1.3 Products of Vectors
1.3.2 The Vector Product
(a)
C2¼C
C¼ðBAÞðBAÞ ¼BBþAA2BASinceB
B¼B2,AA¼A2, andBA¼AB¼BAcosϕBA¼ABcosϕAB, we getC2¼A2þB22ABcosϕAB ! C¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A2þB22ABcosϕAB
p
(b) For the two vectors inExample 1.5
A¼^x þ^y5^z, B¼ ^x þ^y5þ^z: we calculated
A¼3 ffiffiffiffiffi p3,
B¼3 ffiffiffiffiffi p3,
cosϕAB¼0:85185 The distance betweenP2andP1is therefore
d¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A2þB22ABcosϕAB
q
¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 27þ272270:85185
p ¼2:828
Exercise 1.3 An airplane flies with a velocity v¼^x100þ^y500þ^z200: Calculate the aircraft’s velocity in the direction of the vectorA¼^x þ^y þ^z:
Answer vA¼^x800=3þ^y800=3þ^z800=3
The unit vector may be in either direction perpendicular to the plane, and to define it uniquely, we employ the right-hand rule, as shown inFigure1.14. According to this rule, if the right-hand palm is placed on the first vector in the product and rotated toward the second vector through an angleϕAB, the extended thumb shows the correct direction of the cross product.
This rule immediately indicates that moving the palm from vectorBto vectorAgives a direction opposite to that moving from vectorAtoB. Thus, we conclude that the vector product is not commutative:
AB¼ BA ðnoncommutativeÞ ð1:32Þ
In addition to the noncommutative property of the vector product, the following properties are noted:
(1) The vector product is always perpendicular to the plane of the two vectors; that is, it is perpendicular to both vectors.
(2) For two vectors which are perpendicular to each other (ϕAB¼π/2), the magnitude of the vector product is equal to the product of the magnitudes of the two vectors (sinϕAB¼1) and is always positive.
(3) The vector product of two parallel vectors is always zero (sinϕAB¼0).
(4) The vector product of a vector with itself is always zero (sinϕAA¼0).
(5) The vector product is not associative (this will be discussed in the following section because it requires the definition of a triple product).
(6) The vector product is distributive:
AðBþCÞ ¼ABþAC ð1:33Þ
(7) The magnitude of the vector product represents the area bounded by the parallelogram formed by the two vectors and two lines parallel to the vectors, as shown inFigure1.15.
Evaluation of the vector product is performed similarly to that for the scalar product: We write the product explicitly and expand the expression based onEq. (1.33). Using two general vectorsAandBin Cartesian coordinates, we get
AB¼ ^xAxþ^yAyþ^zAz
^
xBxþ^yByþ^zBz
¼ð^x^xÞAxBxþ
^ x^y
AxByþ
^ x^z
AxBz
þð^y^xÞAyBxþ
^ y^y
AyByþ
^ y^z
AyBz
þð^z^xÞAzBxþ
^z^y
AzByþ
^ z^z
AzBz
ð1:34Þ A
B
n A×B
φAB
Figure 1.14 The vector product between vectorsAandB
A B S=|A×B|
Figure 1.15 Interpretation of the magnitude of the vector product as a surface
Because the unit vectors^x,^y,^zare perpendicular to each other, and using the right-hand rule inFigure1.14, we can write
^
x ^x ¼^y ^y ¼^z ^z ¼0 ð1:35Þ
from property 4 above. Similarly, using property 2 and the right-hand rule, we can write
^
x ^y ¼^z, ^y ^z ¼^x, ^z^x ¼^y,
^
y ^x ¼ ^z, ^z ^y ¼ ^x, ^x ^z ¼ ^y ð1:36Þ Substitution of these products and rearranging terms gives
AB¼^x AyBzAzBy
þ^yðAzBxAxBzÞ þ^z AxByAyBx
ð1:37Þ This is a rather straightforward operation, although lengthy. To avoid having to go through this process every time we use the vector product, we note that the expression inEq. (1.37)has the form of the determinant of a 33 matrix:
AB¼
^
x ^y ^z Ax Ay Az
Bx By Bz
¼^x AyBzAzBy
þ^yðAzBxAxBzÞ þ^z AxByAyBx
ð1:38Þ
In the system of coordinates used here (right-hand Cartesian coordinates), the vector product is cyclic; that is, the products inEq. (1.36)are cyclical, as shown inFigure1.16. This is a simple way to generate the signs of the components of the cross product: a cross product performed in the sequence shown by the arrows inFigure1.16ais positive; if it is in the opposite sequence (Figure1.16b), it is negative.
