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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 financial 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 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 √ ,√ 2 , where the first 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, √ , where the first 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 first study some algebra which is motivated by geometric considerations We then use the algebra later to better understand some problems in geometry Chapter : Vectors page of 24 cc Linear Algebra Algebra Linear WW WL L Chen, Chen, 1982, 1982, 2008 2006 W 2006 4.2 Vectors Vectors in in R R22 4.2 A vector vector on on the the plane plane R R22 can can be be described described as as an an ordered ordered pair pair u u= = (u (u11,, u u22), ), where where u u11,, u u22 ∈ ∈ R R A Definition Two Two vectors vectors u u= = (u (u11,, u u22)) and and v v= = (v (v11,, vv22)) in in R R22 are are said said to to be be equal, equal, denoted denoted by by u u= = v, v, if if Definition u11 = = vv11 and and u u22 = = vv22 u Definition For any two vectors u = (u11, u22) and v = (v11, v22) in R2 , we define their sum to be u + v = (u11, u22) + (v11, v22) = (u11 + v11, u22 + v22) −−→ −−→ 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: C u+v u+v A v v B u u The next next diagram diagram demonstrates demonstrates geometrically geometrically that that u u+ +v v= =v v+ + u: u: The u u D v v A C u+v u+v u u v v B PROPOSITION 4A 4A (VECTOR (VECTOR ADDITION) ADDITION) PROPOSITION 2 (a) For every u, v ∈ R , we have u +v v∈ ∈R R222 (a) For every u, v ∈ R , we have u + (b) For For every every u, u, v, v, w w∈ ∈R R22,, we we have have u u+ + (v (v + + w) w) = = (u (u + + v) v) + + w w (b) 2 2 (c) For every u ∈ R , we have u + = u, where = (0, 0) ∈ R (c) For every u ∈ R2, we have u + = u, where = (0, 0) ∈ R2 2 , there exists v ∈ R2 such that u + v = (d) For every u ∈ R (d) For every u ∈ R , 2there exists v ∈ R such that u + v = (e) For For every every u, u, v v∈ ∈R R22,, we we have have u u+ +v v= =v v+ + u u (e) Proof Write Write u u= = (u (u11,, u u22), ), v v= = (v (v11,, vv22)) and and w w= = (w (w11,, w w22), ), where where u u11,, u u22,, vv11,, vv22,, w w11,, w w22 ∈ ∈ R R To To check check Proof part (a), (a), simply simply note note that that u u11 + + vv11,, u u22 + + vv22 ∈ ∈ R R To To check check part part (b), (b), note note that that part u+ + (v (v + + w) w) = = (u (u11,, u u22)) + + (v (v11 + +w w11,, vv22 + +w w22)) = = (u (u11 + + (v (v11 + +w w11), ), u u22 + + (v (v22 + +w w22)) )) u = ((u + v ) + w , (u + v ) + w ) = (u + v , u + v ) + (w , w = ((u11 + v11) + w11, (u22 + v22) + w22) = (u11 + v11, u22 + v22) + (w11, w22)) = (u (u + + v) v) + + w w = Part (c) (c) is is trivial trivial Next, Next, if if v v= = (−u (−u11,, −u −u22), ), then then u u+ +v v= = 0, 0, giving giving part part (d) (d) To To check check part part (e), (e), note note Part that u + v = (u + v , u + v ) = (v + u , v + u ) = v + u that u + v = (u11 + v11, u22 + v22) = (v11 + u11, v22 + u22) = v + u Chapter 4 :: Vectors Vectors Chapter page 2 of of 24 24 page cc Linear Algebra Algebra Linear WW WL L Chen, Chen, 1982, 1982, 2008 2006 W Definition For For any any vector vector u u= = (u (u11,, u u22)) in in R R22 and and any any scalar scalar cc ∈ ∈ R, R, we we define define the the scalar scalar multiple multiple to to be be Definition cu = c(u1 , u2 ) = (cu1 , cu2 ) Example 4.2.1 Suppose that u = (2, 1) Then −2u = (−4, 2) Geometrically, if we represent the two −→ −−→ vectors u and −2u by OA and OB respectively, then we have the diagram below: A u O −2u B PROPOSITION 4B 4B (SCALAR (SCALAR MULTIPLICATION) MULTIPLICATION) PROPOSITION (a) For For every every cc ∈ ∈R R and and u u∈ ∈R R22,, we we have have cu cu ∈ ∈R R22 (a) (b) For For every every cc ∈ ∈R R and and u, u, v v∈ ∈R R2 ,, we we have have c(u c(u + + v) v) = = cu cu + + cv cv (b) (c) For For every every a, a, bb ∈ ∈R R and and u u∈ ∈R R22,, we we have have (a (a + + b)u b)u = = au au + + bu bu (c) (d) For For every every a, a, bb ∈ ∈R R and and u u∈ ∈R R22,, we we have have (ab)u (ab)u = = a(bu) a(bu) (d) 2 , we have 1u = u (e) For every u ∈ R (e) For every u ∈ R , we have 1u = u Proof Write Write u u= = (u (u1,, u u2)) and and v v= = (v (v1,, vv2), ), where where u u1,, u u2,, vv1,, vv2 ∈ ∈ R R To To check check part part (a), (a), simply simply note note Proof 2 2 that cu , cu ∈ R To check part (b), note that that cu11, cu22 ∈ R To check part (b), note that c(u + + v) v) = = c(u c(u1 + + vv1,, u u2 + + vv2)) = = (c(u (c(u1 + + vv1), ), c(u2 + v2 )) c(u 1 2 1 c(u2 + v2 )) = (cu (cu1 + + cv cv1,, cu cu2 + + cv cv2)) = = (cu (cu1,, cu cu2)) + + (cv (cv1,, cv cv2)) = = cu cu + + cv cv = 1 2 2 To check check part part (c), (c), note note that that To (a + + b)u b)u = = ((a ((a + + b)u b)u1,, (a (a + + b)u b)u2)) = = (au (au1 + + bu bu1,, au au2 + + bu bu2)) (a 1 2 = (au , au ) + (bu , bu ) = au + bu 2 = (au , au ) + (bu , bu ) = au + bu 2 To check check part part (d), (d), note note that that To (ab)u = = ((ab)u ((ab)u1,, (ab)u (ab)u2)) = = (a(bu (a(bu1), ), a(bu2 )) = a(bu1 , bu2 ) = a(bu) (ab)u a(bu2 )) = a(bu1 , bu2 ) = a(bu) Finally, to to check check part part (e), (e), note note that that 1u 1u = = (1u (1u1,, 1u 1u2)) = = (u (u1,, u u2)) = = u u Finally, 2 Definition For For any any vector vector u u= = (u (u1,, u u2)) in in R R22,, we we define define the the norm norm of of u u to to be be the the non-negative non-negative real real Definition number number u = = u u2121 + +u u2222 u Remarks (1) (1) The The norm norm of of aa vector vector is is simply simply its its magnitude magnitude or or length length The The definition definition follows follows from from the the Remarks famous theorem of Pythagoras famous theorem of Pythagoras (2) Suppose Suppose that that P P (u (u11,, u u22)) and and Q(v Q(v11,, vv22)) are are two two points points on on the the plane plane R R22 To To calculate calculate the the distance distance (2) d(P, Q) Q) between between the the two two points, points, we we can can first first find find aa vector vector from from P P to to Q Q This This is is given given by by (v (v11 −u −u11,, vv22 −u −u22) ) d(P, The distance distance d(P, d(P, Q) Q) is is then then the the norm norm of of this this vector, vector, so so that that The d(P, Q) Q) = = d(P, Chapter 4 :: Vectors Vectors Chapter (v11 − −u u11))22 + + (v (v22 − −u u22))22 (v page 3 of of 24 24 page cc Linear Algebra Algebra Linear WW WL L Chen, Chen, 1982, 1982, 2008 2006 W (3) It It is is not not difficult difficult to to see see that that for for any any vector vector u u∈ ∈R R22 and and any any scalar scalar cc ∈ ∈ R, R, we we have have cu cu = = |c| |c| u u (3) Definition Any Any vector vector u u∈ ∈R R22 satisfying satisfying u u = = 11 is is called called aa unit unit vector vector Definition Example 4.2.2 4.2.2 The The vector vector (3, (3, 4) 4) has has norm norm 5 Example Example 4.2.3 4.2.3 The The distance distance between between the the points points (6, (6, 3) 3) and and (9, (9, 7) 7) is is Example (9 − − 6) 6)22 + + (7 (7 − − 3) 3)22 = = 5 (9 Example 4.2.4 4.2.4 The The vectors vectors (1, (1, 0) 0) and and (0, (0, −1) −1) are are unit unit vectors vectors in in R R22 Example √ √ √ √ Example 4.2.5 4.2.5 The The unit unit vector vector in in the the direction direction of of the the vector vector (1, (1, 1) 1) is is (1/ (1/ 2, 2, 1/ 1/ 2) 2) Example Example 4.2.6 4.2.6 In In fact, fact, all all unit unit vectors vectors in in R R22 are are of of the the form form (cos (cos θ, θ, sin sin θ), θ), where where θθ ∈ ∈ R R Example Quite often, often, we we may may want want to to find find the the angle angle between between two two vectors vectors The The scalar scalar product product of of the the two two vectors vectors Quite 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 define the scalar product in two ways, one in terms of the angle between the two two vectors vectors and and the the other other not not in in terms terms of of this this angle, angle, and and show show that that the the two two definitions definitions are are in in fact fact the equivalent equivalent Definition Suppose that u = (u11 , u22 ) and v = (v11 , v22 ) are vectors in R22 , and that θ ∈ [0, π] represents the angle between them We define the scalar product u · v of u and v by (1) u v cos θ if u = and v = 0, if u = or v = u·v = (1) Alternatively, we write (2) u · v = u11v11 + u22v22 (2) The definitions (1) and (2) are clearly equivalent if u = or v = On the other hand, we have the following result PROPOSITION 4C Suppose that u = (u 1, u 2) and v = (v 1, v 2) are non-zero vectors in R2 , and that θ ∈ [0, π] represents the angle between them Then u v cos θ = u 1v + u 2v −→ −−→ Proof Geometrically, if we represent the two vectors u and v by OA and OB respectively, then the −−→ difference v − u is represented by AB as shown in the diagram below: B v−u v A O θ u By the the Law Law of of cosines, cosines, we we have have By 2 AB2 = = OA OA2 + + OB OB2 − − 2OA 2OA OB OB cos cos θ; θ; AB Chapter 4 :: Vectors Vectors Chapter page 44 of of 24 24 page c Linear Algebra W W L Chen, 1982, 2008 in other words, we have v−u = u + v −2 u v cos θ, so that u v cos θ = 21 ( u + v − v − u 2) = 21 (u21 + u22 + v12 + v22 − (v1 − u1 )2 − (v2 − u2 )2 ) = u1 v1 + u2 v2 as required Remarks (1) We say that two non-zero vectors in R2 are orthogonal if the angle between them is π/2 It follows immediately from the definition of the scalar product that two non-zero vectors u, v ∈ R2 are orthogonal if and only if u · v = (2) We can calculate the scalar product of any two non-zero vectors u, v ∈ R2 by the formula (2) and then use the formula (1) to calculate the angle between u and v √ √ Example 4.2.7 Suppose that u = ( 3, 1) and v = ( 3, 3) Then by the formula (2), we have u · v = + = Note now that u =2 and √ v = It follows from the formula (1) that cos θ = √ u·v = √ = , u v so that θ = π/6 √ √ Example 4.2.8 Suppose that u = ( 3, 1) and v = (− 3, 3) Then by the formula (2), we have u · v = It follows that u and v are orthogonal PROPOSITION 4D (SCALAR PRODUCT) Suppose that u, v, w ∈ R2 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 = if and only if u = Proof Write u = (u1 , u2 ), v = (v1 , v2 ) and w = (w1 , w2 ), where u1 , u2 , v1 , v2 , w1 , w2 ∈ R Part (a) is trivial To check part (b), note that u · (v + w) = u1 (v1 + w1 ) + u2 (v2 + w2 ) = (u1 v1 + u2 v2 ) + (u1 w1 + u2 w2 ) = u · v + u · w Part (c) is rather simple To check parts (d) and (e), note that u · u = u21 + u22 ≥ 0, and that equality holds precisely when u1 = u2 = Chapter : Vectors page of 24 cc Linear Algebra Algebra Linear WW WL L Chen, Chen, 1982, 1982, 2006 2006 W 2008 Consider the the diagram diagram below: below: Consider P P R R A A (3) (3) u u Q Q v v a a w w O O (3) −→ → − − → − − − → Here we we represent represent the the two two vectors vectors aa and and u u by by OA OA and and OP OP respectively respectively If If we we project project the the vector vector u u on on to to Here − − → − − → 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, we project project the the vector vector u u on on to to aa line line perpendicular perpendicular to to the the line line OA, OA, then then the the image image of of the the projection projection is is ifif we − − → − − → the vector v, represented by OR the vector v, represented by OR Definition In In the the notation notation of of the the diagram diagram (3), (3), the the vector vector w w is is called called the the orthogonal orthogonal projection projection of of the the Definition vector u on the vector a, and denoted by w = proj u vector u on the vector a, and denoted by w = projaaau PROPOSITION 4E 4E (ORTHOGONAL (ORTHOGONAL PROJECTION) PROJECTION) Suppose Suppose that that u, u, aa ∈ ∈R R222 Then Then PROPOSITION u·a projaau u = u · a2 a a proj a = a a − − → − → Remark Note Note that that the the component component of of u u orthogonal orthogonal to to a, a, represented represented by by − OR in the the diagram diagram (3), (3), is is Remark OR in u·a u− − proj projau u= =u u− − u · a2 a a u a aa Proof of of Proposition Proposition 4E 4E Note Note that that w w= = ka ka for for some some kk ∈ ∈ R R It It clearly clearly suffices suffices to to prove prove that that Proof u·a = u · a2 kk = aa It is is easy easy to to see see that that the the vectors vectors u u− −w w and and aa are are orthogonal orthogonal It It follows follows that that the the scalar scalar product product It (u − − w) w) ·· aa = = 0 In In other other words, words, (u (u − − ka) ka) ·· aa = = 0 Hence Hence (u u·a u·a = u·a = = u · a2 kk = aa ·· aa aa as required required as To end end this this section, section, we we shall shall apply apply our our knowledge knowledge gained gained so so far far to to find find aa formula formula that that gives gives the the To perpendicular distance distance of of aa point point (x (x00,, yy00)) from from aa line line ax ax + + by by + + cc = = 0 Consider Consider the the diagram diagram below: below: perpendicular P P D D n= = (a, (a, b) b) n ax+by+c=0 ax+by+c=0 u u Q Q O O Chapter : Vectors Chapter : Vectors page of 24 page of 24 c Linear Algebra W W L Chen, 1982, 2008 Suppose that (x1 , y1 ) is any arbitrary point O on the line ax + by + c = For any other point (x, y) on the line ax + by + c = 0, the vector (x − x1 , y − y1 ) is parallel to the line On the other hand, (a, b) · (x − x1 , y − y1 ) = (ax + by) − (ax1 + by1 ) = −c + c = 0, −−→ so that the vector n = (a, b), in the direction OQ, is perpendicular to the line ax + by + c = Suppose next that the point (x0 , y0 ) is represented by the point P in the diagram Then the vector −−→ −−→ u = (x0 − x1 , y0 − y1 ) is represented by OP , and OQ represents the orthogonal projection projn u of u on the vector n Clearly the