How to draw vectors linear Algebra

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LINEAR ALGEBRA W W L CHEN c W W L Chen, 1982, 2008. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990. It is available free to all individuals, on the understanding that it is not to be used for nancial gain, and may be downloaded and/or photocopied, with or without permission from the author. However, this document may not be kept on any information storage and retrieval system without permission from the author, unless such system is not accessible to any individuals other than its owners. Chapter 4 VECTORS 4.1. Introduction A vector is an object which has magnitude and direction. Example 4.1.1. We may be travelling north-east at 50 kph. In this case, the direction of the velocity is north-east and the magnitude of the velocity is 50 kph. We can describe our velocity in kph as   50 50 p ;p ; 2 2 where the rst coordinate describes the speed with which we are moving east and the second coordinate describes the speed with which we are moving north. Example 4.1.2. An object in the sky may be 100 metres away in the south-east direction 45 degrees upwards. In this case, the direction of its position is south-eastand 45 degrees upwards and the magnitude of its distance is 100 metres. We can describe the position of the object in metres as   100 50;50;p ; 2 where the rst coordinate describes the distance east, the second coordinate describes the distance north and the third coordinate describes the distance up. The purpose of this chapter is to study some relationship between algebra and geometry. We shall rst study some algebra which is motivated by geometric considerations. We then use the algebra later to better understand some problems in geometry. Chapter 4 : Vectors page 1 of 24Linear Algebra c WWL Chen, 1982, 2006 c Linear Algebra c WWL Chen, 1982, 2006 Linear Algebra W W L Chen, 1982, 2008 2 2 2 4.2. Vectors in R 4.2. Vectors in R 4.2. Vectors in R 2 22 Avector on the planeR can be described as an ordered pair u=(u ,u ), where u ,u ∈R. 1 2 1 2 AAv vector ector on on the the pl plane aneRR can can bbee des describ cribed ed as as an an ordered ordered pai pair r uu ==( (uu ;,uu ), ), where whereuu ;,uu 2∈RR.. 1 2 1 2 1 2 1 2 2 22 Definition. Two vectors u=(u ,u ) and v=(v ,v ) inR are said to be equal, denoted by u=v, if Definition. Two vectors u=(u1,u2) and v=(v1,v2) inR are said to be equal, denoted by u=v, if Definition. Two vectors u = (u ;u ) and v = (v ;v ) inR are said to be equal, denoted by u =v, if 11 22 11 22 u =v and u =v . u1 =v1 and u2 =v2. u =v and u =v . 11 11 22 22 2 2 2 Definition. Forany two vectors u=(u ,u ) and v=(v ,v ) inR ,we define their sum to be Definition. Definition. F Fo orra anny y ttwo wo vvec ectors tors uu ==( (uu1;,uu2)) and andvv ==( (vv1;,vv2)) in inRR ,,w wee de ne define their their sum sum to to bbee 11 22 11 22 u+v=(u ,u )+(v ,v )=(u +v ,u +v ). u +v = (u1;u2) + (v1;v2) = (u1 +v1;u2 +v2): u+v=(u ,u )+(v ,v )=(u +v ,u +v ). 11 22 11 22 11 11 22 22 −−→ −−→ −−→ −−→ Geometrically, if we represent the two vectors u and v by AB and BC respectively, then the sum Geometrically, if we represent the two vectors u and v by AB and BC respectively, then the sum Geometrically, if we represent the two vectors u and v by AB and BC respectively, then the sum −→ −→ u+v is represented by AC as shown in the diagram below: u +v is represented by AC as shown in the diagram below: u+v is represented by AC as shown in the diagram below: C C u+v u+v v v AB u AB u The next diagram demonstrates geometrically that u+v =v+u: The The next next diagram diagram demonstrates demonstrates geome geometrically trically that thatuu + +vv = =vv + +uu:: u u DC DC u+v u+v v v v v AB u AB u PROPOSITION 4A. (VECTOR ADDITION) PROPOSITION 4A. (VECTOR ADDITION) PROPOSITION 4A. (VECTOR ADDITION) 2 2 2 2 (a) For every u,v∈R2,we have u+v∈R2. (a) For every u,v∈R ,we have u+v∈R . (a) For every u;v2R , we have u +v2R . 2 2 (b) For every u,v,w∈R2,we have u+(v+w)=(u+v)+w. (b) For every u,v,w∈R ,we have u+(v+w)=(u+v)+w. (b) For every u;v;w2R , we have u + (v +w) = (u +v) +w. 2 2 2 2 (c) For every u∈R2,we have u+0=u, where 0=(0,0)∈R2. (c) For every u∈R ,we have u+0=u, where 0=(0,0)∈R . (c) For every u2R , we have u +0 =u, where 0 = (0; 0)2R . 2 2 2 2 (d) For every u∈R2, there exists v∈R2 such that u+v =0. (d) For every u∈R , there exists v∈R such that u+v =0. (d) For every u2R , there exists v2R such that u +v =0. 2 2 (e) For every u,v∈R2,we have u+v =v+u. (e) For every u,v∈R ,we have u+v =v+u. (e) For every u;v2R , we have u +v =v +u. Proof. Write u=(u ,u ), v=(v ,v ) and w=(w ,w ), where u ,u ,v ,v ,w ,w ∈R. To check 1 2 1 2 1 2 1 2 1 2 1 2 Proof. Write u=(u ,u ), v=(v ,v ) and w=(w ,w ), where u ,u ,v ,v ,w ,w ∈R. To check Proof. Write u = (u1;u2), v = (v1;v2) and w = (w1;w2), where u1;u2;v1;v2;w1;w22 R. To check 1 2 1 2 1 2 1 2 1 2 1 2 part (a), simply note that u +v ,u +v ∈R.Tocheck part (b), note that 1 1 2 2 part (a), simply note that u +v ,u +v ∈R.Tocheck part (b), note that part (a), simply note that u1 +v1;u2 +v22R. To check part (b), note that 1 1 2 2 u+(v+w)=(u ,u )+(v +w ,v +w )=(u +(v +w ),u +(v +w )) u+(v+w)=(u1,u2)+(v1+w1,v2+w2)=(u1+(v1+w1),u2+(v2+w2)) u + (v +w) = (u ;u ) + (v +w ;v +w ) = (u + (v +w );u + (v +w )) 1 2 1 1 2 2 1 1 1 2 2 2 1 2 1 1 2 2 1 1 1 2 2 2 =((u +v )+w ,(u +v )+w )=(u +v ,u +v )+(w ,w ) =((u1+v1)+w1,(u2+v2)+w2)=(u1+v1,u2+v2)+(w1,w2) = ((u +v ) +w ; (u +v ) +w ) = (u +v ;u +v ) + (w ;w ) 1 1 1 2 2 2 1 1 2 2 1 2 1 1 1 2 2 2 1 1 2 2 1 2 =(u+v)+w. =(u+v)+w. = (u +v) +w: Part (c) is trivial. Next, if v=(−u ,−u ), then u+v = 0, giving part (d). To check part (e), note 1 2 Part (c) is trivial. Next, if v=(−u ,−u ), then u+v = 0, giving part (d). To check part (e), note Part (c) is trivial. Next, if v = (u1;u2), then u +v = 0, giving part (d). To check part (e), note 1 2 that u+v=(u +v ,u +v )=(v +u ,v +u )=v+u. 1 1 2 2 1 1 2 2 that u+v=(u +v ,u +v )=(v +u ,v +u )=v+u. that u +v = (u1 +v1;u2 +v2) = (v1 +u1;v2 +u2) =v +u. 1 1 2 2 1 1 2 2 Chapter 4 : Vectors page 2 of 24 Chapter Chapter 44 :: V Veectors ctors page page 22 of of 24 24Linear Algebra c WWL Chen, 1982, 2006 c Linear Algebra W W L Chen, 1982, 2008 2 2 Definition. Forany vectoru=(u ,u ) inR and any scalar c∈R,we define the scalar multiple to be 1 2 Definition. For any vector u = (u ;u ) inR and any scalarc2R, we de ne the scalar multiple to be 1 2 ccuu = =cc((uu ;,uu ))=( = (cu cu ;,cu cu )):. 11 22 11 22 Example Example 4.2.1. 4.2.1. Supp Suppose ose that that uu = = (2 (2;, 1). 1). Then Then−22uu ==( (−44;, 2). 2). Geometrically Geometrically,, ifif w wee represen representt the the ttw woo −→ −− → vvectors ectorsuu and and−22uu bby y O OA A and andO OB B resp respectiv ectively ely,, then then w wee ha havvee the the diagram diagram bbelo elow: w: A u O −2u B PROPOSITION 4B. (SCALAR MULTIPLICATION) PROPOSITION 4B. (SCALAR MULTIPLICATION) 2 2 (a) For every c∈R and u∈R2,we have cu∈R2. (a) For every c2R and u2R , we have cu2R . 