The vector product is used for a number of important operations. They include finding the direction of the vector product, calculation of areas, evaluation of normal unit vectors, and representation of fields.
Example 1.7Application: Vector Normal to a Plane
(a) Find a vector normal to a plane that contains pointsP1(0,1,0),P2(1,0,1), andP3(0,0,1).
(b) Find the normal unit vector.
Solution: This is a common use for the vector product. Because the vector product of two vectors is normal to both vectors, we must first find two vectors that lie in the plane. Their vector product gives the normal vector. Calculation of the normal unit vector can be done either using the definition of the unit vector inEq. (1.2)or through the use of the scalar and vector products.
Two vectors in the plane can be defined using any two pairs of points. UsingP1andP2, we define a vector (fromP1toP2) as A¼^xðx2x1Þ þ^y
y2y1 þ^z
z2z1
¼^xð10Þ þ^y 01
þ^z 10
¼^x1^y1þ^z1 Similarly for a second vector, we choose the vector betweenP1andP3. This gives
B¼^xðx3x1Þ þ^y y3y1
þ^z z3z1
¼^xð00Þ þ^y 01
þ^z 10
¼ ^y1þ^z1 x=y×z
y=z×x z=x×y
−x=z×y
−y=x×z
−z=y×x
z z
y
a b
y
x x
Figure 1.16 The cyclical relations between the various vector products of the unit vectors in Cartesian coordinates. (a) Positive sequence. (b) Negative sequence
1.3 Products of Vectors 15
The cross product,C, is a vector normal to bothAandBand, therefore, to the plane:
AB¼ð^x1^y1þ^z1Þ
^y1þ^z1
¼^x1 ^ð y1Þ þ^x1^z1þ ^y1
^y1
þ ^ð y1Þ ^z1þ^z1 ^y1
þ^z1^z1 Using the identities inEqs. (1.35)and(1.36), we get
C¼AB¼ ^y1^z1
The unit vector can be found fromEq. (1.2)or from the definition of the vector product inEq. (1.31). We use the latter as an example of an alternative method:
^
n ¼ AB ABsinϕAB
j j
The angleϕABcan be most easily calculated from the scalar product inEq. (1.23)as ϕAB¼ cos1 A
BAB
To do so, we need to evaluate the scalar product and the magnitude of the vectors. These are A
B¼ð^x1^y1þ^z1Þð^y1þ^z1Þ ¼2, A¼pffiffiffi3 B¼pffiffiffi2Thus,
ϕAB¼ cos1 2 ffiffiffi6 p
¼35160 The unit normal vector is now
^
n ¼ AB ABsinϕAB
j j¼ ^y1^z1 ffiffiffi6
p sin 35 160
¼^y1^z1
1:4142 ¼ ^y0:7071^z0:7071 The same result is obtained usingEq. (1.2):
^
n ¼ AB AB
j j¼ ^y1^z1
^y1^z1
j j¼^y1^z1 ffiffiffi2
p ¼ ^y0:7071^z0:7071
Exercise 1.4 VectorsA¼^x1^y2þ^z3 andB¼^x3þ^y5þ^z1 are in a plane, not necessarily perpendicular to each other. VectorC¼^x17 +^y8^z11 is perpendicular to the same plane. Show that the vector product between CandA(or betweenCandB) must also be in the plane ofAandB.
Example 1.8Application: Area of a Triangle Find the area of the triangle with vertices at three general pointsP1(x1, y1, z1),P2(x2,y2,z2), andP3(x3,y3,z3) (Figure1.17a).
Solution: In this case, the vector nature of the vector product is irrelevant, but the magnitude of the vector product in Eq. (1.31)is equal to the area of the parallelogram formed by the two vectors. This can be seen from the fact that the magnitude ofABisA(BsinϕAB). This is the area of rectangleabb0c0inFigure1.17b. Since trianglesacc0andbdb0are identical, this is also the area of parallelogramabdc. Since trianglesabcandcbdare identical, the area ofabcis equal to half the area ofabdc. Calculation of the area of triangleabcis done by calculating the magnitude of the cross product of two of the vectors forming the sides of the triangle and dividing by 2:
Sabc¼jABj 2
FromFigure1.17a, vectorsAandBare
A¼^xðx2x1Þ þ^y y2y1
þ^z z2z1
, B¼^xðx3x1Þ þ^y
y3y1 þ^z
z3z1
The vector product is obtained usingEq. (1.38), and from this, the area of the triangle is
Sabc¼jABj
2 ¼1
2
^
x ^y ^z
x2x1 y2y1 z2z1
x3x1 y3y1 z3z1
Exercise 1.5 Find the area of the triangle formed by points (1,3,0), (1,2,1), and (3,5,2).
