Vectors in Cartesian Three-Space Help (page 3)
Introduction to Vectors in Cartesian Three-Space
A vector in Cartesian three-space is the same as a vector in the Cartesian plane, except that there is more “freedom” in terms of direction. This makes the expression of direction in 3D more complicated than is the case in 2D. It also makes vector arithmetic a lot more interesting!
In Cartesian three-space, vectors a and b can be denoted as arrow-tipped line segments from the origin (0,0,0) to points ( x a , y a , z a ) and ( x b , y b , z b ), as shown in Fig. 9-5. This, like all three-space drawings in this chapter, is a perspective illustration. Both vectors in this example point in directions on the reader’s side of the plane containing the page. In a true 3D model, both of them would “stick up out of the paper at an angle.”
In Fig. 9-5, both vectors a and b have their end points at the origin. This is the standard form of a vector in any coordinate system. In order for the following formulas to hold, vectors must be expressed in standard form. If a given vector is not in standard form, it can be converted by subtracting the coordinates ( x 0 , y 0 , z 0 ) of the starting point from the coordinates of the end point ( x 1 , y 1 , z 1 ). For example, if a vector a * starts at (4,7,0) and ends at (1,−3,5), it reduces to an equivalent vector a in standard form:
a = [(1 − 4), (−3 − 7), (5 − 0)]
= (−3, −10, 5)
Any vector a *, which is parallel to a and has the same length as a , is equal to vector a , because a* has the same magnitude and the same direction as a . Similarly, any vector b *, which is parallel to b and has the same length as b , is defined as being equal to b . As in the 2D case, a vector is defined solely on the basis of its magnitude and its direction. Neither of these two properties depends on the location of the end point.
Defining the Magnitude and Direction of a Vector
Defining The Magnitude
When the end point of a vector a is at the origin, the magnitude of a , written | a | or a , can be found by a three-dimensional extension of the Pythagorean theorem for right triangles. The formula looks like this:
The magnitude of any vector a in standard form is simply the distance of the end point from the origin. Note that the above formula is the distance formula for two points, (0,0,0) and ( x a , y a , z a ).
Direction Angles And Cosines
The direction of a vector a in standard form can be defined by specifying the angles θ x , θ y , and θ z that the vector a subtends relative to the positive x , y , and z axes respectively (Fig. 9-6). These angles, expressed in radians as an ordered triple ( θ x , θ y , θ z ), are the direction angles of a .
Sometimes the cosines of these angles are used to define the direction of a vector a in 3D space. These are the direction cosines of a :
dir a = (α, β, γ)
α = cos θ x
β = cos θ y
γ = cos θ z
For any vector a in Cartesian three-space, the sum of the squares of the direction cosines is always equal to 1. That is
α 2 + β 2 + γ 2 = 1
Another way of expressing this is:
cos 2 θ x + cos 2 θ y + cos 2 θ z = 1
where the expression cos 2 θ means (cos θ ) 2.
The sum of vectors a = ( x a , y a , z a ) and b = ( x b , y b , z b ) in standard form is given by the following formula:
a + b = [( x a + x b ), ( y a + y b ), ( z a + z b )]
This sum can, as in the two-dimensional case, be found geometrically by constructing a parallelogram with a and b as adjacent sides. The sum a + b is the diagonal of the parallelogram. This is shown in Fig. 9-7. (The parallelogram appears distorted because of the perspective of the drawing.)
Multiplication By Scalar and The Dot and Cross Product
Multiplication By Scalar
In three-dimensional Cartesian coordinates, let vector a be defined by the coordinates ( x a , y a , z a ) when reduced to standard form. Suppose a is multiplied by a positive real scalar k . Then the following equation holds:
k a = k ( x a , y a , z a ) = ( kx a , ky a , kz a )
If a is multiplied by a negative real scalar − k , then:
− k a = − k ( x a , y a , z a ) = (− kx a , − ky a , − kz a )
Suppose the direction angles of a are represented by the ordered triple ( θ xa , θ ya , θ za ). Then the direction angles of k a are the same; they are also ( θ xa , θ ya , θ za ). The direction angles of − k a are all changed by 180° (π rad). The direction angles of − k a are obtained by adding or subtracting 180° (π rad) to each of the direction angles for k a , so that the results are all at least 0° (0 rad) but less than 360° (2π rad).
The dot product , also known as the scalar product and written a • b , of vectors a = ( x a , y a , z a ) and b = ( x b , y b , z b ) in standard form is a real number given by the formula:
a • b = x a x b + y a y b + z a z b
The dot product can also be found from the magnitudes | a | and | b |, and the angle θ between vectors a and b as measured counterclockwise in the plane containing them both:
a • b = | a || b | cos θ
The cross product , also known as the vector product and written a × b , of vectors a = ( x a , y a , z a ) and b = ( x b , y b , z b ) in standard form is a vector perpendicular to the plane containing a and b . Let θ be the angle between vectors a and b as measured counterclockwise in the plane containing them both, as shown in Fig. 9-8. The magnitude of a × b is given by the formula:
| a × b | = | a || b | sin θ
In the example shown, a × b points upward at a right angle to the plane containing both vectors a and b . If 0° < θ < 180° (0 < θ < π), you can use the right-hand rule to ascertain the direction of a × b . Curl your fingers in the direction that θ , the angle between a and b , is defined. Extend your thumb. Then a × b points in the direction of your thumb.
When 180° < θ < 360° (π rad < θ < 2π rad), the cross-product vector reverses direction because its magnitude becomes negative. This is demonstrated by the fact that, in the above formula, sin θ is positive when 0° < θ < 180° (0 rad < θ < π rad), but negative when 180° < θ < 360° (π rad < θ < 2π rad).
Any vector a, reduced to standard form so its starting point is at the origin, ends up at some point ( x a , y a , z a ). This vector can be broken down into the sum of three mutually perpendicular vectors, each of which lies along one of the coordinate axes as shown in Fig. 9-9:
a = ( x a , y a , z a )
= ( x a ,0,0) + (0, y a ,0) + (0,0, z a )
= x a (1,0,0) + y a (0,1,0) + z a (0,0,1)
The vectors (1,0,0), (0,1,0), and (0,0,1) are called unit vectors because their length is 1. It is customary to name these vectors i , j , and k , because they come in handy:
(1,0,0) = i
(0,1,0) = j
(0,0,1) = k
Therefore, the vector a shown in Fig. 9-9 breaks down this way:
a = ( x a , y a , z a ) = x a i + y a j + z a k
Break the vector b = (−2, 3, −7) down into a sum of multiples of the unit vectors i , j , and k .
This is a simple process—almost trivial—but envisioning it requires a keen “mind’s eye.” If you have any trouble seeing this in your imagination, think of i as “one unit of width going to the right,” j as “one unit of height going up,” and k as “one unit of depth coming towards you.” Here we go:
b = (−2,3, − 7)
= −2 × (1,0,0) + 3 × (0,1,0) + [−7 × (0,0,1)]
= −2 i + 3 j + (−7) k
= −2 i + 3 j − 7 k
Practice problems for these concepts can be found at: Vectors And Cartesian Three-Space Practice Test.
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