2 (b) For every c∈R and u,v∈R2,we have c(u+v)=cu+cv. (b) For every c2R and u;v2R , we have c(u +v) =cu +cv. 2 (c) For every a,b∈R and u∈R2,we have (a+b)u=au+bu. (c) For every a;b2R and u2R , we have (a +b)u =au +bu. 2 (d) For every a,b∈R and u∈R2,we have (ab)u=a(bu). (d) For every a;b2R and u2R , we have (ab)u =a(bu). 2 2 (e) For every u∈R ,we have 1u=u. (e) For every u2R , we have 1u =u. Proof. Write u=(u ,u ) and v=(v ,v ), where u ,u ,v ,v ∈ R.Tocheck part (a), simply note 1 2 1 2 1 2 1 2 Proof. Write u = (u ;u ) and v = (v ;v ), where u ;u ;v ;v 2 R. To check part (a), simply note 1 2 1 2 1 2 1 2 that cu ,cu ∈R.Tocheck part (b), note that 1 2 that cu ;cu 2R. To check part (b), note that 1 2 c(u+v)=c(u +v ,u +v )=(c(u +v ),c(u +v )) 1 1 2 2 1 1 2 2 c(u +v) =c(u +v ;u +v ) = (c(u +v );c(u +v )) 1 1 2 2 1 1 2 2 =(cu +cv ,cu +cv )=(cu ,cu )+(cv ,cv )=cu+cv. 1 1 2 2 1 2 1 2 = (cu +cv ;cu +cv ) = (cu ;cu ) + (cv ;cv ) =cu +cv: 1 1 2 2 1 2 1 2 To check part (c), note that To check part (c), note that (a+b)u= ((a+b)u ,(a+b)u )=(au +bu ,au +bu ) 1 2 1 1 2 2 (a +b)u = ((a +b)u ; (a +b)u ) = (au +bu ;au +bu ) 1 2 1 1 2 2 =(au ,au )+(bu ,bu )=au+bu. 1 2 1 2 = (au ;au ) + (bu ;bu ) =au +bu: 1 2 1 2 To check part (d), note that To check part (d), note that (ab)u= ((ab)u ,(ab)u )=(a(bu ),a(bu ))=a(bu ,bu )=a(bu). 1 2 1 2 1 2 (ab)u = ((ab)u ; (ab)u ) = (a(bu );a(bu )) =a(bu ;bu ) =a(bu): 1 2 1 2 1 2 Finally, to check part (e), note that 1u=(1u ,1u )=(u ,u )=u. 1 2 1 2 Finally, to check part (e), note that 1u = (1u ; 1u ) = (u ;u ) =u. 1 2 1 2 2 Definition. Forany vector u=(u ,u ) in R,we define the norm of u to be the non-negative real 2 1 2 Definition. For any vector u = (u ;u ) in R , we de ne the norm of u to be the non-negative real 1 2 number number q 2 2 2 2 %u%= u +u . kuk = u +u : 1 2 1 2 Remarks. (1) The norm of a vector is simply its magnitude or length. The definition follows from the Remarks. (1) The norm of a vector is simply its magnitude or length. The de nition follows from the famous theorem of Pythagoras. famous theorem of Pythagoras. 2 2 (2) Suppose that P(u ,u ) and Q(v ,v ) are two points on the plane R.To calculate the distance 1 2 1 2 (2) Suppose that P (u ;u ) and Q(v ;v ) are two points on the plane R . To calculate the distance 1 2 1 2 d(P,Q)betweenthetwopoints, wecanfirstfindavectorfromP toQ. Thisisgivenby(v −u ,v −u ). 1 1 2 2 d(P;Q) between the two points, we can rst nd a vector fromP toQ. This is given by (v u ;v u ). 1 1 2 2 The distance d(P,Q)is then the norm of this vector, so that The distance d(P;Q) is then the norm of this vector, so that " p 2 2 d(P,Q)= (v −u )2+(v −u )2. 1 1 2 2 d(P;Q) = (v u ) + (v u ) : 1 1 2 2 Chapter 4 : Vectors page 3 of 24 Chapter 4 : Vectors page 3 of 24Linear Algebra c WWL Chen, 1982, 2006 c Linear Algebra W W L Chen, 1982, 2008 2 (3) It is not difficult to see that for any vectoru∈R2 and any scalar c∈R, we havecu = cu. (3) It is not dicult to see that for any vector u2R and any scalar c2R, we havekcuk =jcjkuk. 2 Definition. Any vectoru∈R2 satisfyingu=1 is called a unit vector. Definition. Any vector u2R satisfyingkuk = 1 is called a unit vector. Example 4.2.2. The vector (3,4) has norm 5. Example 4.2.2. The vector (3; 4) has norm 5. p 2 2 2 2 Example 4.2.3. The distance between the points (6,3) and (9,7) is (9−6) +(7−3) = 5. Example 4.2.3. The distance between the points (6; 3) and (9; 7) is (9 6) + (7 3) = 5. 2 2 Example 4.2.4. The vectors (1,0) and (0,−1) are unit vectors inR . Example 4.2.4. The vectors (1; 0) and (0;1) are unit vectors inR . √ √ p p Example 4.2.5. The unit vector in the direction of the vector (1,1) is (1/ 2,1/ 2). Example 4.2.5. The unit vector in the direction of the vector (1; 1) is (1= 2; 1= 2). 2 2 Example 4.2.6. In fact, all unit vectors inR are of the form (cosθ,sinθ), where θ∈R. Example 4.2.6. In fact, all unit vectors inR are of the form (cos; sin), where 2R. Quite often, we may want to find the angle between two vectors. The scalar product of the two vectors Quite often, we may want to nd the angle between two vectors. The scalar product of the two vectors then comes in handy. We shall define the scalar product in two ways, one in terms of the angle between then comes in handy. We shall de ne the scalar product in two ways, one in terms of the angle between the two vectors and the other not in terms of this angle, and show that the two definitions are in fact the two vectors and the other not in terms of this angle, and show that the two de nitions are in fact equivalent. equivalent. 2 2 Definition. Suppose thatu=(u ,u ) andv=(v ,v ) are vectors inR , and that θ∈ 0,π represents 1 2 1 2 Definition. Suppose thatu = (u ;u ) andv = (v ;v ) are vectors inR , and that2 0; represents 1 2 1 2 the angle between them. We define the scalar productu ·v ofu andv by the angle between them. We de ne the scalar product uv of u and v by " n uvcosθ ifu =& 0 andv&=0, kukkvk cos if u =6 0 and v6=0, (1) u ·v = uv = (1) 0 ifu =0 orv =0. 0 if u =0 or v =0. Alternatively, we write Alternatively, we write (2) u ·v = u v +u v . uv =u1v1 +u2v2: (2) 1 1 2 2 The definitions (1) and (2) are clearly equivalent ifu =0 or v =0.On the other hand, we have the The de nitions (1) and (2) are clearly equivalent if u =0 or v =0. On the other hand, we have the following result. following result. 2 2 PROPOSITION 4C. Suppose that u=(u ,u ) and v=(v ,v ) are non-zero vectors inR , and that PROPOSITION 4C. Suppose that u = (u1;u2) and v = (v1;v2) are non-zero vectors inR , and that 1 2 1 2 θ∈ 0,π represents the angle between them. Then 2 0; represents the angle between them. Then uvcosθ = u v +u v . kukkvk cos =u v +u v : 1 1 2 2 1 1 2 2 −→ −− → Proof. Geometrically, if we represent the two vectors u and v by OA and OB respectively, then the Proof. Geometrically, if we represent the two vectors u and v by OA and OB respectively, then the −−→ differencev−u is represented by AB as shown in the diagram below: di erence vu is represented by AB as shown in the diagram below: B v−u v A u θ O By the Law of cosines, we have By the Law of cosines, we have 2 2 2 2 2 2 AB =OA +OB −2OAOBcosθ; AB =OA +OB 2OAOB cos; Chapter 4 : Vectors page 4 of 24 Chapter 4 : Vectors page 4 of 24c Linear Algebra W W L Chen, 1982, 2008 in other words, we have 2 2 2 kvuk =kuk +kvk 2kukkvk cos; so that 1 2 2 2 kukkvk cos = (kuk +kvk kvuk ) 2 1 2 2 2 2 2 2 = (u +u +v +v (v u ) (v u ) ) 1 1 2 2 1 2 1 2 2 =u v +u v 1 1 2 2 as required. 2 Remarks. (1) We say that two non-zero vectors inR are orthogonal if the angle between them is =2. 2 It follows immediately from the de nition of the scalar product that two non-zero vectors u;v2R are orthogonal if and only if uv = 0. 2 (2) We can calculate the scalar product of any two non-zero vectors u;v2R by the formula (2) and then use the formula (1) to calculate the angle between u and v. p p Example 4.2.7. Suppose that u = ( 3; 1) and v = ( 3; 3). Then by the formula (2), we have uv = 3 + 3 = 6: Note now that p kuk = 2 and kvk = 2 3: It follows from the formula (1) that p uv 6 3 cos = = p = ; kukkvk 2 4 3 so that  ==6. p p Example 4.2.8. Suppose that u = ( 3; 1) and v = ( 3; 3). Then by the formula (2), we have uv = 0. It follows that u and v are orthogonal. 2 PROPOSITION 4D. (SCALAR PRODUCT) Suppose that u;v;w2R and c2R. Then (a) uv =vu; (b) u (v +w) = (uv) + (uw); (c) c(uv) = (cu)v =u (cv); (d) uu 0; and (e) uu = 0 if and only if u =0. Proof. Write u = (u ;u ), v = (v ;v ) and w = (w ;w ), where u ;u ;v ;v ;w ;w 2R. Part (a) is 1 2 1 2 1 2 1 2 1 2 1 2 trivial. To check part (b), note that u (v +w) =u (v +w ) +u (v +w ) = (u v +u v ) + (u w +u w ) =uv +uw: 1 1 1 2 2 2 1 1 2 2 1 1 2 2 2 2 Part (c) is rather simple. To check parts (d) and (e), note that uu =u +u  0, and that equality 1 2 holds precisely when u =u = 0. 1 2 Chapter 4 : Vectors page 5 of 24Linear Algebra c WWL Chen, 1982, 2006 c Linear Algebra c WWL Chen, 1982, 2006 Linear Algebra W W L Chen, 1982, 2008 Consider the diagram below: Consider the diagram below: Consider the diagram below: P P RA RA u (3) u (3) a Q a v Q v w w O O (3) −→ −−→ −→ −−→ Here we represent the two vectorsa andu by OA and OP respectively. If we project the vectoru on to Here we represent the two vectorsa andu by OA and OP respectively. If we project the vectoru on to Here we represent the two vectors a andu byOA andOP respectively. If we project the vector u on to −−→ −−→ the line OA, then the image of the projection is the vector w, represented by OQ.On the other hand, the line OA, then the image of the projection is the vector w, represented by OQ.On the other hand, the line OA, then the image of the projection is the vector w, represented by OQ. On the other hand, if we project the vectoru on to a line perpendicular to the line OA, then the image of the projection is if we project the vectoru on to a line perpendicular to the line OA, then the image of the projection is if we project the vector u on to a line perpendicular to the line OA, then the image of the projection is −−→ −−→ the vectorv, represented by OR. the vectorv, represented by OR. the vector v, represented by OR. Definition. In the notation of the diagram (3), the vectorw is called the orthogonal projection of the Definition. In the notation of the diagram (3), the vectorw is called the orthogonal projection of the Definition. In the notation of the diagram (3), the vector w is called the orthogonal projection of the vectoru on the vectora, and denoted byw = proj u. a vectoru on the vectora, and denoted byw = proj u. vector u on the vector a, and denoted by w = proj u. a a 2 2 PROPOSITION 4E. (ORTHOGONAL PROJECTION) Suppose that u,a∈R2. Then PROPOSITION 4E. (ORTHOGONAL PROJECTION) Suppose that u,a∈R . Then PROPOSITION 4E. (ORTHOGONAL PROJECTION) Suppose that u;a2R . Then u ·a u ·a ua proj u = a. projau = a. proj u = 2a: a a %a%2 2 %a% kak −−→ −−→ Remark. Note that the component ofu orthogonal toa, represented by OR in the diagram (3), is Remark. Note that the component ofu orthogonal toa, represented by OR in the diagram (3), is Remark. Note that the component of u orthogonal to a, represented by OR in the diagram (3), is u ·a u ·a ua u−proj u =u− a. u−projau =u− a. u proj u =u a: 2 a a 2 %a%2 %a% kak Proof of Proposition 4E. Note thatw =ka for some k∈R.It clearly suffices to prove that Proof of Proposition 4E. Note thatw =ka for some k∈R.It clearly suffices to prove that Proof of Proposition 4E. Note that w =ka for some k2R. It clearly suces to prove that u ·a u ·a ua k = . k = . k = : 2 2 %a%2 %a% kak It is easy to see that the vectors u−w and a are orthogonal. It follows that the scalar product It is easy to see that the vectors u−w and a are orthogonal. It follows that the scalar product It is easy to see that the vectors u w and a are orthogonal. It follows that the scalar product (u−w) ·a=0.In other words, (u−ka) ·a=0. Hence (u−w) ·a=0.In other words, (u−ka) ·a=0. Hence (uw)a = 0. In other words, (uka)a = 0. Hence u ·a u ·a uu·aa uu·aa k = = k = = k = = 2 2 a ·a %a%2 a ·a %a% aa kak as required. as as required. required. To end this section, we shall apply our knowledge gained so far to find a formula that gives the To end this section, we shall apply our knowledge gained so far to find a formula that gives the To end this section, we shall apply our knowledge gained so far to nd a formula that gives the perpendicular distance of a point (x ,y ) from a line ax+by+c=0. Consider the diagram below: 0 0 perpendicular distance of a point (x ,y ) from a line ax+by+c=0. Consider the diagram below: perpendicular distance of a point (x ;y ) from a line ax +by +c = 0. Consider the diagram below: 00 00 P P D D n=(a,b) n=(a,b) ax+by+c=0 ax+by+c=0 u u Q Q O O Chapter 4 : Vectors page 6 of 24 Chapter 4 : Vectors page 6 of 24 Chapter 4 : Vectors page 6 of 24c Linear Algebra W W L Chen, 1982, 2008 Suppose that (x ;y ) is any arbitrary point O on the lineax +by +c = 0. For any other point (x;y) on 1 1 the line ax +by +c = 0, the vector (xx ;yy ) is parallel to the line. On the other hand, 1 1 (a;b) (xx ;yy ) = (ax +by) (ax +by ) =c +c = 0; 1 1 1 1 so that the vector n = (a;b), in the direction OQ, is perpendicular to the line ax + by + c = 0. Suppose next that the point (x ;y ) is represented by the point P in the diagram. Then the vector 0 0 u = (x x ;y y ) is represented by OP , and OQ represents the orthogonal projection proj u of u 0 1 0 1 n on the vectorn. Clearly the perpendicular distanceD of the point (x ;y ) from the lineax +by +c = 0 0 0 satis es un j(x x ;y y ) (a;b)j jax +by ax byj jax +by +cj 0 1 0 1 0 0 1 1 0 0 D =kproj uk = n = p = p = p : n 2 2 2 2 2 2 2 knk a +b a +b a +b We have proved the following result. PROPOSITION 4F. The perpendicular distance D of a point (x ;y ) from a line ax +by +c = 0 is 0 0 given by jax +by +cj 0 0 D = p : 2 2 a +b Example 4.2.9. The perpendicular distanceD of the point (5; 7) from the line 2x 3y + 5 = 0 is given by j10 21 + 5j 6 D = p =p : 4 + 9 13 3 4.3. Vectors in R 3 In this section, we consider the same problems as in Section 4.2, but in 3-space R . Any reader who feels con dent may skip this section. 3 A vector on the planeR can be described as an ordered triple u = (u ;u ;u ), where u ;u ;u 2R. 1 2 3 1 2 3 3 Definition. Two vectors u = (u ;u ;u ) and v = (v ;v ;v ) in R are said to be equal, denoted by 1 2 3 1 2 3 u =v, if u =v , u =v and u =v . 