients of the unit vector terms are typically irrational numbers, the magnitude rarely computes to be exactly 1.0. This discrepancy is another example of the importance of significant digits and computational accuracy that I introduce in Chapter 2. For the purposes of the example here, you'll be happy that 0.999 is approximately 1.000.

Another technique that's sometimes handy for creating a unit vector utilizes calculations involving the direction cosines. The direction cosines represent the angles between any two given vectors — even two unit vectors! You can use the three Cartesian axes as the reference because you already have those three special i, j, and k unit vectors to define them. (Flip to the earlier section "Cartesian vector notation" for the details on these unit vectors.)

One of the major difficulties of using direction cosines is in actually determining the angle between the vector or line of action of interest and the Cartesian axis. These angles often occur on geometric planes that aren't perpendicular to any Cartesian axis and therefore may be difficult or cumbersome to compute.

Figure 5-7 earlier in the chapter illustrates the three unit vectors with respect to each of the three principal Cartesian axes. For example, Figure 5-8a shows the x-direction unit vector u_x = i. The direction cosine for this vector, the angle α, is the angle between the line of action of the vector you're working with and the line of action of the x-direction unit vector, contained within the plane of those two vectors. Similarly, Figure 5-8b shows the y-direction unit vector u_y = j. The direction cosine for this vector, the angle β, is the angle between the line of action of your vector and the y-direction unit vector. Figure 5-8c shows the z-direction unit vector u_z = k, and the direction cosine for this vector, the angle γ, is the angle between the line of action of your vector and the z-direction unit vector, contained within the plane of the two vectors.

Figure 5.8. Direction cosines can help create unit vectors: In some texts, you may see the angles for the direction cosines referred to as θ_x, θ_y, and θ_z to represent the angle between the vector's line of action and its corresponding Cartesian axis unit vector. Here I use α, β, and γ, respectively.

The previous example showed how to piece together different parts of a vector in order to compute a different vector. You can use a similar technique with direction cosines to create a unit vector as follows:

Yet another method that you may find useful for creating unit vectors utilizes the proportions or ratios of the dimensions of a vector's or object's line of action.

In some textbooks, you see a vector or line of action with an odd-looking pair of numbers written around a triangle. In the case of the vector, this small triangle is attached to the shaft of the vector. For a line of action, you see the triangle attached somewhere along the line, or you can infer it from the given dimensions of an object or distance between given points. This proportion technique is useful because you eliminate the need to even calculate the angle for the line of action, and as a result you may actually improve issues with accuracy and significant digits, as I mention in Chapter 2.

This proportion triangle represents a horizontal and vertical proportion and is comparable to the slope diagrams you may remember from your algebra class. The horizontal line segment of this proportion triangle represents the horizontal proportion, and similarly the vertical line segment represents the vertical proportion.

The secret to using these proportion values lies in applying the basic trigonometry functions sine, cosine, and tangent (which I touch on in Chapter 2). In Figure 5-9, each of the three proportion triangles shown all have the same angle θ in common. If you draw a right triangle such as the first one shown, you can easily compute θ from

You can then take that angle and plug it back into the direction cosines formula from the preceding section.

For the example of Figure 5-9, you can compute the angle between the horizontal x-axis and the vector is

Note that the horizontal proportion in this example is a negative value because it's measured in the direction opposite to the positive x-axis. The vertical proportion is positive because it's measured in the same direction as the positive y-axis.

Figure 5.9. Creating unit vectors by proportions.

As drawn, the angle θ is a direction cosine with respect to the x-axis. Similarly, you can calculate the remaining direction cosines for the other axes and plug them straight into the unit vector notation from the preceding section. Just remember, by first calculating this angle, you normally end up dealing with an irrational numerical value when you apply a sine, cosine, or tangent function.

In Figure 5-9, the proportion triangle is actually a right triangle. Using the Pythagorean theorem (which I discuss earlier in this chapter), you can quickly compute its hypotenuse:

You can then calculate the direction cosine with respect to the positive x-axis as

and the direction cosine with respect to the positive y-axis as

Note that for the third dimension, the direction cosine for the z-axis is 90 degrees, and consequently cos(90) = 0.

Assembling these into the unit vector equation as follows:

produces a unit vector

So how do you know which of the techniques in the preceding sections is best for your situation? The following checklist shows you some factors that affect your choice: