LINEAR ALGEBRA W W L CHEN

LINEAR ALGEBRA W W L CHEN c W W L Chen, 1982, 2008. This chapter originates from material used by author at Imperial College, University of London, between 1981 and 1990. It is available free to all individuals, on understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied, with or without permission from author. However, this document may not be kept on any information storage and retrieval system without permission from author, unless such system is not accessible to any individuals or 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, direction of velocity is north-east and magnitude of velocity is 50 kph. We can describe our velocity in kph as ( 50, 50 ), 2 2 where first coordinate describes speed with which we are moving east and second coordinate describes speed with which we are moving north. Example 4.1.2. An object in sky may be 100 metres away in south-east direction 45 degrees upwards. In this case, direction of its position is south-eastand 45 degrees upwards and magnitude of its distance is 100 metres. We can describe position of object in metres as ( 50, 50, 100 ), 2 where first coordinate describes distance east, second coordinate describes distance north and third coordinate describes distance up. The purpose of this chapter is to study some relationship between algebra and geometry. We shall first study some algebra which is motivated by geometric considerations. We n use algebra later to better understand some problems in geometry. Chapter 4 : Vectors page 1 of 24

Linear Algebra c WWLChen, W L 1982, 2008 2006 4.2. Vectors in R 2 on plane R 2 can be described as an ordered pair u = =(u 1,,uu 2 ), where u 1,,uu 2 R. Definition. Two vectors u = =(u, u =, v 1,u 2 ) and v =(v 1,v 2 ) in R 2 are said to be equal, denoted by u = v, if u 1 = v 1 and u 2 = v 2. Definition. Forany two vectors u = =(u, u =, v 1,u 2 ) and v =(v 1,v 2 ) in R 2,,we define ir sum to be u + v = =(u 1 u 2 ) )+(v+ 1,,vv 2 = 1 + v 1,,uu 2 + v 2 ). Geometrically, if we represent two vectors u and v by AB and BC respectively, n sum u + v is represented by AC as shown in diagram below: C u+v v A u B The next diagram demonstrates geometrically that u + v = v + u: D u C v u+v v A u B PROPOSITION PROPOSITION 4A. 4A. (VECTOR (VECTOR ADDITION) ADDITION) (a) (a) For For every every u, u, v R 2,,we have have u + v R 2. (b) (b) For every u, u, v, v, w R 2,,we have u + +(v + w) w) = =(u + v)+w. + (c) For every u R 2,,we have u + 0 = u, where 0 = (0, 0) R 2. (d) For every u R 2, re exists v R 2 such that u + v = 0. (e) For every u, v R 2,,we have u + v = v + u. Proof. Write u = =(u 1,,u u 2 ), v = =(v 1,,v v 2 ) and w = =(w 1,,ww 2 ), where u 1,,u u 2,,vv 1,,v v 2,,ww 1,,w w 2 R. To check part (a), simply note that u 1 + v 1,,u u 2 + v 2 R. Tocheck part (b), note that u + +(v + w) = =(u 1,,u u 2 )+(v + 1 + w 1,,vv 2 + w 2 )=(u = 1 + +(v 1 + w 1 ),u u 2 + +(v 2 + w 2 )) = ((u 1 + v 1 )+w + w 1, (u 2 + v 2 ) )+w+ w 2 ) )=(u= 1 + v 1,,uu 2 + v 2 )+(w + 1,,w w 2 ) = =(u + v)+w. + Part (c) is trivial. Next, if v = =( u 1, u 2 ), n u + v = 0, giving part (d). To check part (e), note that u + v = =(u 1 + v 1,u u 2 + v 2 )=(v = 1 + u 1,,vv 2 + u 2 )=v = v + u. Chapter Chapter 4 : Vectors Vectors page 2 of 24

Linear Algebra c WWWLChen, W L 1982, 2008 2006 Definition. Forany vector u = =(u 1,,u u 2 ) in R 2 and any scalar c R, we define scalar multiple to be cu = c(u 1,uu 2 )=(cu= 1, cu 2 ). Example 4.2.1. Suppose = that u = (2, 1). Then 2u =( 4, 2). Geometrically, if we represent two vectors u and 2u by OA and OB respectively, n we have diagram below: A O u 2u B PROPOSITION 4B. (SCALAR MULTIPLICATION) (a) For every c R and u R 2,,we have cu R 2. (b) For every c R and u, v R 2,,we have c(u + v) = =cu + cv. (c) For every a, b R and u R 2,,we have (a + b)u = au + bu. (d) For every a, b R and u R 2,,we have (ab)u = a(bu). (e) For every u R 2,,we have 1u = u. Proof. Write u = =(u,u =(v,v,u,v,v Tocheck 1, u 2 ) and v = 1, v 2 ), where u 1, u 2, v 1, v 2 R. part (a), simply note that cu Tocheck 1, cu 2 R. part (b), note that c(u + v) = =c(u,u )=(c(u ),c(u )) 1 + v 1, u 2 + v 2 ) = 1 + v 1 ), 2 + v 2 )) = =(cu cv cv )=(cu )+(cv cv )=cu 1 + cv 1, cu 2 + cv 2 ) = 1, cu 2 ) + 1, cv 2 ) = + cv. To To check check part part (c), (c), note note that that (a (a + b)u b)u = ((a ((a + b)u b)u (a b)u )=(au bu au bu 1, (a + b)u 2 ) = (au 1 + bu 1, au 2 + bu 2 ) = =(au (au au )+(bu bu )=au bu. 1, au 2 ) + (bu 1, bu 2 ) = au + bu. To To check check part part (d), (d), note note that that (ab)u (ab)u = ((ab)u ((ab)u (ab)u )=(a(bu ),a(bu )) a(bu bu )=a(bu). 1, (ab)u 2 ) = (a(bu 1 ), a(bu 2 )) = a(bu 1, bu 2 ) = a(bu). Finally, Finally, to to check check part part (e), (e), note note that that 1u 1u = (1u (1u 1u )=(u,u )=u. 1, 1u 2 ) = (u 1, u 2 ) = u. Definition. Definition. For Forany any vector vector u = =(u (u,u in,we define norm of to be non-negative real 1, u 2 ) in R 2, we define norm of u to be non-negative real number number u = u 2 1 + u2 2. Remarks. (1) The norm of a vector is simply its magnitude or length. The definition follows from famous orem of Pythagoras. (2) Suppose that P (u,u,v 1, u 2 ) and Q(v 1, v 2 ) are two points on plane R 2. To calculate distance d(p, Q) between two points, we can first find a vector from P to Q. This is given by (v 1 u 1,,v v 2 u 2 ). The distance d(p, Q) is n norm of this vector, so that d(p, Q) = (v +(v 1 u 1 ) 2 + 2 u 2 ) 2. Chapter 4 : Vectors page of of 24

Linear Linear Algebra Algebra c W WWLChen, W L Chen, 1982, 1982, 2008 2006 () It is not difficult to see that for any vector u R 2 and any scalar c R, we have cu = c u. Definition. Any vector u R 2 satisfying u = =1is 1 called a unit vector. Example 4.2.2. The vector (, 4) has norm 5. Example 4.2.. The distance between points (6, ) and (9, 7) is (9 6) 2 + (7 ) 2 = 5. Example 4.2.4. The vectors (1, 0) and (0, 1) are unit vectors in R 2. Example 4.2.5. The unit vector in direction of vector (1, 1) is (1/ 2, 1/ 2). Example 4.2.6. In fact, all unit vectors in R 2 are of form (cos θ, sin θ), where θ R. Quite often, we may want to find angle between two vectors. The scalar product of two vectors n comes in handy. We shall define scalar product in two ways, one in terms of angle between two vectors and or not in terms of this angle, and show that two definitions are in fact equivalent. Definition. Suppose that u = =(u 1,,u u 2 ) and v = =(v 1,,v v 2 ) are vectors in R 2, and that θ [0, π] represents angle between m. We define scalar product u v of u and v by { (1) u v = u v cos θ if u 0 and v 0, 0 if u = 0 or v = 0. Alternatively, we write (1) (2) u v = u 1 v 1 + u 2 v 2. (2) The definitions (1) and (2) are clearly equivalent if u = 0 or v = 0. On or hand, we have following result. PROPOSITION 4C. Suppose that u = =(u, u =, v 1,u 2 ) and v =(v 1,v 2 ) are non-zero vectors in R 2, and that θ [0, π] represents angle between m. Then u v cos θ = u 1 v 1 + u 2 v 2. Proof. Geometrically, if we represent two vectors u and v by OA and OB respectively, n difference v u is represented by AB as shown in diagram below: B v v u A O θ u By Law of cosines, we have AB 2 = OA 2 + OB 2 2OA OB cos θ; Chapter Chapter 4 : Vectors Vectors page page 4 of of 24 24

in or words, we have so that as required. v u 2 = u 2 + v 2 2 u v cos θ, u v cos θ = 1 2 ( u 2 + v 2 v u 2 ) = 1 2 (u2 1 + u 2 2 + v 2 1 + v 2 2 (v 1 u 1 ) 2 (v 2 u 2 ) 2 ) = u 1 v 1 + u 2 v 2 Remarks. (1) We say that two non-zero vectors in R 2 are orthogonal if angle between m is π/2. It follows immediately from definition of scalar product that two non-zero vectors u, v R 2 are orthogonal if and only if u v = 0. (2) We can calculate scalar product of any two non-zero vectors u, v R 2 by formula (2) and n use formula (1) to calculate angle between u and v. Example 4.2.7. Suppose that u = (, 1) and v = (, ). Then by formula (2), we have Note now that It follows from formula (1) that so that θ = π/6. u v = + = 6. u = 2 and v = 2. cos θ = u v u v = 6 4 = 2, Example 4.2.8. Suppose that u = (, 1) and v = (, ). u v = 0. It follows that u and v are orthogonal. Then by formula (2), we have PROPOSITION 4D. (SCALAR PRODUCT) Suppose that u, v, w R 2 and c R. Then (a) u v = v u; (b) u (v + w) = (u v) + (u w); (c) c(u v) = (cu) v = u (cv); (d) u u 0; and (e) u u = 0 if and only if u = 0. Proof. Write u = (u 1, u 2 ), v = (v 1, v 2 ) and w = (w 1, w 2 ), where u 1, u 2, v 1, v 2, w 1, w 2 R. Part (a) is trivial. To check part (b), note that u (v + w) = u 1 (v 1 + w 1 ) + u 2 (v 2 + w 2 ) = (u 1 v 1 + u 2 v 2 ) + (u 1 w 1 + u 2 w 2 ) = u v + u w. Part (c) is rar simple. To check parts (d) and (e), note that u u = u 2 1 + u 2 2 0, and that equality holds precisely when u 1 = u 2 = 0. Chapter 4 : Vectors page 5 of 24

Linear Algebra c W WWLChen, W L Chen, 1982, 2008 2006 Consider diagram below: P () R v u Q a A w O () Here Here we we represent represent two two vectors vectors a and and u by by OA OA and and OP OP respectively. respectively. If If we we project project vector vector u on on to to line line OA, OA, n n image image of of projection projection is is vector vector w, w, represented represented by by OQ. OQ. On On or or hand, hand, if if we we project project vector vector u on on to to a line line perpendicular perpendicular to to line line OA, OA, n n image image of of projection projection is is vector vector v, v, represented represented by by OR. OR. Definition. Definition. In In notation notation of of diagram diagram (), (), vector vector w is is called called orthogonal orthogonal projection projection of of vector vector u on on vector vector a, a, and and denoted denoted by by w = proj proj a u. u. PROPOSITION PROPOSITION 4E. 4E. (ORTHOGONAL (ORTHOGONAL PROJECTION) PROJECTION) Suppose Suppose that that u, u, a R 2. Then Then proj proj a u = u a a a 2a. a. Remark. Note Note that that component of of u orthogonal to to a, a, represented by by OR OR in in diagram diagram (), (), is is u proj a u = u u a a a 2a. a. Proof of of Proposition 4E. 4E. Note that w = ka ka for for some k R. R. It It clearly suffices to to prove that that k = u a a 2. It It is is easy to to see that vectors u w and a are orthogonal. It It follows that scalar product (u (u w) a = =0.In or words, (u (u ka) a = =0. Hence k = u a a a = u a a 2 as as required. To end this section, we shall apply our knowledge gained so so far to to find a formula that gives perpendicular distance of of a point (x 0,,y y 0 ) from a line ax + by by + c = =0. Consider diagram below: D P ax+by+c=0 u n =(a, b) Q O Chapter Vectors page of 24 Chapter 4 : Vectors page 6 of 24

Suppose that (x 1, y 1 ) is any arbitrary point O on line ax + by + c = 0. For any or point (x, y) on line ax + by + c = 0, vector (x x 1, y y 1 ) is parallel to line. On or hand, (a, b) (x x 1, y y 1 ) = (ax + by) (ax 1 + by 1 ) = c + c = 0, so that vector n = (a, b), in direction OQ, is perpendicular to line ax + by + c = 0. Suppose next that point (x 0, y 0 ) is represented by point P in diagram. Then vector u = (x 0 x 1, y 0 y 1 ) is represented by OP, and OQ represents orthogonal projection projn u of u on vector n. Clearly perpendicular distance D of point (x 0, y 0 ) from line ax + by + c = 0 satisfies D = proj n u = u n n 2 n = (x 0 x 1, y 0 y 1 ) (a, b) = ax 0 + by 0 ax 1 by 1 = ax 0 + by 0 + c. a2 + b 2 a2 + b 2 a2 + b 2 We have proved following result. PROPOSITION 4F. The perpendicular distance D of a point (x 0, y 0 ) from a line ax + by + c = 0 is given by D = ax 0 + by 0 + c a2 + b 2. Example 4.2.9. The perpendicular distance D of point (5, 7) from line 2x y + 5 = 0 is given by D = 10 21 + 5 4 + 9 = 6 1. 4.. Vectors in R In this section, we consider same problems as in Section 4.2, but in -space R. Any reader who feels confident may skip this section. A vector on plane R can be described as an ordered triple u = (u 1, u 2, u ), where u 1, u 2, u R. Definition. Two vectors u = (u 1, u 2, u ) and v = (v 1, v 2, v ) in R are said to be equal, denoted by u = v, if u 1 = v 1, u 2 = v 2 and u = v. Definition. For any two vectors u = (u 1, u 2, u ) and v = (v 1, v 2, v ) in R, we define ir sum to be u + v = (u 1, u 2, u ) + (v 1, v 2, v ) = (u 1 + v 1, u 2 + v 2, u + v ). Definition. For any vector u = (u 1, u 2, u ) in R and any scalar c R, we define scalar multiple to be cu = c(u 1, u 2, u ) = (cu 1, cu 2, cu ). The following two results are analogues of Propositions 4A and 4B. The proofs are essentially similar. Chapter 4 : Vectors page 7 of 24

PROPOSITION 4A. (VECTOR ADDITION) (a) For every u, v R, we have u + v R. (b) For every u, v, w R, we have u + (v + w) = (u + v) + w. (c) For every u R, we have u + 0 = u, where 0 = (0, 0, 0) R. (d) For every u R, re exists v R such that u + v = 0. (e) For every u, v R, we have u + v = v + u. PROPOSITION 4B. (SCALAR MULTIPLICATION) (a) For every c R and u R, we have cu R. (b) For every c R and u, v R, we have c(u + v) = cu + cv. (c) For every a, b R and u R, we have (a + b)u = au + bu. (d) For every a, b R and u R, we have (ab)u = a(bu). (e) For every u R, we have 1u = u. Definition. For any vector u = (u 1, u 2, u ) in R, we define norm of u to be non-negative real number u = u 2 1 + u2 2 + u2. Remarks. (1) Suppose that P (u 1, u 2, u ) and Q(v 1, v 2, v ) are two points in R. To calculate distance d(p, Q) between two points, we can first find a vector from P to Q. This is given by (v 1 u 1, v 2 u 2, v u ). The distance d(p, Q) is n norm of this vector, so that d(p, Q) = (v 1 u 1 ) 2 + (v 2 u 2 ) 2 + (v u ) 2. (2) It is not difficult to see that for any vector u R and any scalar c R, we have cu = c u. Definition. Any vector u R satisfying u = 1 is called a unit vector. Example 4..1. The vector (, 4, 12) has norm 1. Example 4..2. The distance between points (6,, 12) and (9, 7, 0) is 1. Example 4... The vectors (1, 0, 0) and (0, 1, 0) are unit vectors in R. Example 4..4. The unit vector in direction of vector (1, 0, 1) is (1/ 2, 0, 1/ 2). The ory of scalar products can be extended to R is natural way. Definition. Suppose that u = (u 1, u 2, u ) and v = (v 1, v 2, v ) are vectors in R, and that θ [0, π] represents angle between m. We define scalar product u v of u and v by u v = { u v cos θ if u 0 and v 0, 0 if u = 0 or v = 0. (4) Alternatively, we write u v = u 1 v 1 + u 2 v 2 + u v. (5) The definitions (4) and (5) are clearly equivalent if u = 0 or v = 0. On or hand, we have following analogue of Proposition 4C. The proof is similar. Chapter 4 : Vectors page 8 of 24

Linear Linear Algebra Algebra c WWLChen, W W L Chen, 1982, 1982, 2006 2008 PROPOSITION 4C. Suppose that u =(u = 1,u, u 2,u, u ) and v =(v = 1,v, v 2,v, v ) are non-zero vectors in R, and that θ [0, π] represents angle between m. Then u v cos θ = u 1 v 1 + u 2 v 2 + u v. Remarks. (1) We say that two non-zero vectors in R are orthogonal if angle between m is π/2. It follows immediately from definition of scalar productthat two non-zero vectors u, v R are orthogonal if and only if u v = 0. (2) We can calculate scalar product of any two non-zero vectors u, v R by formula (5) and n use formula (4) to calculate angle between u and v. Example 4..5. Suppose that u = (2, 0, 0) and v = (1, 1, 2). Then by formula (5), we have u v =2. = Note now that u =2and = 2 v =2.It= follows from formula (4) that cos θ = u v u v = 2 4 = 1 2, so that θ = π/. Example 4..6. Suppose that u = (2,, 5) and v = (1, 1, 1). u v =0.It= follows that u and v are orthogonal. Then by formula (5), we have The following result is analogue of Proposition 4D. The proof is similar. PROPOSITION 4D. (SCALAR PRODUCT) Suppose that u, v, w R and c R. Then (a) u v = v u; (b) u (v + w) =(u = v)+(u + w); (c) c(u v) =(cu) = v = u (cv); (d) u u 0; and (e) u u =0if= 0 and only if u = 0. Suppose now that a and u are two vectors in R. Then since two vectors are always coplanar, we can draw following diagram which represents plane y lie on: P (6) R v u Q a A O w (6) Note that this diagram is essentially same as diagram (), only difference being that while diagram () shows whole.asbefore, of R 2, diagram (6) only shows part of R. we represent two vectors a and u by OA and OP respectively. If we project vector u on to line OA, n image of projection is vector w, represented by OQ. On or hand, if we project vector u on to a line perpendicular to line OA, n image of projection is vector v, represented by OR. Definition. In notation of diagram (6), vector w is called orthogonal projection of vector u on vector a, and denoted by w = proj a u. Chapter Chapter 4 : Vectors Vectors page page 9 of of 24 24

The following result is analogue of Proposition 4E. The proof is similar. PROPOSITION 4E. (ORTHOGONAL PROJECTION) Suppose that u, a R. Then proj a u = u a a 2 a. Remark. Note that component of u orthogonal to a, represented by OR in diagram (6), is u proj a u = u u a a 2 a. 4.4. Vector Products In this section, we shall discuss a product of vectors unique to R. The idea of vector products has wide applications in geometry, physics and engineering, and is motivated by wish to find a vector that is perpendicular to two given vectors. We shall use right hand rule. In or words, if we hold thumb on right hand upwards and close remaining four fingers, n fingers point from x-direction towards y-direction, while thumb points towards z-direction. Alternatively, if we imagine Columbus had never lived and that earth were flat, n taking x-direction as east and y-direction as north, n z-direction is upwards! We shall frequently use three vectors i = (1, 0, 0), j = (0, 1, 0) and k = (0, 0, 1) in R. Definition. Suppose that u = (u 1, u 2, u ) and v = (v 1, v 2, v ) are two vectors in R. Then vector product u v is defined by determinant u v = det i j k u 1 u 2 u. v 1 v 2 v Remarks. (1) Note that i j = (j i) = k, j k = (k j) = i, k i = (i k) = j. (2) Using cofactor expansion by row 1, we have ( ) ( ) ( ) u2 u u v = det u1 u i det u1 u j + det 2 k v 2 v v 1 v v 1 v 2 ( ( ) ( ) ( )) u2 u = det u1 u, det u1 u, det 2 v 2 v v 1 v v 1 v 2 = (u 2 v u v 2, u v 1 u 1 v, u 1 v 2 u 2 v 1 ). We shall first of all show that vector product u v is orthogonal to both u and v. Chapter 4 : Vectors page 10 of 24

PROPOSITION 4G. Suppose that u = (u 1, u 2, u ) and v = (v 1, v 2, v ) are two vectors in R. Then (a) u (u v) = 0; and (b) v (u v) = 0. Proof. Note first of all that u (u v) = (u 1, u 2, u ) ( det ( u2 u = u 1 det v 2 v ( ) ( u2 u u1 u, det v 2 v v 1 v ) ( ) u1 u u 2 det + u det v 1 v in view of cofactor expansion by row 1. On or hand, clearly det u 1 u 2 u u 1 u 2 u = 0. v 1 v 2 v This proves part (a). The proof of part (b) is similar. ) ( )) u1 u, det 2 v 1 v 2 ( ) u1 u 2 = det u 1 u 2 u u v 1 v 1 u 2 u, 2 v 1 v 2 v Example 4.4.1. Suppose that u = (1, 1, 2) and v = (, 0, 2). Then u v = det i j k ( ( ) ( ) ( )) 1 1 2 1 2 1 2 1 1 = det, det, det = ( 2, 4, ). 0 2 2 0 0 2 Note that (1, 1, 2) ( 2, 4, ) = 0 and (, 0, 2) ( 2, 4, ) = 0. PROPOSITION 4H. (VECTOR PRODUCT) Suppose that u, v, w R and c R. Then (a) u v = (v u); (b) u (v + w) = (u v) + (u w); (c) (u + v) w = (u w) + (v w); (d) c(u v) = (cu) v = u (cv); (e) u 0 = 0; and (f) u u = 0. Proof. Write u = (u 1, u 2, u ), v = (v 1, v 2, v ) and w = (w 1, w 2, w ). To check part (a), note that det i j k u 1 u 2 u = det i j k v 1 v 2 v. v 1 v 2 v u 1 u 2 u To check part (b), note that det i j k u 1 u 2 u = det i j k u 1 u 2 u + det i j k u 1 u 2 u. v 1 + w 1 v 2 + w 2 v + w v 1 v 2 v w 1 w 2 w Part (c) is similar. To check part (d), note that c det i j k u 1 u 2 u = det i j k cu 1 cu 2 cu v 1 v 2 v v 1 v 2 v = det i j k u 1 u 2 u cv 1 cv 2 cv To check parts (e) and (f), note that u 0 = det i j k u 1 u 2 u = 0 and u u = det i j k u 1 u 2 u = 0 0 0 0 u 1 u 2 u as required.. Chapter 4 : Vectors page 11 of 24

Linear Algebra c WWLChen, 1982, 2006 Next, we shall discuss an application of vector product to evaluaton of area of parallelogram. Next, we shall discuss an application of vector product to evaluaton of area of a parallelogram. To do this, we shall first establish following result. To do this, we shall first establish following result. PROPOSITION 4J. Suppose that = (u, u, u and = (v, v, v are non-zero vectors in PROPOSITION 4J. Suppose that u =(u 1,u 2,u ) and v =(v 1,v 2,v ) are non-zero vectors in R, and that [0, π] represents angle between m. Then and that θ (a) u v [0, π] represents u v angle (u v) between m. Then and (a) u v 2 = u 2 v 2 (u v) 2 ; and (b) u v u v sin θ. (b) u v = u v sin θ. Proof. Note that Proof. Note that u v 2 = (u 2 2 2 + (u 1 1 2 + (u 1 2 2 1 2 (7) (7) u v 2 =(u 2 v u v 2 ) 2 +(u v 1 u 1 v ) 2 +(u 1 v 2 u 2 v 1 ) 2 and and u 2 v 2 (u v) 2 = (u 2 1 2 2 )(v1 2 2 ) 2 (u 1 1 2 2 2 (8) (8) u 2 v 2 (u v) 2 =(u 2 1 + u 2 2 + u 2 )(v1 2 + v2 2 + v) 2 (u 1 v 1 + u 2 v 2 + u v ) 2. Part (a) follows on expanding right hand sides of (7) and (8) and checking that y are equal. To prove Part (a) part follows (b), recall on expanding that right hand sides of (7) and (8) and checking that y are equal. To prove part (b), recall that u v = u v cos θ. u v = u v cos θ. Combining with part (a), we obtain Combining with part (a), we obtain u v 2 = u 2 v 2 u 2 v 2 cos 2 θ = u 2 v 2 sin 2 θ. u v 2 = u 2 v 2 u 2 v 2 cos 2 θ = u 2 v 2 sin 2 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 andconsider OC respectively. now a parallelogram If we imagine with vertices side O, OA A, to B, represent C. Suppose that base u of and v parallelogram, are represented so by that OA andbase OC has respectively. length u, If n we imagine height of side OA to parallelogram represent is given base of by v parallelogram, sin θ, as shown so in that diagram below: has length u, n height of parallelogram is given by v sin θ, as shown in diagram below: C B v sin θ v O θ u A u It follows from Proposition 4J that area of parallelogram is given by u v. We have proved It follows following from result. Proposition 4J that area of parallelogram is given by u v. We have proved following result. PROPOSITION 4K. Suppose that u, Then parallelogram with and as two of its sides PROPOSITION has area u v. 4K. Suppose that u, v R. Then parallelogram with u and v as two of its sides has area u v. We conclude this section by making remark on vector product of two vectors in We conclude this section by making a remark on vector product u v of two vectors in R Recall that vector product is perpendicular to both and v. Furrmore, it can be shown that. Recall that vector product is perpendicular to both u and v. Furrmore, it can be shown that direction of satisfies right hand rule, in sense that if we hold thumb on right hand direction of u v satisfies right hand rule, in sense that if we hold thumb on right hand outwards and close remaining four fingers, n thumb points towards v-direction when outwards and close remaining four fingers, n thumb points towards u v-direction when fingers point from u-direction towards v-direction. Also, we showed in Proposition 4J that fingers point from u-direction towards v-direction. Also, we showed in Proposition 4J that magnitude of u v depends only on norm of u and v and angle between two vectors. It Chapter 4 : Vectors page 12 of 24 Chapter Vectors page 12 of 24

Linear Algebra c WWLChen, 1982, 2006 magnitude of u v depends only on norm of u and v and angle between two vectors. It follows that 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 on same problem. 4.5. Scalar Triple Products Suppose that u, v, w R do not all lie on same plane. Consider parallelepiped with u, v, w as three of its edges. We are interested in calculating volume of this parallelepiped. Suppose that u, v and w are represented by OA, OB and OC respectively. Consider diagram below: v w P A u C O w v B By Proposition 4K, base of this parallelepiped, with O, B, C as three of vertices, has area v w. Next, note that if OP is perpendicular to base of parallelepiped, n OP is in direction of v w. If PAA is perpendicular to OP,, n height of parallelepiped is equal to norm of orthogonal projection of u on v w. In or words, parallelepiped has height proj v w u = u (v w) (v w) u (v w) v w 2 = v w. Hence volume of parallelepiped is given by V = u (v w). We We have have proved proved following following result. result. PROPOSITION 4L. Suppose that u, v, w R PROPOSITION 4L. Suppose that u, v,. Then parallelepiped with u, v and w as three of Then parallelepiped with u, and as three of its edges has volume u (v w). its edges has volume u (v w). Definition. Suppose that u, v, w R Definition. Suppose that u, v, R. Then u (v w) is called scalar triple product of u, v. Then u (v w) is called scalar triple product of u, v and and w. w. Remarks. (1) It follows immediately from Proposition 4L that three vectors in R Remarks. (1) It follows immediately from Proposition 4L that three vectors in are coplanar if and are coplanar if and only if ir scalar triple product is zero. only if ir scalar triple product is zero. (2) Note that (2) Note that ( ( ) ( ) ( )) ( ( v2 v ) ( ) ( )) u (v w) = (u 1, u 2, u ) det v1 v v2 v u (v w) =(u 1,u 2,u ) det, det v1 v v1 v, det, det 2 v1 v (9) w 2 w w 1 w, det w 1 w 2 ( ) w 2 w ( ) w 1 w ( w 1 ) w 2 ( v2 v ) ( ) ( ) = u 1 det v1 v v2 v = u 1 det u w v1 v u w 2 w 2 det v1 2 w 2 det v1 v + u w 1 w det 2 v + u w 1 w det w 1 w 2 w 1 w 2 = det u = det u 1 u 2 u v 1 vu 2 vu, (9) v v v w 1 w 2 w, w 1 w 2 w in view of cofactor expansion by row 1. in view of cofactor expansion by row 1. Chapter 4 : Vectors page 1 of 24 Chapter 4 : Vectors page 1 of 24

() It follows from identity (9) that u (v w) = v (w u) = w (u v). Note that each of determinants can be obtained from or two by twice interchanging two rows. Example 4.5.1. Suppose that u = (1, 0, 1), v = (2, 1, ) and w = (0, 1, 1). Then so that u, v and w are coplanar. u (v w) = det 1 0 1 2 1 = 0, 0 1 1 Example 4.5.2. The volume of parallelepiped with u = (1, 0, 1), v = (2, 1, 4) and w = (0, 1, 1) as three of its edges is given by u (v w) = det 1 0 1 2 1 4 = 1 = 1. 0 1 1 4.6. Application to Geometry in R In this section, we shall study lines and planes in R by using our results on vectors in R. Consider first of all a plane in R. Suppose that (x 1, y 1, z 1 ) R is a given point on this plane. Suppose furr that n = (a, b, c) is a vector perpendicular to this plane. Then for any arbitrary point (x, y, z) R on this plane, vector (x, y, z) (x 1, y 1, z 1 ) = (x x 1, y y 1, z z 1 ) joins one point on plane to anor point on plane, and so must be parallel to plane and hence perpendicular to n = (a, b, c). It follows that scalar product and so (a, b, c) (x x 1, y y 1, z z 1 ) = 0, a(x x 1 ) + b(y y 1 ) + c(z z 1 ) = 0. (10) If we write d = ax 1 + by 1 + cz 1, n (10) can be rewritten in form ax + by + cz + d = 0. (11) Equation (10) is usually called point-normal form of equation of a plane, while equation (11) is usually known as general form of equation of a plane. Example 4.6.1. Consider plane through point (2, 5, 7) and perpendicular to vector (, 5, 4). Here (a, b, c) = (, 5, 4) and (x 1, y 1, z 1 ) = (2, 5, 7). The equation of plane is given in point-normal form by (x 2) + 5(y + 5) 4(z 7) = 0, and in general form by x + 5y 4z + 7 = 0. Here d = 6 25 28 = 7. Chapter 4 : Vectors page 14 of 24

Example 4.6.2. Consider plane through points (1, 1, 1), (2, 2, 0) and (4, 6, 2). Then vectors (2, 2, 0) (1, 1, 1) = (1, 1, 1) and (4, 6, 2) (1, 1, 1) = (, 7, 1) join point (1, 1, 1) to points (2, 2, 0) and (4, 6, 2) respectively and are refore parallel to plane. It follows that vector product (1, 1, 1) (, 7, 1) = ( 6, 4, 10) is perpendicular to plane. The equation of plane is n given by 6(x 1) 4(y 1) 10(z 1) = 0, or x + 2y + 5z 10 = 0. Consider next a line in R. Suppose that (x 1, y 1, z 1 ) R is a given point on this line. Suppose furr that n = (a, b, c) is a vector parallel to this line. Then for any arbitrary point (x, y, z) R on this line, vector (x, y, z) (x 1, y 1, z 1 ) = (x x 1, y y 1, z z 1 ) joins one point on line to anor point on line, and so must be parallel to n = (a, b, c). It follows that re is some number λ R such that (x x 1, y y 1, z z 1 ) = λ(a, b, c), so that x = x 1 + aλ, y = y 1 + bλ, (12) z = z 1 + cλ, where λ is called a parameter. Suppose furr that a, b, c are all non-zero. Then, eliminating parameter λ, we obtain x x 1 a = y y 1 b = z z 1. (1) c Equations (12) are usually called parametric form of equations of a line, while equations (1) are usually known as symmetric form of equations of a line. Example 4.6.. Consider line through point (2, 5, 7) and parallel to vector (, 5, 4). Here (a, b, c) = (, 5, 4) and (x 1, y 1, z 1 ) = (2, 5, 7). The equations of line are given in parametric form by x = 2 + λ, y = 5 + 5λ, z = 7 4λ, and in symmetric form by x 2 = y + 5 5 = z 7 4. Chapter 4 : Vectors page 15 of 24

Example 4.6.4. Consider line through points (, 0, 5) and (7, 0, 8). Then a vector in direction of line is given by (7, 0, 8) (, 0, 5) = (4, 0, ). The equation of line is n given in parametric form by x = + 4λ, y = 0, z = 5 + λ, and in symmetric form by x 4 = z 5 and y = 0. Consider plane through three fixed points (x 1, y 1, z 1 ), (x 2, y 2, z 2 ) and (x, y, z ), not lying on same line. Let (x, y, z) be a point on plane. Then vectors (x, y, z) (x 1, y 1, z 1 ) = (x x 1, y y 1, z z 1 ), (x, y, z) (x 2, y 2, z 2 ) = (x x 2, y y 2, z z 2 ), (x, y, z) (x, y, z ) = (x x, y y, z z ), each joining one point on plane to anor point on plane, are all parallel to plane. Using vector product, we see that vector (x x 2, y y 2, z z 2 ) (x x, y y, z z ) is perpendicular to plane, and so perpendicular to vector (x x 1, y y 1, z z 1 ). It follows that scalar triple product (x x 1, y y 1, z z 1 ) ((x x 2, y y 2, z z 2 ) (x x, y y, z z )) = 0; in or words, det x x 1 y y 1 z z 1 x x 2 y y 2 z z 2 = 0. x x y y z z This is anor technique to find equation of a plane through three fixed points. Example 4.6.5. We return to plane in Example 4.6.2, through three points (1, 1, 1), (2, 2, 0) and (4, 6, 2). The equation is given by det x 1 y 1 z 1 x 2 y 2 z 0 = 0. x 4 y + 6 z 2 The determinant on left hand side is equal to 6x 4y 10z + 20. Hence equation of plane is given by 6x 4y 10z + 20 = 0, or x + 2y + 5z 10 = 0. We observe that calculation for determinant above is not very pleasant. However, technique can be improved in following way by making less reference to unknown point (x, y, z). Note that vectors (x, y, z) (x 1, y 1, z 1 ) = (x x 1, y y 1, z z 1 ), (x 2, y 2, z 2 ) (x 1, y 1, z 1 ) = (x 2 x 1, y 2 y 1, z 2 z 1 ), (x, y, z ) (x 1, y 1, z 1 ) = (x x 1, y y 1, z z 1 ), Chapter 4 : Vectors page 16 of 24

Linear Algebra c WWLChen, W W L 1982, 2006 2008 each joining one point on plane to anor point on plane, are all parallel to plane. Using vector product, we see that vector (x 2 x 1,y, y 2 y 1,z, z 2 z 1 ) (x x 1,y, y y 1,z, z z 1 ) is perpendicular to plane, and so perpendicular to vector (x x 1,y, y y 1,z, z z 1 ). It follows that scalar triple product (x x 1,y, y y 1,z, z z 1 ) ((x 2 x 1,y, y 2 y 1,z, z 2 z 1 ) (x x 1,y, y y 1,z, z z 1 )) = 0; in or words, det x x 1 y y 1 z z 1 x 2 x 1 y 2 y 1 z 2 z 1 =0. = x x 1 y y 1 z z 1 Example 4.6.6. We return to plane in Examples 4.6.2 and 4.6.5, through three points (1, 1, 1), (2, 2, 0) and (4, 6, 2). The equation is given by det x 1 y 1 z 1 2 1 2 2 1 0 0 1 =0. = 4 1 6 1 2 2 1 The determinant on left hand side is equal to det x 1 y 1 z 1 1 1 1 = 6(x 1) 4(y 1) 10(z 1) = 6x 4y 10z + 20. 7 1 Hence equation of plane is given by 6x 4y 10z + 20 = 0, or x +2y + +5z+ 10 = 0. We next consider problem of dividing a line segment in a given ratio. Suppose that x 1 and x 2 are two given points in R. We wish to divide line segment joining x 1 and x 2 internally in ratio α 1 : α 2, where α 1 and α 2 are positive real numbers. In or words, we wish to find point x on line segment joining x 1 and x 2 such that as shown in diagram below: x x 1 x x 2 = α 1, α 2 x 1 x x 2 x x 1 x x 2 Since x x 1 and x 2 x are both in same direction as x 2 x 1, we must have )=α α 2 (x x 1 ) = α 1 (x 2 x), or x = α 1x 2 + α 2 x 1. α 1 + α 2 We wish next to find point x on line joining x 1 and x 2, but not between x 1 and x 2, such that x x 1 x x 2 = α 1, α 2 Chapter 4 : Vectors page 17 of 24

Linear Algebra c W WWLChen, W L Chen, 1982, 2008 2006 2006 where α 1 and α 2 are positive real numbers, as shown in diagrams below for cases α 1 < α 2 and α 1 > α 2 respectively: x x 1 x 2 x 1 x 2 x Since x x 1 and x x 2 are in same direction as each or, we must have Since x x 1 and x x 2 are in same direction as each or, we must have α 2 (x x 1 )=α 1 (x x 2 ), or x = α 2 (x x 1 ) = α 1 (x x 2 ), or x = α α 1x 2 α 2 x 1 1x 2 x 1. α 1 α 2. α 1 α 2 Example 4.6.7. Let x 1 = (1, 2, ) and x 2 = (7, 11, 6). The point Example 4.6.7. Let x 1 = (1, 2, ) and x 2 = (7, 11, 6). The point x = 2x 2x 2 + x 1 2(7, 11, 6) + (1, 2, ) 2 1 = 2(7, 11, 6) (1, 2, ) = (5, 8, 5) 2 2+1 + 1 (5, 8, 5) divides divides line line segment segment joining joining (1, (1, 2, 2, ) ) and and (7, (7, 11, 11, 6) 6) internally internally in in ratio ratio 2 : 1, 1, whereas whereas point point x = 4x 4x 2 2x 2 2x 1 4(7, 11, 6) 2(1, 2, ) 1 = 4(7, 11, 6) 2(1, 2, ) = (1, 20, 9) 4 2 2 (1, 20, 9) satisfies satisfies x x 1 x x 2 = 4 2. Finally we turn our attention to question of finding distance of a plane from a given point. We shall prove following analogue of Proposition 4F. PROPOSITION 4F. The perpendicular distance D of a plane ax + by + cz + d = =0from 0 a point (x 0,,y y 0,,z z 0 ) is given by D = ax 0 + by 0 + cz 0 + d. a2 + b 2 + c 2 Proof. Consider following diagram: D P u n =(a, b, c) Q ax+by+cz=0 O Chapter Vectors page 18 of 24 Chapter 4 : Vectors page 18 of 24