1 1 2 2 3 3 3 Definition. For any two vectors u = (u ;u ;u ) and v = (v ;v ;v ) inR , we de ne their sum to be 1 2 3 1 2 3 u +v = (u ;u ;u ) + (v ;v ;v ) = (u +v ;u +v ;u +v ): 1 2 3 1 2 3 1 1 2 2 3 3 3 Definition. For any vector u = (u ;u ;u ) in R and any scalar c2R, we de ne the scalar multiple 1 2 3 to be cu =c(u ;u ;u ) = (cu ;cu ;cu ): 1 2 3 1 2 3 The following two results are the analogues of Propositions 4A and 4B. The proofs are essentially similar. Chapter 4 : Vectors page 7 of 24c Linear Algebra W W L Chen, 1982, 2008 PROPOSITION 4A'. (VECTOR ADDITION) 3 3 (a) For every u;v2R , we have u +v2R . 3 (b) For every u;v;w2R , we have u + (v +w) = (u +v) +w. 3 3 (c) For every u2R , we have u +0 =u, where 0 = (0; 0; 0)2R . 3 3 (d) For every u2R , there exists v2R such that u +v =0. 3 (e) For every u;v2R , we have u +v =v +u. PROPOSITION 4B'. (SCALAR MULTIPLICATION) 3 3 (a) For every c2R and u2R , we have cu2R . 3 (b) For every c2R and u;v2R , we have c(u +v) =cu +cv. 3 (c) For every a;b2R and u2R , we have (a +b)u =au +bu. 3 (d) For every a;b2R and u2R , we have (ab)u =a(bu). 3 (e) For every u2R , we have 1u =u. 3 Definition. For any vector u = (u ;u ;u ) inR , we de ne the norm of u to be the non-negative real 1 2 3 number q 2 2 2 kuk = u +u +u : 1 2 3 3 Remarks. (1) Suppose that P (u ;u ;u ) and Q(v ;v ;v ) are two points in R . To calculate the 1 2 3 1 2 3 distance d(P;Q) between the two points, we can rst nd a vector from P to Q. This is given by (v u ;v u ;v u ). The distance d(P;Q) is then the norm of this vector, so that 1 1 2 2 3 3 p 2 2 2 d(P;Q) = (v u ) + (v u ) + (v u ) : 1 1 2 2 3 3 3 (2) It is not dicult to see that for any vector u2R and any scalar c2R, we have kcuk =jcjkuk: 3 Definition. Any vector u2R satisfyingkuk = 1 is called a unit vector. Example 4.3.1. The vector (3; 4; 12) has norm 13. Example 4.3.2. The distance between the points (6; 3; 12) and (9; 7; 0) is 13. 3 Example 4.3.3. The vectors (1; 0; 0) and (0;1; 0) are unit vectors inR . p p Example 4.3.4. The unit vector in the direction of the vector (1; 0; 1) is (1= 2; 0; 1= 2). 3 The theory of scalar products can be extended toR is the natural way. 3 Definition. Suppose that u = (u ;u ;u ) and v = (v ;v ;v ) are vectors in R , and that 2 0; 1 2 3 1 2 3 represents the angle between them. We de ne the scalar product uv of u and v by n kukkvk cos if u6=0 and v =6 0, uv = (4) 0 if u =0 or v =0. Alternatively, we write uv =u v +u v +u v : (5) 1 1 2 2 3 3 The de nitions (4) and (5) are clearly equivalent if u =0 or v =0. On the other hand, we have the following analogue of Proposition 4C. The proof is similar. Chapter 4 : Vectors page 8 of 24c Linear Algebra WWL Chen, 1982, 2006 c Linear Algebra W W L Chen, 1982, 2008 3 3 PROPOSITION 4C’. Suppose that u=(u ,u ,u ) and v=(v ,v ,v ) are non-zero vectors in R , 1 2 3 1 2 3 PROPOSITION 4C'. Suppose that u = (u ;u ;u ) and v = (v ;v ;v ) are non-zero vectors in R , 1 2 3 1 2 3 and that θ∈ 0,π represents the angle between them. Then and that 2 0; represents the angle between them. Then uvcosθ =u v +u v +u v . 1 1 2 2 3 3 kukkvk cos =u v +u v +u v : 1 1 2 2 3 3 3 3 Remarks. (1) We say that two non-zero vectors inR are orthogonal if the angle between them is π/2. Remarks. (1) We say that two non-zero vectors inR are orthogonal if the angle between them is =2. 3 3 It follows immediately from the definition of the scalar productthat two non-zero vectors u,v∈R are It follows immediately from the de nition of the scalar productthat two non-zero vectors u;v2R are orthogonal if and only ifu ·v = 0. orthogonal if and only if uv = 0. 3 3 (2) We can calculate the scalar product of any two non-zero vectorsu,v∈R by the formula (5) and (2) We can calculate the scalar product of any two non-zero vectors u;v2R by the formula (5) and then use the formula (4) to calculate the angle betweenu andv. then use the formula (4) to calculate the angle between u and v. √ p Example 4.3.5. Suppose that u = (2,0,0) and v = (1,1, 2). Then by the formula (5), we have Example 4.3.5. Suppose that u = (2; 0; 0) and v = (1; 1; 2). Then by the formula (5), we have u ·v=2. Note now thatu=2 andv=2.It follows from the formula (4) that uv = 2. Note now thatkuk = 2 andkvk = 2. It follows from the formula (4) that u ·v 2 1 uv 2 1 cosθ = = = , cos = = = ; uv 4 2 kukkvk 4 2 so that θ = π/3. so that  ==3. Example Example 4.3.6. 4.3.6. Supp Suppose ose that that u u = = (2 (2,;3 3,;5) 5) and and vv = = (1 (1,;1 1,;−1). 1). Then Then b by y the the form formula ula (5), (5), w wee ha havvee u u ·vv=0 = 0..I Itt follo follows ws that thatu u and andvv are are orthogonal. orthogonal. The following result is the analogue of Proposition 4D. The proof is similar. The following result is the analogue of Proposition 4D. The proof is similar. 3 3 PR PROPOSITION OPOSITION 4D’. 4D'. (SCALAR (SCALAR PR PRODUCT) ODUCT) Supp Suppose ose that that u u,;vv,;w w∈2RR and and cc∈2RR.. Then Then (a) (a) u u ·vv = =vv ·u u;; (b) (b) u u ·( (vv+ +w w)=( ) = (u u ·vv)+( ) + (u u ·w w) );; (c) (c) cc( (u u ·vv)=( ) = (ccu u) ) ·vv = =u u ·( (ccvv) );; (d) u ·u≥ 0; and (d) uu 0; and (e) u ·u=0 if and only if u =0. (e) uu = 0 if and only if u =0. 3 3 Supp Suppose ose no now w that thataa and andu u are are t tw wo o v vectors ectors in inRR .. Then Then since since t tw wo o v vectors ectors are are alw alwaays ys coplan coplanar, ar, w wee can can dra draw w the the follo following wing diagram diagram whic which h represen represents ts the the plane plane they they lie lie on: on: P RA u (6) a Q v w O (6) Note that this diagram is essentially the same as the diagram (3), the only difference being that while Note that this diagram is essentially the same as the diagram (3), the only di erence being that while 2 3 the diagram (3) shows the whole ofR2, the diagram (6) only shows part ofR3.Asbefore, we represent the diagram (3) shows the whole ofR , the diagram (6) only shows part ofR . As before, we represent −→ −−→ the two vectorsa andu by OA and OP respectively. If we project the vectoru on to the line OA, then the two vectors a and u byOA andOP respectively. If we project the vector u on to the lineOA, then −−→ the image of the projection is the vector w, represented by OQ.On the other hand, if we project the the image of the projection is the vector w, represented by OQ. On the other hand, if we project the vector u on to a line perpendicular to the line OA, then the image of the projection is the vector v, vector u on to a line perpendicular to the line OA, then the image of the projection is the vector v, −−→ represented by OR. represented by OR. Definition. In the notation of the diagram (6), the vectorw is called the orthogonal projection of the Definition. In the notation of the diagram (6), the vector w is called the orthogonal projection of the vectoru on the vectora, and denoted byw = proj u. a vector u on the vector a, and denoted by w = proj u. a Chapter 4 : Vectors page 9 of 24 Chapter 4 : Vectors page 9 of 24c Linear Algebra W W L Chen, 1982, 2008 The following result is the analogue of Proposition 4E. The proof is similar. 