2.2. The dot product of two vectors 23 2.2 The dot product of two vectors The dot product is used to project a vector in a given direction, to reduce a vector to components, to reduce vector equations to scalar equations, to define work and power, and to help solve geometry problems. The dot product of two vectors  A and  B is written  A ·  B (pronounced ‘A dot B’). The dot product of  A and  B is the product of the magnitudes of the two vectors times a number that expresses the degree to which  A and  B are parallel: cos θ AB , where θ AB is the angle between  A and  B. That is,  B  A θ AB B A cos θ AB cos θ AB Figure 2.17: The dot product of  A and  B is ascalar and so is not easily drawn. It is given by  A ·  B = ABcos θ AB which is A times the projection of  B in the A direction and also B times the projection of  A in the B direction. (Filename:tfigure1.11)  A ·  B def =|  A||  B|cos θ AB which is sometimes written more concisely as  A ·  B = ABcos θ . One special case is when cosθ AB = 1,  A and  B are parallel, and  A ·  B = AB. Another is when cos θ AB = 0,  A and  B are perpendicular, and  A ·  B = 0. 1  1  If you don’t know, almost without a thought, that cos0 = 1, cos π/2 = 0, sin 0 = 0, and sin π/2 = 1 now is as good a time as any to draw as many trian- gles and unit circles as it takes to cement these special cases into your head. The dot product of two vectors is a scalar. So the dot product is sometimes called the scalar product. Using the geometric definition of dot product, and the rules for vector addition we have already discussed, you can convince yourself of (or believe) the following properties of dot products. •  A ·  B =  B ·  A commutative law, ABcos θ = BAcosθ • (a  A) ·  B =  A · (a  B) = a(  A ·  B) a distributive law, (aA)B cosθ = A(aB) cos θ •  A · (  B +  C) =  A ·  B +  A ·  C another distributive law, the projection of  B +  C onto  A is the sum of the two separate projections •  A ·  B = 0if  A ⊥  B perpendicular vectors have zero for a dot product, ABcosπ/2 = 0 •  A ·  B =|  A||  B| if  A   B parallel vectors have the product of their magnitudes for a dot product, ABcos 0 = AB. In particular,  A ·  A = A 2 or |  A|= √  A ·  A • ˆ ı · ˆ ı = ˆ  · ˆ  = ˆ k · ˆ k = 1, ˆ ı · ˆ  = ˆ  · ˆ k = ˆ k · ˆ ı = 0 The standard base vectors used with cartesiancoordinates are unitvectors and they are perpendicular to each other. In math language they are ‘or- thonormal.’ • ˆ ı  · ˆ ı  = ˆ   · ˆ   = ˆ k  · ˆ k  = 1, ˆ ı  · ˆ   = ˆ   · ˆ k  = ˆ k  · ˆ ı  = 0 The standard crooked base vectors are orthonormal. The identities above lead to the following equivalent ways of expressing the dot product of  A and  B (see box 2.2 on page 24 to see how the component formula follows from the geometric definition above): 24 CHAPTER 2. Vectors for mechanics  A ·  B =|  A||  B|cos θ AB = A x B x + A y B y + A z B z (component formula for dot product) = A x  B x  + A y  B y  + A z  B z  =|  A|·[projection of  B in the  A direction] =|  B|·[projection of  A in the  B direction] Using the dot product to find components To find the x component of a vector or vector expression one can use the dot product of the vector (or expression) with a unit vector in the x direction as in figure 2.18. In particular, v x =  v · ˆ ı. x y v x v ˆ ı ˆ  Figure 2.18: The dot product with unit vectors gives projection. For example, v x =  v· ˆ ı. (Filename:tfigure1.3.dotprod) This idea can be used for finding components in any direction. If one knows the orientation of the crooked unit vectors ˆ ı  , ˆ   , ˆ k  relative to the standard bases ˆ ı, ˆ , ˆ k then all the angles between the base vectors are known. So one can evaluate the dot products between the standard base vectors and the crooked base vectors. In 2-D 2.3 THEORY Using the geometric definition of the dot product to find the dot product in terms