Answer ffiffiffiffiffi p24
=2¼2:4495=m2:
Example 1.9 Find a unit vector normal to both of the vectors
A¼^x3þ^y1^z2, B¼^x1^y5:
Solution: The vector productsABorBAresult in vectors normal to bothAandB, respectively:
AB¼ð^x3þ^y1^z2Þ
^
x1^y5
¼ð Þ ^x3
^ x1
þ
^ x3
^y5
þ
^ y1
^ x1 þð Þ ^^y1 ð y5Þ þ ^ð z2Þ ð Þ þ ^^x1 ð z2Þ ^ð y5Þ
¼ ^x10^y2^z16 The unit vector is
^
n ¼ ^x10^y2^z16
^x10^y2^z16
j j¼^x10^y2^z16 ffiffiffiffiffiffiffiffi
p360 ¼^x5^y ^z8 3 ffiffiffiffiffi
p10
Using the productBAresults in a unit vector^n as can be shown by application of the right-hand rule.
a b
A B
A B
|B|
|A|
a b
c d
a
b
c'
b'
φAB φAB
BsinφAB
P1
P1
P2 P2
P3 P3
c
Figure 1.17 Area of a triangle. (a) A triangle with two of its sides shown as vectors. (b) The area of the triangle is half the area of the parallelogramabdc
1.3 Products of Vectors 17
Example 1.10 The general equation of a plane in Cartesian coordinates isax+by +cx+ d¼0. The equation of the plane may be found asf(x,y,z)¼n
C¼0 wherenis the normal vector to the plane andCis a general vector in the plane.(a) Given three pointsP1(1,0,2),P2(3,1,–2),P3(2,3,2) in a plane, find the equation of the plane.
(b) Show that the equation of the plane may be written as
f x;ð y;zÞ ¼nxðxx0Þ þnyðyy0Þ þnzðzz0Þ ¼0
where nx, ny, and nz are the scalar components of the normal unit vector to the plane and (x0,y0,z0) are the coordinates of a point in the plane.
Solution: The three points given define two vectors (sayP1toP2andP1toP3). The normal to the plane is obtained through use of the vector product. The vectorAis then defined between a general point (x,y,z) and any of the points given. The form in(b)is found from(a):
(a) Two vectors necessary to calculate the normal vector to the plane are P1toP2:A¼^xð31Þ þ^yð10Þ þ^zð22Þ ¼^x2þ^y1^z4 P1toP3:B¼^xð21Þ þ^yð30Þ þ^zð22Þ ¼^x1þ^y3
These two vectors are in the plane. Therefore, the normal vector to the plane may be written as:
n¼AB¼ð^x2þ^y1^z4Þ ð^x1þ^y3Þ ¼^x12^y4þ^z5 A general vector in the plane may be written as:
C¼^xðx1Þ þ^yðy0Þ þ^zðz2Þ where pointP1was used, arbitrarily.
The equation of the plane is
f x;ð y;zÞ ¼n
C¼ð^x12^y4þ^z5Þð^xðx1Þ þ^yðy0Þ þ^zðz2ÞÞ ¼0or
f x;ð y;zÞ ¼12ðx1Þ 4ðy0Þ þ5ðz2Þ ¼12x4yþ5z22¼0 (b) To show that the formula given produces the same result, we first calculate the normal unit vector:
^
n ¼^x 12 ffiffiffiffiffiffiffiffi p185^y 4
ffiffiffiffiffiffiffiffi p185þ^z 5
ffiffiffiffiffiffiffiffi p185
The point (x0,y0,z0) can be any point in the plane. SelectingP3, for example, the equation of the plane is f x;ð y;zÞ ¼ 12
ffiffiffiffiffiffiffiffi
p185ðx2Þ 4 ffiffiffiffiffiffiffiffi
p185ðy3Þ þ 5 ffiffiffiffiffiffiffiffi
p185ðz2Þ ¼0 Multiplying both sides of the equation by ffiffiffiffiffiffiffiffi
p185
, we get
f x;ð y;zÞ ¼12ðx2Þ 4ðy3Þ þ5ðz2Þ ¼12x4yþ5z22¼0 This is the same as the result obtained in(a).
The formula in(b)is called the scalar equation of the plane whereas the formf(x,y,z)¼n
C¼0 is referred to as the vector equation of the plane.