3 PROPOSITION 4E'. (ORTHOGONAL PROJECTION) Suppose that u;a2R . Then ua proj u = a: a 2 kak Remark. Note that the component of u orthogonal to a, represented by OR in the diagram (6), is ua u proj u =u a: a 2 kak 4.4. Vector Products 3 In this section, we shall discuss a product of vectors unique toR . The idea of vector products has wide applications in geometry, physics and engineering, and is motivated by the wish to nd a vector that is perpendicular to two given vectors. We shall use the right hand rule. In other words, if we hold the thumb on the right hand upwards and close the remaining four ngers, then the ngers point from the x-direction towards they-direction, while the thumb points towards the z-direction. Alternatively, if we imagine Columbus had never lived and that the earth were at, then taking the x-direction as east and the y-direction as north, then the z-direction is upwards 3 We shall frequently use the three vectors i = (1; 0; 0), j = (0; 1; 0) and k = (0; 0; 1) inR . 3 Definition. Suppose that u = (u ;u ;u ) and v = (v ;v ;v ) are two vectors inR . Then the vector 1 2 3 1 2 3 product uv is de ned by the determinant 0 1 i j k A uv = det u u u : 1 2 3 v v v 1 2 3 Remarks. (1) Note that ij =(ji) =k; jk =(kj) =i; ki =(ik) =j: (2) Using cofactor expansion by row 1, we have       u u u u u u 2 3 1 3 1 2 uv = det i det j + det k v v v v v v 2 3 1 3 1 2        u u u u u u 2 3 1 3 1 2 = det ; det ; det v v v v v v 2 3 1 3 1 2 = (u v u v ;u v u v ;u v u v ): 2 3 3 2 3 1 1 3 1 2 2 1 We shall rst of all show that the vector product uv is orthogonal to both u and v. Chapter 4 : Vectors page 10 of 24c Linear Algebra W W L Chen, 1982, 2008 3 PROPOSITION 4G. Suppose that u = (u ;u ;u ) and v = (v ;v ;v ) are two vectors in R . Then 1 2 3 1 2 3 (a) u (uv) = 0; and (b) v (uv) = 0. Proof. Note rst of all that        u u u u u u 2 3 1 3 1 2 u (uv) = (u ;u ;u ) det ; det ; det 1 2 3 v v v v v v 2 3 1 3 1 2 0 1       u u u 1 2 3 u u u u u u 2 3 1 3 1 2 A =u det u det +u det = det u u u ; 1 2 3 1 2 3 v v v v v v 2 3 1 3 1 2 v v v 1 2 3 in view of cofactor expansion by row 1. On the other hand, clearly 0 1 u u u 1 2 3 A det u u u = 0: 1 2 3 v v v 1 2 3 This proves part (a). The proof of part (b) is similar. Example 4.4.1. Suppose that u = (1;1; 2) and v = (3; 0; 2). Then 0 1        i j k 1 2 1 2 1 1 A uv = det 1 1 2 = det ; det ; det = (2; 4; 3): 0 2 3 2 3 0 3 0 2 Note that (1;1; 2) (2; 4; 3) = 0 and (3; 0; 2) (2; 4; 3) = 0. 3 PROPOSITION 4H. (VECTOR PRODUCT) Suppose that u;v;w2R and c2R. Then (a) uv =(vu); (b) u (v +w) = (uv) + (uw); (c) (u +v)w = (uw) + (vw); (d) c(uv) = (cu)v =u (cv); (e) u0 =0; and (f) uu =0. Proof. Write u = (u ;u ;u ), v = (v ;v ;v ) and w = (w ;w ;w ). To check part (a), note that 1 2 3 1 2 3 1 2 3 0 1 0 1 i j k i j k A A det u u u = det v v v : 1 2 3 1 2 3 v v v u u u 1 2 3 1 2 3 To check part (b), note that 0 1 0 1 0 1 i j k i j k i j k A A A det u u u = det u u u + det u u u : 1 2 3 1 2 3 1 2 3 v +w v +w v +w v v v w w w 1 1 2 2 3 3 1 2 3 1 2 3 Part (c) is similar. To check part (d), note that 0 1 0 1 0 1 i j k i j k i j k A A A c det u u u = det cu cu cu = det u u u : 1 2 3 1 2 3 1 2 3 v v v v v v cv cv cv 1 2 3 1 2 3 1 2 3 To check parts (e) and (f), note that 0 1 0 1 i j k i j k A A u0 = det u u u = 0 and uu = det u u u = 0 1 2 3 1 2 3 0 0 0 u u u 1 2 3 as required. Chapter 4 : Vectors page 11 of 24c Linear Algebra W W L Chen, 1982, 2008 c Linear Algebra WWL Chen, 1982, 2006 Next, we shall discuss an application of vector product to the evaluaton of the area of a parallelogram. Next, we shall discuss an application of vector product to the evaluaton of the area of a parallelogram. To do this, we shall rst establish the following result. To do this, we shall first establish the following result. 3 PROPOSITION 4J. Suppose that u = (u ;u ;u ) and v = (v ;v ;v ) are non-zero vectors in R , 1 2 3 1 2 3 3 PROPOSITION 4J. Suppose that u=(u ,u ,u ) and v=(v ,v ,v ) are non-zero vectors in R , 1 2 3 1 2 3 and that 2 0; represents the angle between them. Then and that θ∈ 0,π represents the angle between them. Then 2 2 2 2 (a) kuvk =kukkvk (uv) ; and 2 2 2 2 (a) u×v =u v −(u ·v) ; and (b) kuvk =kukkvk sin. (b) u×v =uvsinθ. Proof. Note that Proof. Note that 2 2 2 2 kuvk = (u v u v ) + (u v u v ) + (u v u v ) (7) 2 2 3 3 2 2 3 1 1 3 2 1 2 2 1 2 (7) u×v =(u v −u v ) +(u v −u v ) +(u v −u v ) 2 3 3 2 3 1 1 3 1 2 2 1 and and 2 2 2 2 2 2 2 2 2 2 kukkvk (uv) = (u +u +u )(v +v +v ) (u v +u v +u v ) : (8) 1 1 2 2 3 3 1 2 3 1 2 3 2 2 2 2 2 2 2 2 2 2 (8) u v −(u ·v) =(u +u +u )(v +v +v )−(u v +u v +u v ) . 1 1 2 2 3 3 1 2 3 1 2 3 Part (a) follows on expanding the right hand sides of (7) and (8) and checking that they are equal. To Part (a) follows on expanding the right hand sides of (7) and (8) and checking that they are equal. To prove part (b), recall that prove part (b), recall that uv =kukkvk cos: u ·v =uvcosθ. Combining with part (a), we obtain Combining with part (a), we obtain 2 2 2 2 2 2 2 2 2 kuvk =kukkvk kukkvk cos  =kukkvk sin : 2 2 2 2 2 2 2 2 2 u×v =u v −u v cos θ =u v sin θ. Part (b) follows. Part (b) follows. Consider now a parallelogram with vertices O;A;B;C. Suppose that u and v are represented by OA −→ Consider now a parallelogram with vertices O,A,B,C. Suppose thatu andv are represented by OA and OC respectively. If we imagine the side OA to represent the base of the parallelogram, so that −−→ and OC respectively. If we imagine the side OA to represent the base of the parallelogram, so that the base has lengthkuk, then the height of the the parallelogram is given bykvk sin, as shown in the the base has lengthu, then the height of the the parallelogram is given byvsinθ, as shown in the diagram below: diagram below: CB v vsinθ θ OA u u It follows from Proposition 4J that the area of the parallelogram is given byu×v. We have proved It follows from Proposition 4J that the area of the parallelogram is given bykuvk. We have proved the following result. the following result. 3 PROPOSITION4K. Suppose thatu,v∈R . Then the parallelogram withu andv as two of its sides 3 PROPOSITION 4K. Suppose that u;v2R . Then the parallelogram with u and v as two of its sides has areau×v. has areakuvk. 3 We conclude this section by making a remark on the vector product u×v of two vectors in R . 