perpendicular distance D of the point (x0 , y0 ) from the line ax + by + c = satisfies D = projn u = u·n |(x0 − x1 , y0 − y1 ) · (a, b)| |ax0 + by0 − ax1 − by1 | |ax0 + by0 + c| √ √ √ = = n = n a2 + b2 a2 + b2 a2 + b2 We have proved the following result PROPOSITION 4F The perpendicular distance D of a point (x0 , y0 ) from a line ax + by + c = is given by D= |ax0 + by0 + c| √ a2 + b2 Example 4.2.9 The perpendicular distance D of the point (5, 7) from the line 2x − 3y + = is given by D= |10 − 21 + 5| √ =√ 4+9 13 4.3 Vectors in R3 In this section, we consider the same problems as in Section 4.2, but in 3-space R3 Any reader who feels confident may skip this section A vector on the plane R3 can be described as an ordered triple u = (u1 , u2 , u3 ), where u1 , u2 , u3 ∈ R Definition Two vectors u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) in R3 are said to be equal, denoted by u = v, if u1 = v1 , u2 = v2 and u3 = v3 Definition For any two vectors u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) in R3 , we define their sum to be u + v = (u1 , u2 , u3 ) + (v1 , v2 , v3 ) = (u1 + v1 , u2 + v2 , u3 + v3 ) Definition For any vector u = (u1 , u2 , u3 ) in R3 and any scalar c ∈ R, we define the scalar multiple to be cu = c(u1 , u2 , u3 ) = (cu1 , cu2 , cu3 ) The following two results are the analogues of Propositions 4A and 4B The proofs are essentially similar Chapter : Vectors page of 24 c Linear Algebra W W L Chen, 1982, 2008 PROPOSITION 4A’ (VECTOR ADDITION) (a) For every u, v ∈ R3 , we have u + v ∈ R3 (b) For every u, v, w ∈ R3 , we have u + (v + w) = (u + v) + w (c) For every u ∈ R3 , we have u + = u, where = (0, 0, 0) ∈ R3 (d) For every u ∈ R3 , there exists v ∈ R3 such that u + v = (e) For every u, v ∈ R3 , we have u + v = v + u PROPOSITION 4B’ (SCALAR MULTIPLICATION) (a) For every c ∈ R and u ∈ R3 , we have cu ∈ R3 (b) For every c ∈ R and u, v ∈ R3 , we have c(u + v) = cu + cv (c) For every a, b ∈ R and u ∈ R3 , we have (a + b)u = au + bu (d) For every a, b ∈ R and u ∈ R3 , we have (ab)u = a(bu) (e) For every u ∈ R3 , we have 1u = u Definition For any vector u = (u1 , u2 , u3 ) in R3 , we define the norm of u to be the non-negative real number u = u21 + u22 + u23 Remarks (1) Suppose that P (u1 , u2 , u3 ) and Q(v1 , v2 , v3 ) are two points in R3 To calculate the distance d(P, Q) between the two points, we can first find a vector from P to Q This is given by (v1 − u1 , v2 − u2 , v3 − u3 ) The distance d(P, Q) is then the norm of this vector, so that d(P, Q) = (v1 − u1 )2 + (v2 − u2 )2 + (v3 − u3 )2 (2) It is not difficult to see that for any vector u ∈ R3 and any scalar c ∈ R, we have cu = |c| u Definition Any vector u ∈ R3 satisfying u = 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 Example 4.3.3 The vectors (1, 0, 0) and (0, −1, 0) are unit vectors in R3 √ √ Example 4.3.4 The unit vector in the direction of the vector (1, 0, 1) is (1/ 2, 0, 1/ 2) The theory of scalar products can be extended to R3 is the natural way Definition Suppose that u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) are vectors in R3 , and that θ ∈ [0, π] represents the angle between them We define the scalar product u · v of u and v by u·v = u v cos θ if u = and v = 0, if u = or v = (4) Alternatively, we write u · v = u1 v1 + u2 v2 + u3 v3 (5) The definitions (4) and (5) are clearly equivalent if u = or v = On the other hand, we have the following analogue of Proposition 4C The proof is similar Chapter : Vectors page of 24 cc Linear Linear Algebra Algebra W WW WL L Chen, Chen, 1982, 1982, 2006 2008 PROPOSITION PROPOSITION 4C’ 4C’ Suppose Suppose that that u u= = (u (u11 ,, u u22 ,, u u33 )) and and v v= = (v (v11 ,, vv22 ,, vv33 )) are are non-zero non-zero vectors vectors in in R R3 ,, and and that that θθ ∈ ∈ [0, [0, π] π] represents represents the the angle angle between between them them Then Then u u v v cos cos θθ = =u u11 vv11 + +u u22 vv22 + +u u33 vv33 Remarks Remarks (1) (1) We We say say that that two two non-zero non-zero vectors vectors in in R R3 are are orthogonal orthogonal if if the the angle angle between between them them is is π/2 π/2 It It follows follows immediately immediately from from the the definition definition of of the the scalar scalar productthat productthat two two non-zero non-zero vectors vectors u, u, v v∈ ∈R R3 are are orthogonal orthogonal if if and and only only if if u u ·· v v= = 0 (2) We can calculate the scalar product of any two non-zero vectors u, v ∈ R33 by the formula (5) and then use the formula (4) to calculate the angle between u and v √ Example 4.3.5 Suppose that u = (2, 0, 0) and v = (1, 1, 2) Then by the formula (5), we have u · v = Note now that u = and v = It follows from the formula (4) that cos θ = u·v = = , u v so that θ = π/3 Example 4.3.6 Suppose that u = (2, 3, 5) and v = (1, 1, −1) Then by the formula (5), we have u · v = It follows that u and v are orthogonal The following result is the analogue of Proposition 4D The proof is similar PROPOSITION 4D’ (SCALAR PRODUCT) Suppose that u, v, w ∈ R3 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 = if and only if u = Suppose now that a and u are two vectors in R3 Then since two vectors are always coplanar, we can draw the following diagram which represents the plane they lie on: P R A (6) u Q v a w O (6) Note Note that that this this diagram diagram is is essentially essentially the the same same as as the the diagram diagram (3), (3), the the only only difference difference being being that that while while 2 , the diagram (6) only shows part of R3 As before, we represent the diagram (3) shows the whole of R the diagram (3) shows the whole of R , the diagram (6) only shows part of R As before, we represent − − − → −→ → − − → the the the two two vectors vectors a a and and u u by by OA OA and and OP OP respectively respectively If If we we project project the vector vector u u on on to to the the line line OA, OA, then then − −− −→ → On the other hand, if we project the the image of the projection is the vector w, represented by OQ the image of the projection is the vector w, represented by OQ On the other hand, if we project the vector aa line vector u u on on to to − line perpendicular perpendicular to to the the line line OA, OA, then then the the image image of of the the projection projection is is the the vector vector v, v, − → − − → represented by OR represented by OR Definition Definition In In the the notation notation of of the the diagram diagram (6), (6), the the vector vector w w is is called called the the orthogonal orthogonal projection projection of of the the vector u on the vector a, and denoted by w = proj u vector u on the vector a, and denoted by w = proja u a Chapter Chapter 4 :: Vectors Vectors page page 9 of of 24 24 c Linear Algebra W W L Chen, 1982, 2008 The following result is the analogue of Proposition 4E The proof is similar PROPOSITION 4E’ (ORTHOGONAL PROJECTION) Suppose that u, a ∈ R3 Then proja u = u·a a a −−→ Remark Note that the component of u orthogonal to a, represented by OR in the diagram (6), is u − proja u = u − u·a a a 4.4 Vector Products In this section, we shall discuss a product of vectors unique to R3 The idea of vector products has wide applications in geometry, physics and engineering, and is motivated by the wish to find 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 fingers, then the fingers point from the x-direction towards the y-direction, while the thumb points towards the z-direction Alternatively, if we imagine Columbus had never lived and that the earth were flat, then taking the x-direction as east and the y-direction as north, then the z-direction is upwards! We shall frequently use the three vectors i = (1, 0, 0), j = (0, 1, 0) and k = (0, 0, 1) in R3 Definition Suppose that u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) are two vectors in R3 Then the vector product u × v is defined by the determinant  i u × v = det  u1 v1  k u3  v3 j u2 v2 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 u × v = det = det u2 v2 u2 v2 u3 v3 u3 v3 i − det , − det u1 v1 u1 v1 u3 v3 u3 v3 j + det , det u1 v1 u1 v1 u2 v2 k u2 v2 = (u2 v3 − u3 v2 , u3 v1 − u1 v3 , u1 v2 − u2 v1 ) We shall first of all show that the vector product u × v is orthogonal to both u and v Chapter : Vectors page 10 of 24 c Linear Algebra W W L Chen, 1982, 2008 PROPOSITION 4G Suppose that u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) are two vectors in R3 Then (a) u · (u × v) = 0; and (b) v · (u × v) = Proof Note first of all that u · (u × v) = (u1 , u2 , u3 ) · det = u1 det u2 v2 u3 v3 u2 v2 u3 v3 − u2 det u1 v1 , − det u1 v1 u3 v3 in view of cofactor expansion by row On the other  u1 u2 det  u1 u2 v1 v2 u3 v3 + u3 det , det u1 v1 u1 v1 u2 v2  u1 = det  u1 v1 u2 v2  u3 u3  , v3 u2 u2 v2 hand, clearly  u3 u3  = v3 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   i j k −1 2 u × v = det  −1  = det , − det , det 3 −1 = (−2, 4, 3) Note that (1, −1, 2) · (−2, 4, 3) = and (3, 0, 2) · (−2, 4, 3) = PROPOSITION 4H (VECTOR PRODUCT) Suppose that u, v, w ∈ R3 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; and (f ) u × u = Proof Write u = (u1 , u2 , u3 ), v = (v1 , v2 , v3 ) and w = (w1 , w2 , w3 ) To check part (a), note that     i j k i j k det  u1 u2 u3  = − det  v1 v2 v3  v1 v v3 u1 u2 u3 To check part (b), note that  i j det  u1 u2 v1 + w1 v2 + w2   k i u3  = det  u1 v3 + w v1 Part (c) is similar To check part (d), note that    i j k i c det  u1 u2 u3  = det  cu1 v1 v2 v3 v1 To check parts (e) and (f), note that   i j k u × = det  u1 u2 u3  = 0 0 j cu2 v2 j u2 v2   k i u3  + det  u1 v3 w1   k i cu3  = det  u1 v3 cv1  and i u × u = det  u1 u1 j u2 cv2 j u2 u2 j u2 w2  k u3  w3  k u3  cv3  k u3  = u3 as required Chapter : Vectors page 11 of 24 c c Linear Algebra Linear Algebra W W L Chen, 1982, 2008 W W L 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 this, we shall first establish the following result To this, we shall first establish the following result PROPOSITION 4J Suppose that u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) are non-zero vectors in R33, PROPOSITION 4J Suppose that u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) are non-zero vectors in R , and that θ ∈ [0, π] represents the angle between them Then and that θ ∈2[0, π] represents the angle between them Then (a) u × v = u 22 v 22 − (u · v)22; and (a) u × v = u v − (u · v) ; and (b) u × v = u v sin θ (b) u × v = u v sin θ Proof Note that Proof Note that = (u2 v3 − u3 v2 )22 + (u3 v1 − u1 v3 )22 + (u1 v2 − u2 v1 )22 = (u2 v3 − u3 v2 ) + (u3 v1 − u1 v3 ) + (u1 v2 − u2 v1 ) (7) (u · v)2 = (u2 + u2 + u2 )(v + v + v ) − (u1 v1 + u2 v2 + u3 v3 )2 − (u · v)2 = (u121 + u22 + u323 )(v112 + v222 + v332 ) − (u1 v1 + u2 v2 + u3 v3 )2 (8) u×v u×v (7) 2 and and (8) u u 2 v v 2− 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 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 22 = u u×v = u Part (b) follows Part (b) follows 2 v v − u − u 2 v v cos22 θ = u cos θ = u 2 v v sin22 θ sin θ −→ 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 that u and v are represented by OA and OC −−→ respectively If we imagine the side OA to represent the base of the parallelogram, so that and base OC has respectively we imagine the ofside represent the base of the length u If, then the height theOA the to parallelogram is given bythev parallelogram, sin θ, as shownsointhat the the base has length u , then the height of the the parallelogram is given by v sin θ, as shown in the diagram below: diagram below: C B v v sin θ O θ u A u It follows from Proposition 4J that the area of the parallelogram is given by u × v We have proved It fromresult Proposition 4J that the area of the parallelogram is given by u × v We have proved thefollows following the following result PROPOSITION 4K Suppose that u, v ∈ R33 Then the parallelogram with u and v as two of its sides PROPOSITION has area u × v 4K Suppose that u, v ∈ R Then the parallelogram with u and v as two of its sides has area u × v We conclude this section by making a remark on the vector product u × v of two vectors in R3 We conclude this section by making a remark on the vector product u × v of two vectors in R3 Recall that the vector product is perpendicular to both u and v Furthermore, it can be shown that the Recall that the vector product is perpendicular to both u and v Furthermore, it can be shown that the direction of u × v satisfies the right hand rule, in the sense that if we hold the thumb on the right hand direction of u × v satisfies 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 the u × v-direction when outwards and close the remaining four fingers, then the thumb points towards the u × v-direction when the fingers point from the u-direction towards the v-direction Also, we showed in Proposition 4J that the fingers point from the u-direction towards the v-direction Also, we showed in Proposition 4J that the magnitude of u × v depends only on the norm of u and v and the angle between the two vectors It Chapter : Vectors Chapter : Vectors page 12 of 24 page 12 of 24 c c Linear Algebra Linear Algebra W W L Chen, 1982, 2008 W W L Chen, 1982, 2006 the magnitude of u × v depends only on the norm of u and v 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 an important consideration in physics and engineering, where we may use different coordinate systems on the same problem 4.5 Scalar Triple Products Suppose that u, v, w ∈ R3 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 → − −→ and − −→ respectively Consider the diagram below: and w are represented by − OA, OB OC v×w P A C u w O v B By Proposition 4K, the base of this parallelepiped, with O, B, C as three of the vertices, has area v×w −→ is in the direction of Next, note that if OP is perpendicular to the base of the parallelepiped, then − OP v × w If P A 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 