of components Vectors are essentially a geometric concept and we have conse- quently defined the dot product geometrically as  A ·  B = ABcosθ . Almost 400 years ago Ren´e Descartes discovered that you could do geometry by doing algebra on the coordinates of points. So we should be able tofigure out thedot product of two vectors by knowing their components. The central key to finding this com- ponentformula isthe distributive law(  A·(  B+  C) =  A·  B+  A·  C). If we write  A = A x ˆ ı + A y ˆ  + A z ˆ k and  B = B x ˆ ı + B y ˆ  + B z ˆ k then we just repeatedly use the distributive law as follows.  A ·  B = (A x ˆ ı + A y ˆ  + A z ˆ k) · (B x ˆ ı + B y ˆ  + B z ˆ k) = (A x ˆ ı + A y ˆ  + A z ˆ k) · B x ˆ ı + (A x ˆ ı + A y ˆ  + A z ˆ k) · B y ˆ  + (A x ˆ ı + A y ˆ  + A z ˆ k) · B z ˆ k = A x B x ˆ ı · ˆ ı + A y B x ˆ  · ˆ ı + A z B x ˆ k · ˆ ı + A x B y ˆ ı · ˆ  + A y B y ˆ  · ˆ  + A z B y ˆ k · ˆ  + A x B z ˆ ı · ˆ k + A y B z ˆ  · ˆ k + A z B z ˆ k · ˆ k = A x B x (1) + A y B x (0) + A z B x (0) + A x B y (0) + A y B y (1) + A z B y (0) + A x B z (0) + A y B z (0) + A z B z (1) ⇒  A ·  B = A x B x + A y B y + A z B z (3D). ⇒  A ·  B = A x B x + A y B y (2D). The demonstration above could have been carried out using a different orthogonal coordinate system x  y  z  that was crooked with respect to the xyz system. By identical reasoning we would find that  A ·  B = A x  B x  + A y  B y  + A z  B z  . Even though all of the numbersin thelist [A x , A y , A z ]might bedifferentfromthe numbers in the list [A x  , A y  , A z  ] and similarly all the list [  B] xyz might be different than the list [  B] x  y  z  , so (somewhat remarkably), A x B x + A y B y + A z B z = A x  B x  + A y  B y  + A z  B z  . If we call our coordinate x 1 , x 2 , and x 3 ; and our unit base vectors ˆ e 1 , ˆ e 2 , and ˆ e 3 we would have  A = A 1 ˆ e 1 + A 2 ˆ e 2 + A 3 ˆ e 3 and  B = B 1 ˆ e 1 + B 2 ˆ e 2 + B 3 ˆ e 3 and the dot product has the tidy form:  A ·  B = A 1 B 1 + A 2 B 2 + A 3 B 3 = 3  i=1 A i B i . 2.2. The dot product of two vectors 25 assume that the dot products between the standard base vectors and the vector ˆ   (i.e., . ˆ ı · ˆ   , ˆ  · ˆ   ) are known. One can then use the dot product to find the x  y  components (A x  , A y  ) from the xycoordinates (A x , A y ). For example, as shown in 2-D in figure 2.19, we can start with the obvious equation x x' y y' A x A y A x' A y'  A = A x ˆ ı + A y ˆ   A = A x  ˆ ı  + A y  ˆ   θ ˆ ı ˆ  ˆ ı  ˆ   Figure 2.19: The dot product helps find components in terms of crooked unit vectors. For example, A y  =  A· ˆ   = A x ( ˆ ı· ˆ   ) + A y ( ˆ ı· ˆ   ) = A x (−sin θ) + A y (cos θ). (Filename:tfigure1.3.dotprod.a)  A =  A and dot both sides with ˆ   to get:  A · ˆ   =  A · ˆ   (A x  ˆ ı  + A y  ˆ   )     A · ˆ   = (A x ˆ ı + A y ˆ )     A · ˆ   A x  ˆ ı  · ˆ    0 +A y  ˆ   · ˆ    1 = A x ˆ ı · ˆ   + A y ˆ  · ˆ   A y  = A x ( ˆ ı · ˆ   )    −sin θ +A y ( ˆ  · ˆ   )    cos θ Similarly, one could find the component A x  using a dot product with ˆ ı  . This technique of finding components is useful when one problem uses more than one base vector system. Using dot products with other than ˆı, ˆ,or ˆ k It is often useful to use dot products to get scalar equations using vectors other than ˆ ı, ˆ , and ˆ k. Example: Getting scalar equations without dotting with ˆ ı, ˆ ,or ˆ k Given the vector equation. −mg ˆ  + N ˆ n = ma ˆ λ where it is known that the unit vector ˆ n is perpendicular to the unit vector ˆ λ, we can get a scalar equation by dotting both sides with ˆ λ which we write as follows  (−mg ˆ  + N ˆ n) = (ma ˆ λ)  · ˆ λ (−mg ˆ  + N ˆ n)· ˆ λ = (ma ˆ λ)· ˆ λ −mg ˆ · ˆ λ + N ˆ n· ˆ λ  0 = ma ˆ λ· ˆ λ  1 −mg ˆ · ˆ λ = ma. Then we find ˆ · ˆ λ as the cosine of the angle between ˆ  and ˆ λ.Wehave thus turned our vector equation into a scalar equation and eliminated the unknown N at the same time. ✷ 26 CHAPTER 2. Vectors for mechanics Using dot products to solve geometry problems We have seen how a vector can be broken down into a sum of components each parallel to one of the orthogonal base vectors. Another useful decomposition is this: Given any vector  A and a unit vector ˆ λ the vector  A can be written as the sum of two parts,  A =  A  +  A ⊥ where  A  is parallelto ˆ λ and  A ⊥ is perpendicularto ˆ λ (seefig. 