3 We conclude this section by making a remark on the vector product uv of two vectors in R . Recall that the vector product is perpendicular to bothu andv.Furthermore, it can be shown that the Recall that the vector product is perpendicular to both u andv. Furthermore, it can be shown that the direction ofu×v satisfies the right hand rule, in the sense that if we hold the thumb on the right hand direction of uv satis es the right hand rule, in the sense that if we hold the thumb on the right hand outwards and close the remaining four fingers, then the thumb points towards theu×v-direction when outwards and close the remaining four ngers, then the thumb points towards the uv-direction when the fingers point from the u-direction towards the v-direction. Also, we showed in Proposition 4J that the ngers point from the u-direction towards the v-direction. Also, we showed in Proposition 4J that the magnitude ofu×v depends only on the norm ofu andv and the angle between the two vectors. It Chapter 4 : Vectors page 12 of 24 Chapter 4 : Vectors page 12 of 24c Linear Algebra W W L Chen, 1982, 2008 c Linear Algebra WWL Chen, 1982, 2006 the magnitude of uv depends only on the norm of u andv and the angle between the two vectors. It follows that the vector product is unchanged as long as we keep a right hand coordinate system. This is follows that the vector product is unchanged as long as we keep a right hand coordinate system. This is an important consideration in physics and engineering, where we may use different coordinate systems an important consideration in physics and engineering, where we may use di erent coordinate systems on the same problem. on the same problem. 4.5. Scalar Triple Products 4.5. Scalar Triple Products 3 3 Suppose that u,v,w∈R do not all lie on the same plane. Consider the parallelepiped with u,v,w as Suppose that u;v;w2R do not all lie on the same plane. Consider the parallelepiped with u;v;w as three of its edges. We are interested in calculating the volume of this parallelepiped. Suppose that u, v three of its edges. We are interested in calculating the volume of this parallelepiped. Suppose that u, v −→ −−→ −−→ and w are represented by OA, OB and OC respectively. Consider the diagram below: and w are represented by OA, OB and OC respectively. Consider the diagram below: v×w PA u C w OB v ByProposition4K,thebaseofthisparallelepiped,withO,B,C asthreeofthevertices,hasarea%v×w%. By Proposition 4K, the base of this parallelepiped, withO;B;C as three of the vertices, has areakvwk. −−→ Next, note that if OP is perpendicular to the base of the parallelepiped, then OP is in the direction of Next, note that if OP is perpendicular to the base of the parallelepiped, then OP is in the direction of v×w. IfPA is perpendicular to OP, then the height of the parallelepiped is equal to the norm of the vw. If PA is perpendicular to OP , then the height of the parallelepiped is equal to the norm of the orthogonal projection of u on v×w.In other words, the parallelepiped has height orthogonal projection of u on vw. In other words, the parallelepiped has height uu· ((vv×w w)) juu· ((vv×w w))j k%pro projj uuk% = = ((vv×w w)) = = :. vw v×w 22 kvwk kvwk %v×w% %v×w% Hence the volume of the parallelepiped is given by Hence the volume of the parallelepiped is given by V =ju (vw)j: V = u ·(v×w). We have proved the following result. We have proved the following result. 3 PROPOSITION 4L. Suppose that u;v;w2R . Then the parallelepiped with u, v and w as three of 3 PROPOSITION 4L. Suppose that u,v,w∈R . Then the parallelepiped with u, v and w as three of its edges has volumeju (vw)j. its edges has volume u ·(v×w). 3 Definition. Suppose that u;v;w2 R . Then u (vw) is called the scalar triple product of u, v 3 Definition. Suppose that u,v,w∈R . Then u ·(v×w)is called the scalar triple product of u, v and and w. w. 3 Remarks. (1) It follows immediately from Proposition 4L that three vectors in R are coplanar if and 3 Remarks. (1) It follows immediately from Proposition 4L that three vectors inR are coplanar if and only if their scalar triple product is zero. only if their scalar triple product is zero. (2) Note that (2) Note that        v v v v v v " " " " 2 3 1 3 1 2 u (vw) = (u ;u ;u ) det ; det ; det 1 2 3 v v v v v v 2 3 1 3 1 2 w w w w w w 2 3 1 3 1 2 (9) u ·(v×w)=(u ,u ,u ) · det ,−det ,det 1 2 3       w w w w w w 2 3 1 3 1 2 "v v "v v "v v 2 3 1 3 1 2 =u det u det +u det 1 2 3 v v v v v v 2 3 1 3 1 2 w w w w w w 2 3 1 3 1 2 =u det −u det +u det 1 2 3 0 w w 1 w w w w 2 3 1 3 1 2 u u u  1 2 3 u u u 1 2 3A = det v v v ; (9) 1 2 3   = det v v v , 1 2 3 w w w 1 2 3 w w w 1 2 3 in view of cofactor expansion by row 1. in view of cofactor expansion by row 1. Chapter 4 : Vectors page 13 of 24 Chapter 4 : Vectors page 13 of 24c Linear Algebra W W L Chen, 1982, 2008 (3) It follows from identity (9) that u (vw) =v (wu) =w (uv): Note that each of the determinants can be obtained from the other two by twice interchanging two rows. Example 4.5.1. Suppose that u = (1; 0; 1), v = (2; 1; 3) and w = (0; 1; 1). Then 0 1 1 0 1 A u (vw) = det 2 1 3 = 0; 0 1 1 so that u, v and w are coplanar. Example 4.5.2. The volume of the parallelepiped with u = (1; 0; 1), v = (2; 1; 4) and w = (0; 1; 1) as three of its edges is given by 0 1 1 0 1 A ju (vw)j = det 2 1 4 =j 1j = 1: 0 1 1 3 4.6. Application to Geometry in R 3 3 In this section, we shall study lines and planes inR by using our results on vectors inR . 3 3 Consider rst of all a plane in R . Suppose that (x ;y ;z )2 R is a given point on this plane. 1 1 1 Suppose further that n = (a;b;c) is a vector perpendicular to this plane. Then for any arbitrary point 3 (x;y;z)2R on this plane, the vector (x;y;z) (x ;y ;z ) = (xx ;yy ;zz ) 1 1 1 1 1 1 joins one point on the plane to another point on the plane, and so must be parallel to the plane and hence perpendicular to n = (a;b;c). It follows that the scalar product (a;b;c) (xx ;yy ;zz ) = 0; 1 1 1 and so a(xx ) +b(yy ) +c(zz ) = 0: (10) 1 1 1 If we writed =ax +by +cz , then (10) can be rewritten in the form 1 1 1 ax +by +cz +d = 0: (11) Equation (10) is usually called the point-normal form of the equation of a plane, while equation (11) is usually known as the general form of the equation of a plane. Example 4.6.1. Consider the plane through the point (2;5; 7) and perpendicular to the vector (3; 5;4). Here (a;b;c) = (3; 5;4) and (x ;y ;z ) = (2;5; 7). The equation of the plane is given 1 1 1 in point-normal form by 3(x 2) + 5(y + 5) 4(z 7) = 0, and in general form by 3x + 5y 4z + 37 = 0. Hered = 6 25 28 =37. Chapter 4 : Vectors page 14 of 24c Linear Algebra W W L Chen, 1982, 2008 Example4.6.2. Consider the plane through the points (1; 1; 1), (2; 2; 0) and (4;6; 2). Then the vectors (2; 2; 0) (1; 1; 1) = (1; 1;1) and (4;6; 2) (1; 1; 1) = (3;7; 1) join the point (1; 1; 1) to the points (2; 2; 0) and (4;6; 2) respectively and are therefore parallel to the plane. It follows that the vector product (1; 1;1) (3;7; 1) = (6;4;10) is perpendicular to the plane. The equation of the plane is then given by6(x1)4(y1)10(z1) = 0, or 3x + 2y + 5z 10 = 0. 