projv×w u = u · (v × w) |u · (v × w)| (v × w) = v×w v×w Hence the volume of the parallelepiped is given by V V = = |u |u ·· (v (v × × w)| w)| We We have have proved proved the the following following result result PROPOSITION 4L Suppose that u, v, w ∈ R33 Then the parallelepiped with u, v and w as three of PROPOSITION 4L Suppose that u, v, w ∈ R Then the parallelepiped with u, v and w as three of its edges has volume |u · (v × w)| its edges has volume |u · (v × w)| Definition Suppose that u, v, w ∈ R33 Then u · (v × w) is called the scalar triple product of u, v Definition Suppose that u, v, w ∈ R Then u · (v × w) is called the scalar triple product of u, v and and w w Remarks (1) It follows immediately from Proposition 4L that three vectors in R33 are coplanar if and Remarks (1) It follows immediately from Proposition 4L that three vectors in R 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 u · (v × w) = (u1 , u2 , u3 ) · det v22 v33 w w u · (v × w) = (u1 , u2 , u3 ) · det (9) w2 w3 v v = u1 det v22 v33 − u2 det w w = u1 det − u det  w2 w3   u1 u2 u3  = det  vu11 vu22 vu33  , = det  w v11 w v22 w v33  , w1 w2 w3 in view of cofactor expansion by row in view of cofactor expansion by row Chapter : Vectors Chapter : Vectors v v v v , − det v11 v33 , det v11 v22 w w w w , − det , det w1 w3 w1 w2 v1 v3 v1 v2 v33 + u3 det w v22 v11 w v11 w w + u3 det w1 w3 w1 w2 (9) page 13 of 24 page 13 of 24 c Linear Algebra W W L Chen, 1982, 2008 (3) It follows from identity (9) that u · (v × w) = v · (w × u) = w · (u × v) 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  u · (v × w) = det  0 1   = 0, 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  |u · (v × w)| = det  0 1   = | − 1| = 1 4.6 Application to Geometry in R3 In this section, we shall study lines and planes in R3 by using our results on vectors in R3 Consider first of all a plane in R3 Suppose that (x1 , y1 , z1 ) ∈ R3 is a given point on this plane Suppose further that n = (a, b, c) is a vector perpendicular to this plane Then for any arbitrary point (x, y, z) ∈ R3 on this plane, the vector (x, y, z) − (x1 , y1 , z1 ) = (x − x1 , y − y1 , z − z1 ) 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) · (x − x1 , y − y1 , z − z1 ) = 0, and so a(x − x1 ) + b(y − y1 ) + c(z − z1 ) = (10) If we write −d = ax1 + by1 + cz1 , then (10) can be rewritten in the form ax + by + cz + d = (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 (x1 , y1 , z1 ) = (2, −5, 7) The equation of the plane is given in point-normal form by 3(x − 2) + 5(y + 5) − 4(z − 7) = 0, and in general form by 3x + 5y − 4z + 37 = Here −d = − 25 − 28 = −37 Chapter : Vectors page 14 of 24 c Linear Algebra W W L Chen, 1982, 2008 Example 4.6.2 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 by −6(x−1)−4(y−1)−10(z−1) = 0, or 3x + 2y + 5z − 10 = Consider next a line in R3 Suppose that (x1 , y1 , z1 ) ∈ R3 is a given point on this line Suppose further that n = (a, b, c) is a vector parallel to this line Then for any arbitrary point (x, y, z) ∈ R3 on this line, the vector (x, y, z) − (x1 , y1 , z1 ) = (x − x1 , y − y1 , z − z1 ) 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 λ ∈ R such that (x − x1 , y − y1 , z − z1 ) = λ(a, b, c), so that x = x1 + aλ, y = y1 + bλ, (12) z = z1 + cλ, where λ is called a parameter Suppose further that a, b, c are all non-zero Then, eliminating the parameter λ, we obtain y − y1 z − z1 x − x1 = = a b c (13) 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 Example 4.6.3 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 (x1 , y1 , z1 ) = (2, −5, 7) The equations of the line are given in parametric form by x = + 3λ, y = −5 + 5λ, z = − 4λ, and in symmetric form by x−2 y+5 z−7 = =− Chapter : Vectors page 15 of 24 c Linear Algebra W W L Chen, 1982, 2008 Example 4.6.4 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 = + 4λ, y = 0, z = + 3λ, and in symmetric form by x−3 z−5 = and y = Consider the plane through three fixed points (x1 , y1 , z1 ), (x2 , y2 , z2 ) and (x3 , y3 , z3 ), not lying on the same line Let (x, y, z) be a point on the plane Then the vectors (x, y, z) − (x1 , y1 , z1 ) = (x − x1 , y − y1 , z − z1 ), (x, y, z) − (x2 , y2 , z2 ) = (x − x2 , y − y2 , z − z2 ), (x, y, z) − (x3 , y3 , z3 ) = (x − x3 , y − y3 , z − z3 ), 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 (x − x2 , y − y2 , z − z2 ) × (x − x3 , y − y3 , z − z3 ) is perpendicular to the plane, and so perpendicular to the vector (x − x1 , y − y1 , z − z1 ) It follows that the scalar triple product (x − x1 , y − y1 , z − z1 ) · ((x − x2 , y − y2 , z − z2 ) × (x − x3 , y − y3 , z − z3 )) = 0; in other words,  x − x1 det  x − x2 x − x3 y − y1 y − y2 y − y3  z − z1 z − z2  = z − z3 This is another technique to find the equation of a plane through three fixed points Example 4.6.5 We return to the plane in Example and (4, −6, 2) The equation is given by  x−1 y−1 det  x − y − x−4 y+6 4.6.2, through the three points (1, 1, 1), (2, 2, 0)  z−1 z −  = z−2 The determinant on the left hand side is equal to −6x − 4y − 10z + 20 Hence the equation of the plane is given by −6x − 4y − 10z + 20 = 0, or 3x + 2y + 5z − 10 = 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) − (x1 , y1 , z1 ) = (x − x1 , y − y1 , z − z1 ), (x2 , y2 , z2 ) − (x1 , y1 , z1 ) = (x2 − x1 , y2 − y1 , z2 − z1 ), (x3 , y3 , z3 ) − (x1 , y1 , z1 ) = (x3 − x1 , y3 − y1 , z3 − z1 ), Chapter : Vectors page 16 of 24 cc Linear Linear Algebra Algebra W WW WL L Chen, Chen, 1982, 1982, 2006 2008 each each joining joining one one point point on on the the plane plane to to another another point point on on the the plane, plane, are are all all parallel parallel to to the the plane plane Using Using the the vector vector product, product, we we see see that that the the vector vector (x (x22 − − xx11,, yy22 − − yy11,, zz22 − − zz11)) × × (x (x33 − − xx11,, yy33 − − yy11,, zz33 − − zz11)) is is perpendicular perpendicular to to the the plane, plane, and and so so perpendicular perpendicular to to the the vector vector (x (x − − xx11,, yy − − yy11,, zz − − zz11) ) It It follows follows that that the the scalar scalar triple triple product product (x (x − − xx11,, yy − − yy11,, zz − − zz11)) ·· ((x ((x22 − − xx11,, yy22 − − yy11,, zz22 − − zz11)) × × (x (x33 − − xx11,, yy33 − − yy11,, zz33 − − zz11)) )) = = 0; 0; in in other other words, words,     xx − − xx11 yy − − yy11 zz − − zz11  xx22 − = det det  − xx11 yy22 − − yy11 zz22 − − zz11  = 0 xx33 − − xx11 yy33 − − yy11 zz33 − − zz11 Example Example 4.6.6 4.6.6 We We return return to to the the plane plane in in Examples Examples 4.6.2 4.6.2 and and 4.6.5, 4.6.5, through through the the three three points points (1, (1, 1, 1, 1), 1), (2, (2, 2, 2, 0) 0) and and (4, (4, −6, −6, 2) 2) The The equation equation is is given given by by     xx − − 11 yy − − 11 zz − − 11  22 − = det det  − 11 22 − − 11 00 − − 11  = 0 44 − − 11 −6 −6 − − 11 22 − − 11 The The determinant determinant on on the the left left hand hand side side is is equal equal to to     xx − − 11 yy − − 11 zz − − 11  11 = det 11 −1 det  = −6(x −6(x − − 1) 1) − − 4(y 4(y − − 1) 1) − − 10(z 10(z − − 1) 1) = = −6x −6x − − 4y 4y − − 10z 10z + + 20 20 −1  33 −7 11 −7 Hence Hence the the equation equation of of the the plane plane is is given given by by −6x −6x − − 4y 4y − − 10z 10z + + 20 20 = = 0, 0, or or 3x 3x + + 2y 2y + + 5z 5z − − 10 10 = = 0 We We next next consider consider the the problem problem of of dividing dividing aa line line segment segment in in aa given given ratio ratio Suppose Suppose that that xx11 and and xx22 are are two two given given points points in in R R33 We We wish wish to to divide divide the the line line segment segment joining joining xx11 and and xx22 internally internally in in the the ratio ratio α α11 :: α α22,, where where α α11 and and α α22 are are positive positive real real numbers numbers In In other other words, words, we we wish wish to to find find the the point point xx on on the the line line segment segment joining joining xx11 and and xx22 such such that that − xx11 α11 xx − α = ,, = x − x α x − x22 α22 as shown shown in in the the diagram diagram below: below: as | x1 | x | x2 ←−−−−−− x − x1 −−−−−−→←−− x − x2 −−→ Since Since xx − − xx11 and and xx22 − − xx are are both both in in the the same same direction direction as as xx22 − − xx11,, we we must must have have α α22(x (x − − xx11)) = =α α11(x (x22 − − x), x), or or xx = = α α11xx22 + +α α22xx11 α α11 + +α α22 We We wish wish next next to to find find the the point point xx on on the the line line joining joining xx11 and and xx22,, but but not not between between xx11 and and xx22,, such such that that xx − α − xx11 α11 = = ,, xx − α − xx22 α22 Chapter Chapter 44 :: Vectors Vectors page page 17 17 of of 24 24 cc Linear Algebra Algebra Linear WW WL L Chen, Chen, 1982, 1982, 2006 2006 W 2008 where α α1 and and α α2 are are positive positive real real numbers, numbers, as as shown shown in in the the diagrams diagrams below below for for the the cases cases α α1 < α respectively: α1 > α2 respectively: || x x || x x11 || x x22 || x x11 || x x22 || x x Since x x− −x x11 and and x x− −x x22 are are in in the the same same direction direction as as each each other, other, we we must must have have Since Since x − x1 and x − x2 are in the same direction as each other, we must have α11x x22 − −α α22x x11 α α22(x (x − −x x11)) = =α α11(x (x − −x x22), ), or x= =α α or x α 1x 2− x1 α − α α2 (x − x1 ) = α1 (x − x2 ), or x = α11 − α22 α1 − α2 Example 4.6.7 4.6.7 Let Let x x1 = = (1, (1, 2, 2, 3) 3) and and x x2 = = (7, (7, 11, 11, 6) 6) The The point point Example Example 4.6.7 Let x11 = (1, 2, 3) and x22 = (7, 11, 6) The point 2x22 + +x x11 2(7, 11, 11, 6) 6) + + (1, (1, 2, 2, 3) 3) 2x 2(7, x= = 2x = 2(7, = (5, (5, 8, 8, 5) 5) 11, 6) + (1, 2, 3) + x1 = x = + = (5, 8, 5) x= 2+1 = 2+1 33 divides the the line line segment segment joining joining (1, (1, 2, 2, 3) 3) and and (7, (7, 11, 11, 6) 6) internally internally in in the the ratio ratio : 1, 1, whereas whereas the the point point divides divides the line segment joining (1, 2, 3) and (7, 11, 6) internally in the ratio 22 :: 1, whereas the point 4x22 − − 2x 2x11 4(7, 11, 11, 6) 6) − − 2(1, 2(1, 2, 2, 3) 3) 4x 4(7, 11, 6) − 2(1, 2, 3) x= = 4x = 4(7, = (13, (13, 20, 20, 9) 9) − 2x1 x = = x = 44 − = = (13, 20, 9) −2 − 22 22 satisfies satisfies x− −x x11 x = = x − x x − x22 22 Finally we we turn turn our our attention attention to to the the question question of of finding finding the the distance distance of of aa plane plane from from aa given given point point Finally We shall prove the following analogue of Proposition 4F We shall prove the following analogue of Proposition 4F PROPOSITION 4F’ 4F’ The The perpendicular perpendicular distance distance D D of of aa plane plane ax ax + + by by + + cz cz + + dd = = 00 from from aa point point PROPOSITION (x , y , z ) is given by (x000, y000, z000) is given by |ax0 + + by + cz + d| √ by00 + cz00 + d| D= = |ax0√ D + b22 + c22 aa2 + b +c Proof Consider Consider the the following following diagram: diagram: Proof P P D D n= = (a, (a, b, b, c) c) n u u Q Q ax+by+cz=0 ax+by+cz=0 Chapter : Vectors Chapter : Vectors O O page 18 of 24 page 18 of 24 c Linear Algebra W W L Chen, 1982, 2008 Suppose that (x1 , x2 , x3 ) is any arbitrary point O on the plane ax + by + cz + d = For any other point (x, y, z) on the plane ax + by + cz + d = 0, the vector (x − x1 , y − y1 , z − z1 ) is parallel to the plane On the other hand, (a, b, c) · (x − x1 , y − y1 , z − z1 ) = (ax + by + cz) − (ax1 + by1 + cz1 ) = −d + d = 0, −−→ so that the vector n = (a, b, c), in the direction OQ, is perpendicular to the plane ax + by + cz + d = Suppose next that the point (x0 , y0 , z0 ) is represented by the point P in the diagram Then the vector −−→ −−→ u = (x0 − x1 , y0 − y1 , z0 − z1 ) is represented by OP , and OQ represents the orthogonal projection projn u of u on the vector n Clearly the perpendicular distance D of the point (x0 , y0 , z0 ) from the plane ax + by + cz + d = satisfies |(x0 − x1 , y0 − y1 , z0 − z1 ) · (a, b, c)| u·n √ n = n a2 + b2 + c2 |ax0 + by0 + cz0 + d| |ax0 + by0 + cz0 − ax1 − by1 − cz1 | √ √ = = a2 + b2 + c2 a2 + b2 + c2 D = projn u = as required A special case of Proposition 4F’ is when (x0 , y0 , z0 ) = (0, 0, 0) is the origin This show that the perpendicular distance of the plane ax + by + cz + d = from the origin is √ |d| a2 + b2 + c2 Example 4.6.8 Consider the plane 3x + 5y − 4z + 37 = The distance of the point (1, 2, 3) from the plane is √ 38 19 |3 + 10 − 12 + 37| √ =√ = + 25 + 16 50 The distance of the origin from the plane is √ |37| 37 =√ + 25 + 16 50 Example 4.6.9 Consider also the plane 3x+5y−4z−1 = Note that this plane is also perpendicular to the vector (3, 5, −4) and is therefore parallel to the plane 3x+5y−4z+37 = It is therefore reasonable to find 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 − = from the plane 3x + 5y − 4z + 37 = Note now that (1, 2, 3) lies on the plane 3x + 5y − 4z − = It follows √ from Example 4.6.8 that the distance between the two planes is 19 2/5 4.7 Application to Mechanics Let u = (ux , uy ) denote a vector in R2 , where the components ux and uy are functions of an independent variable t Then the derivative of u with respect to t is given by du = dt Chapter : Vectors dux , dt dt page 19 of 24 c 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 v= dr = dt dx dy , dt dt represents the velocity of the particle, and its derivative a= d2 x d2 y , dt2 dt2 dv = dt represents the acceleration of the particle We often write r = r , v = v and a = a Suppose that w = (wx , wy ) is another vector in R2 Then it is not difficult to see that d dw du (u · w) = u · + · w dt dt dt (14) Example 4.7.2 Consider a particle moving at constant speed along a circular path centred at the origin Then r = r is constant More precisely, the