2.20). The part parallel to ˆ λ is a vector pointed in the ˆ λ direction that has the magnitude of the projection of  A in that direction,  A  = (  A · ˆ λ) ˆ λ. The perpendicular part of  A is just what you get when you subtract out the parallel part, namely,  A ⊥ =  A −  A  =  A − (  A · ˆ λ) ˆ λ The claimed properties of the decomposition can now be checked, namely that  A =  A ⊥  A ˆ λ  A || Figure 2.20: For any  A and ˆ λ,  A can be decomposed into a part parallel to ˆ λ and a part perpendicular to ˆ λ . (Filename:tfigure.Graham1)  A  +  A ⊥ (just add the 2 equations above and see), that  A  is in the direction of ˆ λ (its a scalar multiple), and that  A ⊥ is perpendicular to ˆ λ (evaluate  A ⊥ · ˆ λ and find 0). Example. Given the positions of three points  r A ,  r B , and  r C what is the position of the point D on the line AB that is closest to C? The answer is,  r D =  r A +  r C/A  where  r C/A  is the part of  r C/A that is parallel to the line segment AB. Thus,  r D =  r A + (  r C −  r A ) ·  r B −  r A |  r B −  r A | . ✷ Likewise we could find the parts of a vector  A in and perpendicular to a given plane. If the plane is defined by two vectors that are not necessarily orthogonal we could follow these steps. First find two vectors in the plane that are orthogonal, using the method above. Next subtract from  A the part of it that is parallel to each of the two orthogonal vectors in the plane. In math lingo the execution of this process goes by the intimidating name ‘Graham Schmidt orthogonalization.’ A Given vector can be written as various sums and products A vector  A has many representations. The equivalence of different representations of a vector is partially analogous to the case of a dimensional scalar which has the same value no matter what units are used (e.g., the mass m = 4.41 lbm is equal to m = 2kg). Here are some common representations of vectors. Scalar times a unit vector in the vector’s direction.  F = F ˆ λ means the scalar F multiplied by the unit vector ˆ λ. 2.2. The dot product of two vectors 27 Sum of orthogonal component vectors.  F =  F x +  F y is a sum of two vectors parallel to the x and y axis, respectively. In three dimensions,  F =  F x +  F y +  F z . Components times unit base vectors.  F = F x ˆ ı + F y ˆ  or  F = F x ˆ ı + F y ˆ  + F z ˆ k in three dimensions. One way to think of this sum is to realize that  F x = F x ˆ ı,  F y = F y ˆ  and  F z = F z ˆ k. Components times rotated unit base vectors.  F = F  x i  + F  y j  or  F = F  x i  + F  y j  + F  z k  in three dimensions. Here the base vectors marked with primes, i  , j  and k  , are unit vectors parallel to some mutually orthogonal x  , y  , and z  axes. These x  , y  , and z  axes may be crooked in relation to the x, y, and z axis. That is, the x  axis need not be parallel to the x axis, the y  not parallel to the y axis, and the z  axis not parallel to the z axis. Components times other unit base vectors. If you use polar or cylindrical coordi- nates the unit base vectors are ˆ e θ and ˆ e R , so in 2-D ,  F = F R ˆ e R + F θ ˆ e θ and in 3-D,  F = F R ˆ e R + F θ ˆ e θ + F z ˆ k. If you use ‘path’ coordinates, you will use the path-defined unit vectors ˆ e t , ˆ e n , and ˆ e b so in 2-D  F = F t ˆ e t + F n ˆ e n . In 3-D  F = F t ˆ e t + F n ˆ e n + F b ˆ e b . A list of components. [  F ] xy = [F x , F y ]or[  F ] xyz = [F x , F y , F z ] in three dimen- sions. This form coincides best with the way computers handle vectors. The row vector [F x , F y ] coincides with F x ˆ ı + F y ˆ  and the row vector [F x , F y , F z ] coincides with F x ˆ ı + F y ˆ  + F z ˆ k. In summary:  A =  A =|  A| ˆ λ A = A ˆ λ A , where ˆ λ A   A, A =|  A| and | ˆ λ A |=1 =  A x +  A y +  A z where  A x ,  A y ,  A z are parallel to the x, y, z axis = A x ˆ ı + A y ˆ  + A z ˆ k, where ˆ ı, ˆ , ˆ k are parallel to the x, y, z axis = A x  ˆ ı  + A y  ˆ   + A z  ˆ k  , where ˆ ı  , ˆ   , ˆ k  are  to skewed x  , y  , z  axes = A R ˆ e R + A θ ˆ e θ + A z ˆ k, using polar coordinate basis vectors. [  A] xyz = [A x , A y , A z ][  A] xyz stands for the component list in xyz [  A] x  y  z  = [A x  , A y  , A z  ][  A] x  y  z  stands for the component list in x  y  z  Vector algebra Vectors are algebraic quantities and manipulated algebraically in equations. Therules for vector algebra are similar to the rules for ordinary (scalar) algebra. For example, if vector  A is the same as the vector  B,  A =  B. For any scalar a and any vector  C, we then  A +  C =  B +  C, a  A = a  B, and  A ·  C =  B ·  C, because performing the same operation on equal quantities maintains the equality. The vectors  A,  B, and  C might themselves be expressions involving other vectors. The equations above show the allowable manipulations of vector equations: adding a common term to both sides, multiplying both sides by a common scalar, taking the dot product of both sides with a common vector. All the distributive, associative, and commutative laws of ordinary addition and multiplication hold. 1  . 1  Caution: Butyou cannot divide a vector by a vector or a scalar by a vector: 7/ ˆ ı and  A/  C are nonsense expressions. And it does not make sense to add a vector and a scalar, 7 +  A is a nonsense expression. 28 CHAPTER 2. Vectors for mechanics Vector calculations on the computer Most computer programs deal conveniently with lists of numbers, but not with vec- tor notation and units. Thus our computer calculations will be in terms of vector components with the units left off. For example, when we write on the computer F=[35-7] we take that to be the plain computer typing for [  F ] xyz = [3N, 5N, −7 N]. This assumes that we are clear about what units and what coordinate system we are using. In particular, at this point in the course, you should only use one coordinate system in one problem in computer calculations. Most computer languages will allow vector addition by a sequence of lines some- thing like this: A=[125] B=[-2 419] C=A+B scaling (stretching) like this: A=[125] C = 3*A and dot products like this: A =[125] B =[-2 419] D = A(1)*B(1) + A(2)*B(2) + A(3)*B(3). In our pseudo code we write D = A dot B. Many computer languages have a shorter way to write the dot product like dot(A,B). In a language built for linear algebra D = A*B’ 1  will work because the rules of matrix multiplication are then 1  B’ is a common notation for the trans- pose of B, which means, in this case, to turn the rowof numbers B into a column of num- bers. the same as the component formula for the dot product. 2.2. The dot product of two vectors 29 SAMPLE 2.12 Calculating dot products: Find the dot product of the two vectors  a = 2 ˆ ı +3 ˆ  − 2 ˆ k and  r = 5m ˆ ı −2m ˆ . Solution The dot product of the two vectors is  a ·  r = (2 ˆ ı +3 ˆ  − 2 ˆ k) · (5m ˆ ı −2m ˆ ) = (2 · 5m) ˆ ı · ˆ ı  1 −(2 · 2m) ˆ ı · ˆ   0 +(3 · 5m) ˆ  · ˆ ı  0 −(3 · 2m) ˆ  · ˆ    1 −(2 · 5m) ˆ k · ˆ ı  0 +(2 · 2m) ˆ k · ˆ    0 = 10 m −6m = 4m.  a ·  r = 4m Comments: Note that with just a little bit of foresight, we could totally ignore the ˆ k component of  a since  r has no ˆ k component, i.e., ˆ k ·  r = 0. Also, if we keep in mind that ˆ ı · ˆ  = ˆ  · ˆ ı = 0, we could compute the above dot product