Transcription of Dot product and vector projections (Sect. 12.3) There are ...
1 Dot product and vector projections (Sect. )ITwo definitions for the dot definition of dot product and orthogonal of the dot product in vector and vector projection are two main ways to introduce the dot productGeometricaldefinition Properties Expression Properties Definition choose the first way, the textbook chooses the second product and vector projections (Sect. )ITwo definitions for the dot definition of dot product and orthogonal of the dot product in vector and vector projection dot product of two vectors is a scalarDefinitionLetv,wbe vectors inRn, withn= 2,3, having length|v|and|w|with angle in between , where 0 . Thedot productofvandw, denoted byv w, is given byv w=|v||w|cos( ).
2 OVWI nitial points dot product of two vectors is a scalarExampleComputev wknowing thatv,w R3, with|v|= 2,w= 1,2,3 and the angle in between is = :We first compute|w|, that is,|w|2= 12+ 22+ 32= 14 |w|= now use the definition of dot product :v w=|v||w|cos( ) = (2) 14 22 v w= 2 angle between two vectors is a usually not know will be convenient to obtain a formula for the dot productinvolving the vector product and vector projections (Sect. )ITwo definitions for the dot definition of dot product and orthogonal of the dot product in vector and vector projection vectors have zero dot vectors areperpendicular, also calledorthogonal, iff the anglein between is = = / 2 VWTheoremThe non-zero vectorsvandware perpendicular iffv w= =v w=|v||w|cos( )|v|6= 0,|w|6= 0} {cos( ) = 006 6 = dot product ofi,jandkis simple to computeExampleCompute all dot products involving the vectorsi,j, :Recall:i= 1,0,0 ,j= 0,1,0 ,k= 0,0,1.
3 Yijkxzi i= 1,j j= 1,k k= 1,i j= 0,j i= 0,k i= 0,i k= 0,j k= 0,k j= product and vector projections (Sect. )ITwo definitions for the dot definition of dot product and orthogonal of the dot product in vector and vector projection dot product and orthogonal dot product is closely related to orthogonal projections of onevector onto the other. Recall:v w=|v||w|cos( ).V W = |V| cos(O) OVW|W||V| OVWV W = |W| cos(O)Dot product and vector projections (Sect. )ITwo definitions for the dot definition of dot product and orthogonal of the dot product in vector and vector projection of the dot (a)v w=w v,(symmetric);(b)v (aw) =a(v w),(linear);(c)u (v+w) =u v+u w,(linear);(d)v v=|v|2>0, andv v= 0 v=0,(positive);(e)0 v= (a), (b), (d), (e) are simple to obtain from thedefinition of dot productv w=|v||w|cos( ).
4 For example, the proof of (b) fora>0:v (aw) =|v||aw|cos( ) =a|v||w|cos( ) =a(v w).Properties of the dot product .(c),u (v+w) =u v+u w, is non-trivial. The proof is:VWw|V+W| cos(0)V+WU0V00W|W| cos(0 ) |V| cos(0 ) VW|v+w|cos( ) =u (v+w)|u|,|w|cos( w) =u w|u|,|v|cos( v) =u v|u|, u (v+w) =u v+u wDot product and vector projections (Sect. )ITwo definitions for the dot definition of dot product and orthogonal of the dot product in vector and vector projection dot product in vector components (CaseR2)TheoremIfv= vx,vy andw= wx,wy , thenv wis given byv w=vxwx+ :v=vxi+vyjandw=wxi+wyj. The linear property ofthe dot product impliesv w= (vxi+vyj) (wxi+wyj)=vxwxi i+vxwyi j+vywxj i+vywyj :i i=j j= 1 andi j=j i= 0.
5 We conclude thatv w=vxwx+ dot product in vector components (CaseR3)TheoremIfv= vx,vy,vz andw= wx,wy,wz , thenv wis given byv w=vxwx+vywy+ proof is similar to the case dot product is simple to compute from the vectorcomponent formulav w=vxwx+vywy+ geometrical meaning of the dot product is simple to seefrom the formulav w=|v||w|cos( ).ExampleFind the cosine of the angle betweenv= 1,2 andw= 2,1 Solution:v w=|v||w|cos( ) cos( ) =v w|v||w|.Furthermore,v w= (1)(2) + (2)(1)|v|= 12+ 22= 5,|w|= 22+ 12= 5, cos( ) = product and vector projections (Sect. )ITwo definitions for the dot definition of dot product and orthogonal of the dot product in vector and vector projection and vector projection scalar projection of vectorvalong the vectorwis the numberpw(v)given bypw(v) =v w|w|.