position vector r = (x, y) satisfies x2 + y = c1 , where c1 is a positive constant, so that r · r = (x, y) · (x, y) = c1 (15) On the other hand, v = v is constant More precisely, the velocity vector v= dx dy , dt dt dx dt satisfies + dy dt = c2 , where c2 is a positive constant, so that v·v = dx dy , dt dt · dx dy , dt dt = c2 (16) Differentiating (15) and (16) with respect to t, and using the identity (14), we obtain respectively r·v =0 and v · a = (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, differentiating the first equation in (17), we obtain r · a + v · v = 0, or r · a = −v < Let θ denote the 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 = 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 of magnitude v /r Let u = (ux uy , uz ) denote a vector in R3 , where the components ux , uy and uz are functions of an independent variable t Then the derivative of u with respect to t is given by du = dt Chapter : Vectors dux duz , , dt dt dt page 20 of 24 c Linear Algebra W W L Chen, 1982, 2008 Suppose that w = (wx , wy , wz ) is another vector in R3 Then it is not difficult to see that d dw du (u · w) = u · + · w dt dt dt (18) Example 4.7.3 When discussing particle motion in 3-dimensional space, we often let r = (x, y, z) denote the position of a particle at time t Then the components x, y and z are functions of t The derivative v= dx dy dz , , dt dt dt dr = dt = (x, ˙ y, ˙ z) ˙ represents the velocity of the particle, and its derivative a= dv = dt d2 x d2 y d2 z , , dt2 dt2 dt2 = (¨ x, y¨, z¨) represents the acceleration of the particle Example 4.7.4 For a particle of mass m, the kinetic energy is given by T = 12 m(x˙ + y˙ + z˙ ) = 21 m(x, ˙ y, ˙ z) ˙ · (x, ˙ y, ˙ z) ˙ = 12 mv · v Using the identity (18), we have dT = ma · v = F · v, dt where F = ma denotes the force On the other hand, suppose that the potential energy is given by V Using knowledge on functions of several real variables, we can show that ∂V dx ∂V dy ∂V dz dV = + + = dt ∂x dt ∂y dt ∂z dt ∂V ∂V ∂V , , ∂x ∂y ∂z · v = ∇V · v, where ∂V ∂V ∂V , , ∂x ∂y ∂z ∇V = is called the gradient of V The law of conservation of energy says that T + V is constant, so that dV dT + = (F + ∇V ) · v = dt dt 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, then the work done is defined as the product of the distance moved and the magnitude of the force in the direction of motion Suppose that a force F acts on a particle with displacement r Then the component of the force in the direction of the motion is given by F · u, where u= r r is a unit vector in the direction of the vector r It follows that the work done is given by r Chapter : Vectors F· r r = F · r page 21 of 24 L 1982, 2006 WW WChen, L Chen, Chen, 1982, 2006 c cW W L 1982, 2008 Linear Algebra Linear Algebra Linear Algebra instance, that work done moving a particle along a vector r= −2, with applied ForFor instance, wewe seesee that thethe work done in in moving a particle along a vector r= (3,(3, −2, 4) 4) with applied force F= −1, is ·Fr ·=r = −1, · (3, −2, force F= (2,(2, −1, 1) 1) is F (2,(2, −1, 1) 1) · (3, −2, 4) 4) == 12.12 Example 4.7.6 also resolve a force into components.Consider Consider a weight mass hanging Example 4.7.6 WeWe cancan also resolve a force into components a weight of of mass mm hanging from ceiling a rope shown picture below: from thethe ceiling onon a rope as as shown in in thethe picture below: m ◦◦ Here rope makes angle with vertical wish find tension rope Here thethe rope makes anan angle of of 60◦60with thethe vertical WeWe wish to to find thethe tension T T onon thethe rope ToTo find this, note that the tension on the rope is a force, and we have the following picture of forces: find this, note that the tension on the rope is a force, and we have the following picture of forces: T11 60◦◦ T22 magnitude mg The force magnitudeT1T11= = Let z be a unit vector pointing vertically upwards Using scalar 11 has The force T1Thas magnitude T TLet z be a unit vector pointing vertically upwards Using scalar products, we see that the component of the force T in the vertical direction is 1 products, we see that the component of the force T1 in the vertical direction is ◦◦ 11 z =T1T11z zcoscos T1T· 11z·= 60◦60= = 22 T T Similarly, force magnitudeT2T22= = , and component it in vertical direction 22 has Similarly, thethe force T2Thas magnitude T ,Tand thethe component of of it in thethe vertical direction is is ◦◦ 11 z =T2T22z zcoscos T2T· 22z·= 60◦60= = 22 T T 11 T + 11 T − mg = Hence T = mg Since weight is stationary, total force upwards it is Since thethe weight is stationary, he he total force upwards onon it is T22+ T22− mg = Hence T = mg Chapter :: Vectors Chapter Vectors Chapter : 44Vectors page of page 22 24 of 24 24 page 22 22 of c Linear Algebra W W L Chen, 1982, 2008 Problems for Chapter For each of the following pairs of vectors in R2 , calculate u + 3v, u · v, u − v and find the angle between u and v: a) u = (1, 1) and v = (−5, 0) b) u = (1, 2) and v = (2, 1) For each of the following pairs of vectors in R2 , calculate 2u − 5v, u − 2v , u · v and the angle between u and v (to the nearest degree): a) u = (1, 3) and v = (−2, 1) b) u = (2, 0) and v = (−1, 2) For the two vectors u = (2, 3) and v = (5, 1) in the 2-dimensional euclidean space R2 , determine each of the following: a) u − v b) u c) u · (u − v) d) the angle between u and u − v For each of the following pairs of vectors in R3 , calculate u + 3v, u · v, u − v , find the 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, 3) and v = (3, 2, 1) Find vectors v and w such that v is parallel to (1, 2, 3), v + w = (7, 3, 5) and w is orthogonal to (1, 2, 3) Let ABCD be a quadrilateral Show that the quadrilateral obtained by joining the midpoints of adjacent sides of ABCD is a parallelogram [Hint: Let a, b, c and d be vectors representing the four sides of ABCD.] Suppose that u, v and w are vectors in R3 such that the scalar triple porduct u · (v × w) = Let u = v×w , u · (v × w) v = w×u , u · (v × w) w = u×v u · (v × w) a) Show that u · u = b) Show that u · v = u · w = c) Use the properties of the scalar triple product to find v · v and w · w, as well as v · u, v · w, w · u and w · v Suppose that u, v, w, u , v and w are vectors in R3 such that u · u = v · v = w · w = and u · v = u · w = v · u = v · w = w · u = w · v = Show that if u · (v × w) = 0, then u = v×w , u · (v × w) v = w×u , u · (v × w) w = u×v u · (v × w) Suppose that u, v and w are vectors in R3 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 the three points P (2, 3, 1), Q(4, 2, 5) and R(1, 6, −3) a) Find the equation of the line through P and Q b) Find the equation of the plane perpendicular to the line in part (a) and passing through R c) Find the distance between R and the line in part (a) d) Find the area of the parallelogram with the three points as vertices e) Find the equation of the plane through the three points f) Find the distance of the origin (0, 