in one line:  a ·  r = (2 ˆ ı +3 ˆ ) · (5m ˆ ı −2m ˆ ) = (2 · 5m) ˆ ı · ˆ ı  1 −(3 · 2m) ˆ  · ˆ    1 = 4m. SAMPLE 2.13 What is the y-component of  F = 5N ˆ ı +3N ˆ  + 2N ˆ k ? Solution Although it is perhaps obvious that the y-component of  F is 3 N, the scalar multiplying the unit vector ˆ , we calculate it below in a formal way using the dot product between two vectors. We will use this method later to find components of vectors in arbitrary directions. F y =  F · (a unit vector along y-axis) = (5N ˆ ı +3N ˆ  + 2N ˆ k) · ˆ  = 5N ˆ ı · ˆ   0 +3N ˆ  · ˆ    1 +2N ˆ k · ˆ    0 = 3N. F y =  F · ˆ  = 3N. 30 CHAPTER 2. Vectors for mechanics SAMPLE 2.14 Finding angle between two vectors using dot product: Find the angle between the vectors  r 1 = 2 ˆ ı +3 ˆ  and  r 2 = 2 ˆ ı − ˆ . Solution From the definition of dot product between two vectors  r 1 ·  r 2 =|  r 1 ||  r 2 |cos θ or cos θ =  r 1 ·  r 2 |  r 1 ||  r 2 | = (2 ˆ ı +3 ˆ ) · (2 ˆ ı − ˆ ) ( √ 2 2 + 3 2 )( √ 2 2 + 1 2 ) = 4 − 3 √ 13 √ 5 = 0.124 Therefore, θ = cos −1 (0.124) = 82.87 o . θ = 83 o SAMPLE 2.15 Finding direction cosines from unit vectors: Find the angles (from direction cosines) between  F = 4N ˆ ı +6N ˆ  + 7N ˆ k and each of the three axes. Solution  F = F ˆ λ ˆ λ =  F F = 4N ˆ ı +6N ˆ  + 7N ˆ k √ 4 2 + 6 2 + 7 2 N = 0.4 ˆ ı +0.6 ˆ  + 0.7 ˆ k. Let the angles between ˆ λ and the x, y, and z axes be θ, φ and ψ respectively. Then cos θ = ˆ ı · ˆ λ | ˆ ı|| ˆ λ| = 0.4 |1||1| = 0.4. ⇒ θ = cos −1 (0.4) = 66.4 o . Similarly, cos φ = 0.6orφ = 53.1 o cos ψ = 0.7orψ = 45.6 o . θ = 66.4 o ,φ= 53.1 o ,ψ= 45.6 o Comments: The components of a unit vector give the direction cosines with the respective axes. That is, if the angle between the unit vector and the x, y, and z axes are θ, φ and ψ, respectively (as above), then ˆ λ = cosθ   λ x ˆ ı +cosφ   λ y ˆ  + cosψ   λ z ˆ k. 2.2. The dot product of two vectors 31 SAMPLE 2.16 Projection of a vector in the direction of another vector: Find the component of  F = 5N ˆ ı +3N ˆ  + 2N ˆ k along the vector  r = 3m ˆ ı −4m ˆ . Solution The dot product of a vector  a with a unit vector ˆ λ gives the projection of the vector  a in the direction of the unit vector ˆ λ. Therefore, to find the component of  F along  r ,wefirst find a unit vector ˆ λ r along  r and dot it with  F . ˆ λ r =  r |  r | = 3m ˆ ı −4m ˆ  √ 3 2 + 4 2 m = 0.6 ˆ ı −0.8 ˆ  F r =  F · ˆ λ r = (5N ˆ ı +3N ˆ  + 2N ˆ k) · (0.6 ˆ ı −0.8 ˆ ) = 3.0N+2.4N= 5.4N. F r = 5.4N SAMPLE 2.17 Assume that after writing the equation   F = m  a in a particular problem, astudent finds   F = (20 N−P 1 ) ˆ ı +7N ˆ  −P 2 ˆ k and  a = 2.4m/s 2 ˆ ı +a 3 ˆ . Separate the scalar equations in the ˆ ı, ˆ , and ˆ k directions. Solution   F = m  a Taking the dot product of both sides of this equation with ˆ ı, we write ˆ ı ·   F = ˆ ı ·m  a ˆ ı ·  (20 N − P 1 ) ˆ ı +7N ˆ  − P 2 ˆ k = m(2.4m/s 2 ˆ ı +a 3 ˆ )  ⇒ (20 N − P 1 )    F x ˆ ı · ˆ ı  1 +7N ˆ  · ˆ ı  0 −P − 2 ˆ k · ˆ ı  0 = m(2.4m/s 2    a x ˆ ı · ˆ ı  1 +a 3 ˆ  · ˆ ı  0 ) ⇒  F x = ma x ⇒ 20 N − P 1 = m(2.4m/s 2 ) Similarly, ˆ  ·    F = m  a  ⇒  F y = ma y (2.1) ˆ k ·    F = m  a  ⇒  F z = ma z . (2.2) Substituting the given components of  F and  a in the remaining Eqns. (2.1) and (2.2) we get 7N = ma y −P 2 = 0. Comments: Aslongasbothsides ofavectorequationareinthe samebasis, separating the scalar equations is trivial—simply equate the respective components from both sides. The technique of taking the dot product of both sides with a vector is quite general and powerful. It gives a scalar equation valid in any direction that one desires. You will appreciate this technique more if the vector equation uses more than one basis. 