6 The vector projection of vectorvalong the vectorwis the vectorpw(v)given bypw(v) =(v w|w|)w|w|.P (V) = V W = |V| cos(O) OVWW|W|P (V) = V W W OVWW|W||W|ExampleFind the scalar projection ofb= 4,1 ontoa= 1,2 .Solution:The scalar projection ofbontoais the numberpa(b) =|b|cos( ) =b a|a|=( 4)(1) + (1)(2) 12+ therefore obtainpa(b) = 2 (b)abExampleFind the vector projection ofb= 4,1 ontoa= 1,2 .Solution:The vector projection ofbontoais the vectorpa(b) =(b a|a|)a|a|=( 2 5)1 5 1,2 ,we therefore obtainpa(b) = 25,45 .ap (b)abExampleFind the vector projection ofa= 1,2 ontob= 4,1 .Solution:The vector projection ofaontobis the vectorpb(a) =(a b|b|)b|b|=( 2 17)1 17 4,1 ,we therefore obtainpa(b) = 817, 217.
7 Bbap (a) cross product and determinants (Sect. )ITwo definitions for the cross definition of cross of the cross product in vector to compute cross product and are two main ways to introduce the cross productGeometricaldefinition Properties Expression Properties Definition choose the first way, like the product and determinants (Sect. )ITwo definitions for the cross definition of cross of the cross product in vector to compute cross product and cross product of two vectors is another vectorDefinitionLetv,wbe vectors inR3having length|v|and|w|with angle inbetween , where 0 . Thecross productofvandw,denoted asv w, is a vector perpendicular to bothvandw,pointing in the direction given by the right hand rule, with norm|v w|=|v||w|sin( ).
8 VWW x VV x WOCross product vectors are perpendicular to the original vectors.|v w|is the area of a parallelogramTheorem|v w|is the area of the parallelogram formed by |V| sin(O)OThe areaAof the parallelogram formed byvandwis given byA=|w|(|v|sin( ))=|v w|. cross product and determinants (Sect. )ITwo definitions for the cross definition of cross of the cross product in vector to compute cross product and vectors have zero cross vectors areparalleliff the angle in between them is = non-zero vectorsvandware parallel iffv w= : Vectorv w=0iff its length|v w|= 0, then|v||w|sin( ) = 0|v|6= 0,|w|6= 0} {sin( ) = 006 6 = 0,or = .Recall:|v w|is the area of a parallelogramExampleThe closer the vectorsv,ware to be parallel, the smaller is thearea of the parallelogram they form, hence the shorter is their crossproduct vectorv OV x WV12 VWV x WOCE xampleCompute all cross products involving the vectorsi,j, :Recall:i= 1,0,0 ,j= 0,1,0 ,k= 0,0,1.
9 Yijkxzi j=k,j k=i,k i=j,i i=0,j j=0,k k=0,i k= j,j i= k,k j= product and determinants (Sect. )ITwo definitions for the cross definition of cross of the cross product in vector to compute cross product and properties of the cross productTheorem(a)v w= (w v),(Skew-symmetric);(b)v v=0;(c)(av) w=v (aw) =a(v w),(linear);(d)u (v+w) =u v+u w,(linear);(e)u (v w)6= (u v) w,(not associative). (a) results from the right hand rule. Part (b) comes from part(a). Parts (b) and (c) are proven in a similar ways as the linearproperty of the dot product . Part (d) is proven by giving cross product is not associative, that is,u (v w)6= (u v) thati (i k) = kand (i i) k= :i (i k) =i ( j) = (i j) = k i (i k) = k,(i i) k=0 j=0 (i i) k= : The cross product of two vectors vanishes when thevectors are parallelCross product and determinants (Sect.)
10 ITwo definitions for the cross definition of cross of the cross product in vector to compute cross product and cross product vector in vector the vector components ofvandwin a Cartesian coordinatesystem arev= v1,v2,v3 andw= w1,w2,w3 , then holdsv w= (v2w3 v3w2),(v3w1 v1w3),(v1w2 v2w1) .For the proof, recall the non-zero cross productsi j=k,j k=i,k i=j,and their skew-symmetric products, while all the other crossproducts vanish, and then use the properties of the cross product in vector :v=v1i+v2j+v3k,w=w1i+w2j+ , it holdsv w= (v1i+v2j+v3k) (w1i+w2j+w3k).Use the linearity property. The only non-zero terms are those withproductsi j=kandj k=iandk i=j. The result isv w= (v2w3 v3w2)i+ (v3w1 v1w3)j+ (v1w2 v2w1) product in vector wforv= 1,2,0 andw= 3,2,1 ,Solution:We use the formulav w= (v2w3 v3w2),(v3w1 v1w3),(v1w2 v2w1) = [(2)(1) (0)(2)],[(0)(3) (1)(1)],[(1)(2) (2)(3)] = (2 0),( 1),(2 6) v w= 2, 1, 4.