3 3 Consider next a line inR . Suppose that (x ;y ;z )2R is a given point on this line. Suppose further 1 1 1 3 that n = (a;b;c) is a vector parallel to this line. Then for any arbitrary point (x;y;z)2R on this line, the vector (x;y;z) (x ;y ;z ) = (xx ;yy ;zz ) 1 1 1 1 1 1 joins one point on the line to another point on the line, and so must be parallel to n = (a;b;c). It follows that there is some number 2R such that (xx ;yy ;zz ) =(a;b;c); 1 1 1 so that x =x +a; 1 y =y +b; (12) 1 z =z +c; 1 where  is called a parameter. Suppose further that a;b;c are all non-zero. Then, eliminating the parameter , we obtain xx yy zz 1 1 1 = = : (13) a b c Equations (12) are usually called the parametric form of the equations of a line, while equations (13) are usually known as the symmetric form of the equations of a line. Example4.6.3. Consider the line through the point (2;5; 7) and parallel to the vector (3; 5;4). Here (a;b;c) = (3; 5;4) and (x ;y ;z ) = (2;5; 7). The equations of the line are given in parametric form 1 1 1 by x = 2 + 3; y =5 + 5; z = 7 4; and in symmetric form by x 2 y + 5 z 7 = = : 3 5 4 Chapter 4 : Vectors page 15 of 24c Linear Algebra W W L Chen, 1982, 2008 Example4.6.4. Consider the line through the points (3; 0; 5) and (7; 0; 8). Then a vector in the direction of the line is given by (7; 0; 8) (3; 0; 5) = (4; 0; 3): The equation of the line is then given in parametric form by x = 3 + 4; y = 0; z = 5 + 3; and in symmetric form by x 3 z 5 = and y = 0: 4 3 Consider the plane through three xed points (x ;y ;z ), (x ;y ;z ) and (x ;y ;z ), not lying on the 1 1 1 2 2 2 3 3 3 same line. Let (x;y;z) be a point on the plane. Then the vectors (x;y;z) (x ;y ;z ) = (xx ;yy ;zz ); 1 1 1 1 1 1 (x;y;z) (x ;y ;z ) = (xx ;yy ;zz ); 2 2 2 2 2 2 (x;y;z) (x ;y ;z ) = (xx ;yy ;zz ); 3 3 3 3 3 3 each joining one point on the plane to another point on the plane, are all parallel to the plane. Using the vector product, we see that the vector (xx ;yy ;zz ) (xx ;yy ;zz ) 2 2 2 3 3 3 is perpendicular to the plane, and so perpendicular to the vector (xx ;yy ;zz ). It follows that 1 1 1 the scalar triple product (xx ;yy ;zz ) ((xx ;yy ;zz ) (xx ;yy ;zz )) = 0; 1 1 1 2 2 2 3 3 3 in other words, 0 1 xx yy zz 1 1 1 A det xx yy zz = 0: 2 2 2 xx yy zz 3 3 3 This is another technique to nd the equation of a plane through three xed points. Example 4.6.5. We return to the plane in Example 4.6.2, through the three points (1; 1; 1), (2; 2; 0) and (4;6; 2). The equation is given by 0 1 x 1 y 1 z 1 A det x 2 y 2 z 0 = 0: x 4 y + 6 z 2 The determinant on the left hand side is equal to6x 4y 10z + 20. Hence the equation of the plane is given by6x 4y 10z + 20 = 0, or 3x + 2y + 5z 10 = 0. We observe that the calculation for the determinant above is not very pleasant. However, the technique can be improved in the following way by making less reference to the unknown point (x;y;z). Note that the vectors (x;y;z) (x ;y ;z ) = (xx ;yy ;zz ); 1 1 1 1 1 1 (x ;y ;z ) (x ;y ;z ) = (x x ;y y ;z z ); 2 2 2 1 1 1 2 1 2 1 2 1 (x ;y ;z ) (x ;y ;z ) = (x x ;y y ;z z ); 3 3 3 1 1 1 3 1 3 1 3 1 Chapter 4 : Vectors page 16 of 24c Linear Algebra c WWL Chen, 1982, 2006 Linear Algebra W W L Chen, 1982, 2008 each joining one point on the plane to another point on the plane, are all parallel to the plane. Using each joining one point on the plane to another point on the plane, are all parallel to the plane. Using the vector product, we see that the vector the vector product, we see that the vector (x − x ,y − y ,z − z )×(x − x ,y − y ,z − z ) (x2x1;y2y1;z2z1 ) (x3x1;y3y1;z3z1 ) 2 1 2 1 2 1 3 1 3 1 3 1 is perpendicular to the plane, and so perpendicular to the vector (x−x ,y−y ,z−z ). It follows that is perpendicular to the plane, and so perpendicular to the vector (xx ;yy ;zz ). It follows that 1 1 1 1 1 1 the scalar triple product the scalar triple product (x− x ,y− y ,z− z ) ·((x − x ,y − y ,z − z )×(x − x ,y − y ,z − z )) = 0; (xx ;yy ;zz ) ((x x ;y y ;z z ) (x x ;y y ;z z )) = 0; 1 1 1 1 1 1 2 2 1 1 2 2 1 1 2 2 1 1 3 3 1 1 3 3 1 1 3 3 1 1 in inother otherw words, ords,   0 1 x− x y− y z− z xx1 yy1 zz1 1 1 1   A det x − x y − y z − z =0. det x2x1 y2y1 z2z1 = 0: 2 1 2 1 2 1 x − x y − y z − z x3x1 y3y1 z3z1 3 1 3 1 3 1 Example 4.6.6. We return to the plane in Examples 4.6.2 and 4.6.5, through the three points (1,1,1), Example 4.6.6. We return to the plane in Examples 4.6.2 and 4.6.5, through the three points (1; 1; 1), (2,2,0) and (4,−6,2). The equation is given by (2; 2; 0) and (4;6; 2). The equation is given by   0 1 xx−1 1 yy−1 1 zz−1 1   A det det 2 2−12 1 2−10 1 0−1 1 =0 = 0.: 4−1 −6−12−1 4 1 6 1 2 1 The determinant on the left hand side is equal to The determinant on the left hand side is equal to 0 1 xx−1 1 yy−1 1 zz−1 1  A det det 11 1 1 −1 1 = =−6( 6(xx−1) 1)−4( 4(yy−1) 1)−10( 10(zz−1) 1)= =−6 6xx−4 4yy−10 10zz+ +20 20.: 3 3 −71 7 1 Hence Hencethe theequation equationof ofthe theplane planeis isgiv given enb by y−6 6xx−4 4yy−10 10zz+ +20 20= =0, 0,or or3 3xx+2 + 2yy+5 + 5zz−10 10= =0. 0. We Wenext nextconsider considerthe theproblem problemof ofdividing dividinga aline linesegmen segment tin ina agiv given enratio. ratio. Supp Suppose osethat thatxx and andxx are are 1 1 2 2 3 3 two twogiv given enp poin oints tsin inRR .. We Wewish wishto todivide dividethe theline linesegmen segment tjoining joiningxx and andxx in internally ternallyin inthe theratio ratioα ::α ,,where whereα and andα 1 1 2 2 1 1 2 2 1 1 2 2 are arep positiv ositive e real real n num umb bers. ers. In In other other w words, ords, w we e wish wish to to find nd the the p poin oint txxon onthe theline linesegmen segmentt joining joiningxx 1 1 and andxx suc such hthat that 2 2 kxx−xxk α 1 1 1 1 = = ,; kxxk x−x α 2 2 2 2 as assho shown wnin inthe thediagram diagramb belo elow: w: x xx 1 2 ←−−−−−−x−x−−−−−−→←−−x−x−−→ 1 2 Since x−x and x −x are both in the same direction as x −x , we must have Since xx1 and x2x are both in the same direction as x2x1 , we must have 1 2 2 1 α x +α x 1x2 + 2x1 1 2 2 1 α (x−x )=α (x −x), or x = . 2 (xx1 ) = 1 (x2x); or x = : 2 1 1 2 α +α 1 + 2 1 2 We wish next to find the point x on the line joining x and x , but not between x and x , such that We wish next to nd the point x on the line joining x1 and x2 , but not between x1 and x2 , such that 1 2 1 2 x−x α kxx1k 1 1 1 = , = ; x−x α kxxk 2 2 2 2 Chapter 4 : Vectors page 17 of 24 Chapter 4 : Vectors page 17 of 24Linear Algebra c WWL Chen, 1982, 2006 c Linear Algebra c WWL Chen, 1982, 2006 Linear Algebra W W L Chen, 1982, 2008 where α and α are positive real numbers, as shown in the diagrams below for the cases α α and 1 2 1 2 where α and α are positive real numbers, as shown in the diagrams below for the cases α α and where 1 and 2 are positive real numbers, as shown in the diagrams below for the cases 1 2 and 1 2 1 2 α α respectively: 1 2 α α respectively: 1 2 respectively: 1 2 xx x x x x 1 2 1 2 xx x x x x 1 2 1 2 Since x−x and x−x are in the same direction as each other, we must have 1 2 Since x−x and x−x are in the same direction as each other, we must have 1 2 Since xx and xx are in the same direction as each other, we must have 1 2 α x −α x 1 2 2 1 α x −α x 1 2 2 1 α (x−x )=α (x−x ), or x = . 