0, 0) from the plane in part (e) g) Are the planes in parts (b) and (e) perpendicular? Justify your assertion Chapter : Vectors page 23 of 24 c W W LWChen, L Chen, 1982, 2006 c W 1982, 2008 Linear Algebra Linear Algebra Consider points 2, 3), 2, 4) 1, 3) in3 R3 11.11 Consider thethe points (1, (1, 2, 3), (0, (0, 2, 4) andand (2, (2, 1, 3) in R Find area a parallelogram with these points three of its vertices a) a) Find thethe area of aofparallelogram with these points as as three of its vertices b) Find the perpendicular distance between (1, 2, 3) and the line passing through 2, and 4) and 1, 3) b) Find the perpendicular distance between (1, 2, 3) and the line passing through (0, (0, 2, 4) (2, (2, 1, 3) Consider points 2, 3), 2, 4) 1, 3) in3 R3 12.12 Consider thethe points (1, (1, 2, 3), (0, (0, 2, 4) andand (2, (2, 1, 3) in R Find a vector perpendicular to the plane containing these points a) a) Find a vector perpendicular to the plane containing these points b) Find the equation of this plane and its perpendicular distance from origin b) Find the equation of this plane and its perpendicular distance from thethe origin c) Find the equation of the line perpendicular to this plane and passing through point 6, 9) c) Find the equation of the line perpendicular to this plane and passing through thethe point (3, (3, 6, 9) Find equation of the plane through points 2, −3), (−3, 1, 2) 13.13 Find thethe equation of the plane through thethe points (1, (1, 2, −3), (2, (2, −3,−3, 4) 4) andand (−3, 1, 2) Find equation of the plane through points 2, −1) (−1, 3, 2) 14.14 Find thethe equation of the plane through thethe points (2, (2, −1,−1, 1), 1), (3, (3, 2, −1) andand (−1, 3, 2) Find volume a parallelepiped with points 2, 3), 2, 4), 1, and 3) and 6, as 9) four as four 15.15 Find thethe volume of aofparallelepiped with thethe points (1, (1, 2, 3), (0, (0, 2, 4), (2, (2, 1, 3) (3, (3, 6, 9) of of its vertices its vertices Consider a weight mass hanging from ceiling supported ropes shown 16.16 Consider a weight of of mass mm hanging from thethe ceiling supported by by twotwo ropes as as shown in in thethe picture below: picture below: m ◦ Here rope makes angle of ◦45with with vertical, while rope right makes Here thethe rope on on thethe leftleft makes an an angle of 45 thethe vertical, while thethe rope on on thethe right makes ◦ ◦ angle of 60with with vertical Find tension ropes an an angle of 60 thethe vertical Find thethe tension on on thethe twotwo ropes Chapter : Vectors Chapter : 4Vectors page 2424 of 24 page 24 of [...]... − xx11,, yy22 − − yy11,, zz22 − − zz11)) × × (x (x33 − − xx11,, yy33 − − yy11,, zz33 − − zz11)) )) = = 0; 0; in in other other words, words,     xx − − xx11 yy − − yy11 zz − − zz11  xx22 − = det det  − xx11 yy22 − − yy11 zz22 − − zz11  = 0 0 xx33 − − xx11 yy33 − − yy11 zz33 − − zz11 Example Example 4. 6.6 4. 6.6 We We return return to to the the plane plane in in Examples Examples 4. 6.2 4. 6.2... 11, 11, 6) 6) internally internally in in the the ratio ratio 2 : 1, 1, whereas whereas the the point point divides divides the line segment joining (1, 2, 3) and (7, 11, 6) internally in the ratio 22 :: 1, whereas the point 4x22 − − 2x 2x11 4( 7, 11, 11, 6) 6) − − 2(1, 2(1, 2, 2, 3) 3) 4x 4( 7, 11, 6) − 2(1, 2, 3) x= = 4x = 4( 7, = (13, (13, 20, 20, 9) 9) 2 − 2x1 x = = x = 44 − = = (13, 20, 9) −2 2 4. .. plane in in Examples Examples 4. 6.2 4. 6.2 and and 4. 6.5, 4. 6.5, through through the the three three points points (1, (1, 1, 1, 1), 1), (2, (2, 2, 2, 0) 0) and and (4, (4, −6, −6, 2) 2) The The equation equation is is given given by by     xx − − 11 yy − − 11 zz − − 11  22 − = det det  − 11 22 − − 11 00 − − 11  = 0 0 44 − − 11 −6 −6 − − 11 22 − − 11 The The determinant determinant on on the the... left hand hand side side is is equal equal to to     xx − − 11 yy − − 11 zz − − 11  11 = det 11 −1 det  = −6(x −6(x − − 1) 1) − − 4( y 4( y − − 1) 1) − − 10(z 10(z − − 1) 1) = = −6x −6x − − 4y 4y − − 10z 10z + + 20 20 −1  33 −7 11 −7 Hence Hence the the equation equation of of the the plane plane is is given given by by −6x −6x − − 4y 4y − − 10z 10z + + 20 20 = = 0, 0, or or 3x 3x + + 2y 2y + +... x1 α − α α2 (x − x1 ) = α1 (x − x2 ), or x = 11 − α22 α1 − α2 Example 4. 6.7 4. 6.7 Let Let x x1 = = (1, (1, 2, 2, 3) 3) and and x x2 = = (7, (7, 11, 11, 6) 6) The The point point Example Example 4. 6.7 Let x11 = (1, 2, 3) and x22 = (7, 11, 6) The point 2x22 + +x x11 2(7, 11, 11, 6) 6) + + (1, (1, 2, 2, 3) 3) 2x 2(7, x= = 2x = 2(7, = (5, (5, 8, 8, 5) 5) 11, 6) + (1, 2, 3) 2 + x1 = x = 2 + 1 3 = (5,... − xx11,, yy22 − − yy11,, zz22 − − zz11)) × × (x (x33 − − xx11,, yy33 − − yy11,, zz33 − − zz11)) is is perpendicular perpendicular to to the the plane, plane, and and so so perpendicular perpendicular to to the the vector vector (x (x − − xx11,, yy − − yy11,, zz − − zz11) ) It It follows follows that that the the scalar scalar triple triple product product (x (x − − xx11,, yy − − yy11,, zz − − zz11))... (x22 − − x), x), or or xx = = α α11xx22 + +α α22xx11 α 11 + +α α22 We We wish wish next next to to find find the the point point xx on on the the line line joining joining xx11 and and xx22,, but but not not between between xx11 and and xx22,, such such that that xx − α − xx11 11 = = ,, xx − α − xx22 α22 Chapter Chapter 44 :: Vectors Vectors page page 17 17 of of 24 24 cc Linear Algebra Algebra Linear... 2  u1 u2 u3  = det  vu11 vu22 vu33  , = det  w v11 w v22 w v33  , w1 w2 w3 in view of cofactor expansion by row 1 in view of cofactor expansion by row 1 Chapter 4 : Vectors Chapter 4 : Vectors v v v v , − det v11 v33 , det v11 v22 w w w w 1 3 1 2 , − det , det w1 w3 w1 w2 v1 v3 v1 v2 v33 + u3 det w v22 v11 w v11 w w + u3 det w1 w3 w1 w2 (9) page 13 of 24 page 13 of 24 c Linear Algebra W W L Chen,... line Example 4. 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 (x1 , y1 , z1 ) = (2, −5, 7) The equations of the line are given in parametric form by x = 2 + 3λ, y = −5 + 5λ, z = 7 − 4 , and in symmetric form by x−2 y+5 z−7 = =− 3 5 4 Chapter 4 : Vectors page 15 of 24 c Linear Algebra W W L Chen, 1982, 2008 Example 4. 6 .4 Consider... 5) − 4( z − 7) = 0, and in general form by 3x + 5y − 4z + 37 = 0 Here −d = 6 − 25 − 28 = −37 Chapter 4 : Vectors page 14 of 24 c Linear Algebra W W L Chen, 1982, 2008 Example 4. 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, ... equation is is given given by by     xx − − 11 yy − − 11 zz − − 11  22 − = det det  − 11 22 − − 11 00 − − 11  = 0 44 − − 11 −6 −6 − − 11 22 − − 11 The The determinant determinant on on the... (1, 2, 3) and (7, 11, 6) internally in the ratio 22 :: 1, whereas the point 4x22 − − 2x 2x11 4( 7, 11, 11, 6) 6) − − 2(1, 2(1, 2, 2, 3) 3) 4x 4( 7, 11, 6) − 2(1, 2, 3) x= = 4x = 4( 7, = (13, (13,...   xx − − 11 yy − − 11 zz − − 11  11 = det 11 −1 det  = −6(x −6(x − − 1) 1) − − 4( y 4( y − − 1) 1) − − 10(z 10(z − − 1) 1) = = −6x −6x − − 4y 4y − − 10z 10z + + 20 20 −1  33 −7 11 −7 Hence