Suppose that (x 1, x 2, x ) is any arbitrary point O on plane ax + by + cz + d = 0. For any or point (x, y, z) on plane ax + by + cz + d = 0, vector (x x 1, y y 1, z z 1 ) is parallel to plane. On or hand, (a, b, c) (x x 1, y y 1, z z 1 ) = (ax + by + cz) (ax 1 + by 1 + cz 1 ) = d + d = 0, so that vector n = (a, b, c), in direction OQ, is perpendicular to plane ax + by + cz + d = 0. Suppose next that point (x 0, y 0, z 0 ) is represented by point P in diagram. Then vector u = (x 0 x 1, y 0 y 1, z 0 z 1 ) is represented by OP, and OQ represents orthogonal projection projn u of u on vector n. Clearly perpendicular distance D of point (x 0, y 0, z 0 ) from plane ax + by + cz + d = 0 satisfies D = proj n u = u n n 2 n = (x 0 x 1, y 0 y 1, z 0 z 1 ) (a, b, c) a2 + b 2 + c 2 = ax 0 + by 0 + cz 0 ax 1 by 1 cz 1 a2 + b 2 + c 2 = ax 0 + by 0 + cz 0 + d a2 + b 2 + c 2 as required. A special case of Proposition 4F is when (x 0, y 0, z 0 ) = (0, 0, 0) is origin. perpendicular distance of plane ax + by + cz + d = 0 from origin is This show that d a2 + b 2 + c 2. Example 4.6.8. Consider plane x + 5y 4z + 7 = 0. The distance of point (1, 2, ) from plane is The distance of origin from plane is + 10 12 + 7 = 8 = 19 2. 9 + 25 + 16 50 5 7 9 + 25 + 16 = 7 50. Example 4.6.9. Consider also plane x+5y 4z 1 = 0. Note that this plane is also perpendicular to vector (, 5, 4) and is refore parallel to plane x+5y 4z+7 = 0. It is refore reasonable to find perpendicular distance between se two parallel planes. Note that perpendicular distance between two planes is equal to perpendicular distance of any point on x + 5y 4z 1 = 0 from plane x + 5y 4z + 7 = 0. Note now that (1, 2, ) lies on plane x + 5y 4z 1 = 0. It follows from Example 4.6.8 that distance between two planes is 19 2/5. 4.7. Application to Mechanics Let u = (u x, u y ) denote a vector in R 2, where components u x and u y are functions of an independent variable t. Then derivative of u with respect to t is given by ( du dt = dux dt, du ) y. dt Chapter 4 : Vectors page 19 of 24

Example 4.7.1. When discussing planar particle motion, we often let r = (x, y) denote position of a particle at time t. Then components x and y are functions of t. The derivative v = dr ( dx dt = dt, dy ) dt represents velocity of particle, and its derivative a = dv ( d 2 ) dt = x dt 2, d2 y dt 2 represents acceleration of particle. We often write r = r, v = v and a = a. Suppose that w = (w x, w y ) is anor vector in R 2. Then it is not difficult to see that d dw (u w) = u dt dt + du w. (14) dt Example 4.7.2. Consider a particle moving at constant speed along a circular path centred at origin. Then r = r is constant. More precisely, position vector r = (x, y) satisfies x 2 + y 2 = c 1, where c 1 is a positive constant, so that r r = (x, y) (x, y) = c 1. (15) On or hand, v = v is constant. More precisely, velocity vector v = ( dx dt, dy ) dt where c 2 is a positive constant, so that v v = satisfies ( ) 2 dx + dt ( ) 2 dy = c 2, dt ( dx dt, dy ) ( dx dt dt, dy ) = c 2. (16) dt Differentiating (15) and (16) with respect to t, and using identity (14), we obtain respectively r v = 0 and v a = 0. (17) Using properties of scalar product, we see that equations in (17) show that vector v is perpendicular to both vectors r and a, and so a must be in same direction as or opposite direction to r. Next, differentiating first equation in (17), we obtain r a + v v = 0, or r a = v 2 < 0. Let θ denote angle between a and r. Then θ = 0 or θ = 180. Since r a = r a cos θ, it follows that cos θ < 0, and so θ = 180. We also obtain ra = v 2, so that a = v 2 /r. This is a vector proof that for circular motion at constant speed, acceleration is towards centre of circle and of magnitude v 2 /r. Let u = (u x.u y, u z ) denote a vector in R, where components u x, u y and u z are functions of an independent variable t. Then derivative of u with respect to t is given by ( du dt = dux dt, du y dt, du ) z. dt Chapter 4 : Vectors page 20 of 24

Suppose that w = (w x, w y, w z ) is anor vector in R. Then it is not difficult to see that d dw (u w) = u dt dt + du w. (18) dt Example 4.7.. When discussing particle motion in -dimensional space, we often let r = (x, y, z) denote position of a particle at time t. Then components x, y and z are functions of t. The derivative v = dr ( dx dt = dt, dy dt, dz ) = (ẋ, ẏ, ż) dt represents velocity of particle, and its derivative represents acceleration of particle. a = dv ( d 2 ) dt = x dt 2, d2 y dt 2, d2 z dt 2 = (ẍ, ÿ, z) Example 4.7.4. For a particle of mass m, kinetic energy is given by Using identity (18), we have T = 1 2 m(ẋ2 + ẏ 2 + ż 2 ) = 1 2 m(ẋ, ẏ, ż) (ẋ, ẏ, ż) = 1 2mv v. dt dt = ma v = F v, where F = ma denotes force. On or hand, suppose that potential energy is given by V. Using knowledge on functions of several real variables, we can show that dv dt = V dx x dt + V dy y dt + V ( dz V z dt = x, V y, V ) v = V v, z where V = ( V x, V y, V ) z is called gradient of V. The law of conservation of energy says that T + V is constant, so that dt dt + dv dt = (F + V ) v = 0 holds for all vectors v, so that F(r) = V (r) for all vectors r. Example 4.7.5. If a force acts on a moving particle, n work done is defined as product of distance moved and magnitude of force in direction of motion. Suppose that a force F acts on a particle with displacement r. Then component of force in direction of motion is given by F u, where is a unit vector in direction of vector r. It follows that work done is given by ( ) r r F = F r. r u = r r Chapter 4 : Vectors page 21 of 24