2 1 1 2 α (x−x )=α (x−x ), or x = . x x 2 1 1 2 1 2 2 1 α −α 1 2 (xx ) = (xx ); or x = α −α : 2 1 1 2 1 2 1 2 Example 4.6.7. Let x = (1,2,3) and x = (7,11,6). The point 1 2 Example 4.6.7. Let x = (1,2,3) and x = (7,11,6). The point 1 2 Example 4.6.7. Let x = (1; 2; 3) and x = (7; 11; 6). The point 1 2 2x +x 2(7,11,6)+(1,2,3) 2 1 2x +x 2(7,11,6)+(1,2,3) 2 1 x = 2x +x = 2(7; 11; 6) + (1; 2; 3) = (5,8,5) x = 2 1 = = (5,8,5) x = 2+1 = 3 = (5; 8; 5) 2+1 3 2 + 1 3 divides the line segment joining (1,2,3) and (7,11,6) internally in the ratio 2 : 1, whereas the point divides the line segment joining (1,2,3) and (7,11,6) internally in the ratio 2 : 1, whereas the point divides the line segment joining (1; 2; 3) and (7; 11; 6) internally in the ratio 2 : 1, whereas the point 4x −2x 4(7,11,6)−2(1,2,3) 2 1 4x −2x 4(7,11,6)−2(1,2,3) 2 1 4x 2x 4(7; 11; 6) 2(1; 2; 3) x = 2 1 = = (13,20,9) x = = = (13,20,9) x = = = (13; 20; 9) 4−2 2 4−2 2 4 2 2 satisfies satisfies satis es x−x 4 1 kxx−xxk 44 11 = . = = :. x−x 2 2 kxx−xxk 22 2 2 Finally we turn our attention to the question of finding the distance of a plane from a given point. Finally we turn our attention to the question of nding the distance of a plane from a given point. Finally we turn our attention to the question of finding the distance of a plane from a given point. We shall prove the following analogue of Proposition 4F. We shall prove the following analogue of Proposition 4F. We shall prove the following analogue of Proposition 4F. PROPOSITION 4F’. The perpendicular distance D of a plane ax +by +cz +d=0 from a point PROPOSITION 4F'. The perpendicular distance D of a plane ax +by +cz +d = 0 from a point PROPOSITION 4F’. The perpendicular distance D of a plane ax +by +cz +d=0 from a point (x ,y ,z ) is given by (x ;y ;z ) is given by 0 0 0 (x0,y0,z0) is given by 0 0 0 ax +by +cz +d 0 0 0 jax ax + +by by + +cz cz + +ddj 0 0 0 0 0 0 D = √ . D = p : D = √ . 2 2 2 a22 +b22 +c22 a +b +c a +b +c Pr Proof. oof. Consider Consider the the follo following wing diagram: diagram: Proof. Consider the following diagram: P P D D n=(a,b,c) n=(a,b,c) u u Q Q ax+by+cz=0 ax+by+cz=0 O O Chapter 4 : Vectors page 18 of 24 Chapter Chapter 44 :: V Veectors ctors page page 18 18 of of 24 24c Linear Algebra W W L Chen, 1982, 2008 Suppose that (x ;x ;x ) is any arbitrary pointO on the planeax +by +cz +d = 0. For any other point 1 2 3 (x;y;z) on the planeax +by +cz +d = 0, the vector (xx ;yy ;zz ) is parallel to the plane. On 1 1 1 the other hand, (a;b;c) (xx ;yy ;zz ) = (ax +by +cz) (ax +by +cz ) =d +d = 0; 1 1 1 1 1 1 so that the vector n = (a;b;c), in the direction OQ, is perpendicular to the plane ax +by +cz +d = 0. Suppose next that the point (x ;y ;z ) is represented by the point P in the diagram. Then the vector 0 0 0 u = (x x ;y y ;z z ) is represented byOP , andOQ represents the orthogonal projection proj u 0 1 0 1 0 1 n of u on the vector n. Clearly the perpendicular distance D of the point (x ;y ;z ) from the plane 0 0 0 ax +by +cz +d = 0 satis es un j(x x ;y y ;z z ) (a;b;c)j 0 1 0 1 0 1 p D =kproj uk = n = n 2 2 2 2 knk a +b +c jax +by +cz ax by czj jax +by +cz +dj 0 0 0 1 1 1 0 0 0 = p = p 2 2 2 2 2 2 a +b +c a +b +c as required. A special case of Proposition 4F' is when (x ;y ;z ) = (0; 0; 0) is the origin. This show that the 0 0 0 perpendicular distance of the plane ax +by +cz +d = 0 from the origin is jdj p : 2 2 2 a +b +c Example 4.6.8. Consider the plane 3x + 5y 4z + 37 = 0. The distance of the point (1; 2; 3) from the plane is p j3 + 10 12 + 37j 38 19 2 p =p = : 5 9 + 25 + 16 50 The distance of the origin from the plane is j37j 37 p =p : 9 + 25 + 16 50 Example4.6.9. Consider also the plane 3x+5y4z1 = 0. Note that this plane is also perpendicular to the vector (3; 5;4) and is therefore parallel to the plane 3x+5y4z+37 = 0. It is therefore reasonable to nd the perpendicular distance between these two parallel planes. Note that the perpendicular distance between the two planes is equal to the perpendicular distance of any point on 3x + 5y 4z 1 = 0 from the plane 3x + 5y 4z + 37 = 0. Note now that (1; 2; 3) lies on the plane 3x + 5y 4z 1 = 0. It follows p from Example 4.6.8 that the distance between the two planes is 19 2=5. 4.7. Application to Mechanics 2 Letu = (u ;u ) denote a vector inR , where the componentsu andu are functions of an independent x y x y variable t. Then the derivative of u with respect to t is given by   du du du x y = ; : dt dt dt Chapter 4 : Vectors page 19 of 24c Linear Algebra W W L Chen, 1982, 2008 Example 4.7.1. When discussing planar particle motion, we often let r = (x;y) denote the position of a particle at time t. Then the components x and y are functions of t. The derivative   dr dx dy v = = ; dt dt dt represents the velocity of the particle, and its derivative   2 2 dv d x d y a = = ; 2 2 dt dt dt represents the acceleration of the particle. We often write r =krk, v =kvk and a =kak. 2 Suppose that w = (w ;w ) is another vector inR . Then it is not dicult to see that x y d dw du (uw) =u + w: (14) dt dt dt Example 4.7.2. Consider a particle moving at constant speed along a circular path centred at the 2 2 origin. Then r =krk is constant. More precisely, the position vector r = (x;y) satis es x +y = c , 1 where c is a positive constant, so that 1 rr = (x;y) (x;y) =c : (15) 1 On the other hand, v =kvk is constant. More precisely, the velocity vector       2 2 dx dy dx dy v = ; satis es + =c ; 2 dt dt dt dt where c is a positive constant, so that 2     dx dy dx dy vv = ;  ; =c : (16) 2 dt dt dt dt Di erentiating (15) and (16) with respect to t, and using the identity (14), we obtain respectively rv = 0 and va = 0: (17) Using the properties of the scalar product, we see that the equations in (17) show that the vector v is perpendicular to both vectors r and a, and so a must be in the same direction as or the opposite direction to r. Next, di erentiating the rst equation in (17), we obtain 2 ra +vv = 0; or ra =v 0:   Let  denote the angle between a and r. Then  = 0 or  = 180 . Since ra =krkkak cos;  2 2 it follows that cos 0, and so  = 180 . We also obtain ra = v , so that a = v =r. This is a vector proof that for circular motion at constant speed, the acceleration is towards the centre of the circle and 2 of magnitude v =r. 3 Let u = (u :u ;u ) denote a vector in R , where the components u , u and u are functions of an x y z x y z independent variable t. Then the derivative of u with respect to t is given by   du du du du x y z = ; ; : dt dt dt dt Chapter 4 : Vectors page 20 of 24

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