32 CHAPTER 2. Vectors for mechanics 2.3 Cross product, moment, and mo- ment about an axis When you try to move something you can push it and you can turn it. In mechanics, the measure of your pushing is the net force you apply. The measure of your turning is the net moment, also sometimes called the net torque or net couple. In this section we will define the moment of a force intuitively, geometrically, and finally using vector algebra. We will do this first in 2 dimensions and then in 3. The main mathematical tool here is the vector cross product, a second way of multiplying vectors together. The cross product is used to define (and calculate) moment and to calculate various quantities in dynamics. The cross product also sometimes helps solve three-dimensional geometry problems. Although concepts involving moment(and rotation) areoften harder forbeginners than force (and translation), they were understood first. The ancient principle of the lever is the basic idea incorporated by moments. The principle of the lever can be viewed as the root of all mechanics. Ultimately you can take on faith the vector definition of moment (given opposite the inside cover) and its role in eqs. II. But we can more or less deduce the definition by generalizing from common experience. Teeter totter mechanics (not a free body diagram) Figure 2.21: On a balanced teeter totter the bigger person gets the short end of the stick. A sideways force directed towards the hinge has no effect on the balance. (Filename:tfigure.teeter) The two people weighing down on the teeter totter in Fig. 2.21 tend to rotate it about its hinge, the right one clockwise and the left one counterclockwise. We will now cook up a measure of the tendency of each force to cause rotation about the hinge and call it the moment of the force about the hinge. As is verified a million times a year by young future engineering students, to balance a teeter-totter the smaller person needs to be further from the hinge. If two people are on one side then the teeter totter is balanced by two similar people an equal distance from the hinge on the other side. Two people can balance one similar person by scooting twice as close to the hinge. These proportionalities generalize to this: the tendency of a force to cause rotation is proportional to the size of the force and to its distance from the hinge (for forces perpendicular to the teeter totter). If someone standingnearbyadds a forcethat is directedtowards the hingeitcauses no tendency to rotate. Because any force can be decomposed into a sum of forces, one perpendicular to the teeter totter and the other towards the hinge, and because we assume that the affect of the sum of these forces is the sum of the affects of each separately, and because the force towards the hinge has no tendency to rotate, we have deduced: The moment of a force about a hinge is the product of its distance from the hinge and the component of the force perpendicular to the line from the hinge to the force. Here then is the formula for 2D moment about C or moment with respect to C. 1  1  The ‘/’ in the subscript of  M reads as ‘relative to’ or ‘about’. For simplicity we often leave the / out and just write  M C . M /C =|  r | (|  F |sinθ) = (|  r |sinθ) |  F |.(2.3) Here, θ is the angle between  r (the position of the point of force application relative to the hinge) and  F (see fig. 2.22). This formula for moment has all the teeter totter deduced properties. Moment is proportional to r, and to the part of  F that is perpendicular to  r . The re-grouping as (|  r |sinθ) shows that a force  F has the same effect if it is applied at a new location that is displaced in the direction of  F . That is, . two vectors 23 2.2 The dot product of two vectors The dot product is used to project a vector in a given direction, to reduce a vector to components, to reduce. reduce vector equations to scalar equations, to define work and power, and to help solve geometry problems. The dot product of two vectors  A and  B is