Linear Linear Algebra Algebra c Wc WWLChen, L 1982, 1982, 2008 2006 For For instance, we we see see that that work work done done in moving moving a a particle along along a vector a vector r = r (, = (, 2, 2, 4) 4) with with applied applied force force F = F (2, = (2, 1, 1, 1) 1) is F is F r = r (2, = (2, 1, 1, 1) 1) (, (, 2, 2, 4) 4) = 12. = 12. Example 4.7.6. 4.7.6. We We can can also also resolve resolve a force a force into into components. Consider a weight a weight of of mass mass m m hanging from from ceiling ceiling on on a rope a rope as as shown shown in in picture picture below: below: m Here Here rope rope makes makes an an angle angle of of 60 60with with vertical. vertical. We We wish wish to to find find tension tension T on T on rope. rope. To To find find this, this, note note that that tension tension on on rope rope is a isforce, a force, and and we we have have following following picture picture of of forces: forces: T 1 60 T 2 magnitude mg The The force force T 1 Thas 1 has magnitude magnitude T T 1 1 = T =. T Let. Let z be z be a unit a unit vector vector pointing pointing vertically vertically upwards. upwards. Using Using scalar scalar products, products, we we see see that that component component of of force force T 1 Tin 1 in vertical vertical direction direction is is T T 1 z = T 1 z cos 60 = 1 2 T. 1 z = T 1 z cos 60 = 1 2 T. Similarly, Similarly, force force T 2 Thas 2 has magnitude magnitude T T 2 2 = T =, T and, and component component of of it in it in vertical vertical direction direction is is T T 2 z = T 2 z cos 60 = 1 2 T. 2 z = T 2 z cos 60 = 1 2 T. Since Since weight weight is stationary, is stationary, he he total total force force upwards upwards on on it is it is 1 1 2 T + 1 2T mg =0. Hence T = mg. 2 T + 1 2T mg = Hence T = mg. Chapter Chapter 4 : 4 Vectors : Vectors page page 22 of 2224 of 24

Problems for Chapter 4 1. For each of following pairs of vectors in R 2, calculate u + v, u v, u v and find angle between u and v: a) u = (1, 1) and v = ( 5, 0) b) u = (1, 2) and v = (2, 1) 2. For each of following pairs of vectors in R 2, calculate 2u 5v, u 2v, u v and angle between u and v (to nearest degree): a) u = (1, ) and v = ( 2, 1) b) u = (2, 0) and v = ( 1, 2). For two vectors u = (2, ) and v = (5, 1) in 2-dimensional euclidean space R 2, determine each of following: a) u v b) u c) u (u v) d) angle between u and u v 4. For each of following pairs of vectors in R, calculate u + v, u v, u v, find angle between u and v, and find a unit vector perpendicular to both u and v: a) u = (1, 1, 1) and v = ( 5, 0, 5) b) u = (1, 2, ) and v = (, 2, 1) 5. Find vectors v and w such that v is parallel to (1, 2, ), v + w = (7,, 5) and w is orthogonal to (1, 2, ). 6. Let ABCD be a quadrilateral. Show that quadrilateral obtained by joining midpoints of adjacent sides of ABCD is a parallelogram. [Hint: Let a, b, c and d be vectors representing four sides of ABCD.] 7. Suppose that u, v and w are vectors in R such that scalar triple porduct u (v w) 0. Let u = v w u (v w), v = w u u (v w), u v w = u (v w). a) Show that u u = 1. b) Show that u v = u w = 0. c) Use properties of scalar triple product to find v v and w w, as well as v u, v w, w u and w v. 8. Suppose that u, v, w, u, v and w are vectors in R such that u u = v v = w w = 1 and u v = u w = v u = v w = w u = w v = 0. Show that if u (v w) 0, n u = v w u (v w), v = w u u (v w), u v w = u (v w). 9. Suppose that u, v and w are vectors in R. a) Show that u (v w) = (u w)v (u v)w. b) Deduce that (u v) w = (u w)v (v w)u. 10. Consider three points P (2,, 1), Q(4, 2, 5) and R(1, 6, ). a) Find equation of line through P and Q. b) Find equation of plane perpendicular to line in part (a) and passing through R. c) Find distance between R and line in part (a). d) Find area of parallelogram with three points as vertices. e) Find equation of plane through three points. f) Find distance of origin (0, 0, 0) from plane in part (e). g) Are planes in parts (b) and (e) perpendicular? Justify your assertion. Chapter 4 : Vectors page 2 of 24

Linear Linear Algebra Algebra c Wc WWWLChen, L 1982, 1982, 20082006 11. 11. Consider Consider points points (1, (1, 2, ), 2, ), (0, (0, 2, 4) 2, 4) and and (2, (2, 1, ) 1, ) in Rin. R. a) a) Find Find area area of aofparallelogram a parallelogram with with se se points points as as three three of its of its vertices. vertices. b) b) Find Find perpendicular perpendicular distance distance between between (1, (1, 2, ) 2, and ) and line line passing passing through through (0, (0, 2, 4) 2, and 4) and (2, (2, 1, ). 1, ). 12. 12. Consider Consider points points (1, (1, 2, ), 2, ), (0, (0, 2, 4) 2, 4) and and (2, (2, 1, ) 1, ) in Rin. R. a) a) Find Find a vector a vector perpendicular perpendicular to to plane plane containing containing se se points. points. b) b) Find Find equation equation of this of this plane plane and and its its perpendicular distance distance from from origin. origin. c) c) Find Find equation equation of of line line perpendicular perpendicular to to this this plane plane and and passing passing through through point point (, (, 6, 9). 6, 9). 1. 1. Find Find equation equation of of plane plane through through points points (1, (1, 2, ), 2, ), (2, (2,,, 4) 4) and and (, (, 1, 2). 1, 2). 14. 14. Find Find equation equation of of plane plane through through points points (2, (2, 1, 1, 1), 1), (, (, 2, 1) 2, 1) and and ( 1, ( 1,, 2)., 2). 15. 15. Find Find volume volume of aof a parallelepiped with with points points (1, (1, 2, ), 2, ), (0, (0, 2, 4), 2, 4), (2, (2, 1, ) 1, and ) and (, (, 6, 9) 6, as 9) four as four of of its its vertices. vertices. 16. 16. Consider Consider a weight a weight of of mass mass m mhanging hanging from from ceiling ceiling supported supported by by two two ropes ropes as as shown shown in in picture picture below: below: m Here Here rope rope on on left left makes makes an angle angle of 45 of 45with wi vertical, vertical, while while rope rope on on right right makes makes an angle angle of 60 of 60with with vertical. vertical. Find Find tension tension on two two ropes. ropes. Chapter Chapter 4 : 4 Vectors : Vectors page page 24 of 24 24 of 24