This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Dr. Doom Lin Alg MATH1850/2050 Sections 3.45 Page 1 of 7 Some stuff from Section 3.4. Cross Product Deﬁnition: CrossProduct. Ifu = (u1,u2,u3) and V = (711,712,713) are vectors in 3—space7 then the cross product u x V is the vector
deﬁned by U. >< V = (U203 — U3027 U301 — U1037 U102 — U201) or in determinant notation m. _,
CC“) _ , k 1,: (1,0,0)
1 A
_ U2 U3 U1 U3 U1 U2 _ J PM“)
uXV— ,— 7 — U1 U2 U3 1::(Q0l)
02 03 01 03 01 02 ’ '
711 U2 U3 The cross product u >< V is a vector that is perpendicular to both
u and V. The cross product is only deﬁned for vectors in R3. Example: With u = (1,1,0) and V = (—1,2, 1) .‘ ". “
\. k
a» K 3 = l 1 O = QMCu "' Q“,ch +qt3C\1 —1 '2. I
\0 N11 1am; 711+ei*‘\;;\f : =(\,~\,%) Dr. Doom Lin Alg MATH1850/2050 Theorem 3.4.2: Properties of the Cross Product. If u7 V7 and W are any vectors in R3 and k is any scalar7 then:
(a) u><V=—(V><u) (b) u >< (V + W) = (u >< V) + (u X W) “(smurlw
(c) (u+V) ><W=(u><w)+(V><w)
(d) k(u >< V) = (ku) >< V = u >< (kV) (e)u><0=0><u=0 ; A  5 k A e .5 .5
uxu— . L. =
\u‘ub‘b :ObFOsKO 0
UN M1 “'5 Sections 3.45 Page 2 of 7 Dr. Doom Lin Alg MATHl850/2050 Sections 3.4,5 Page 3 of 7 Deﬁnition: Unit Vectors and Standard Unit Vectors. The
vector u is a unit vector if = 1. In other words, a unit vector is
any vector with unit length. The standard unit vectors in R” are
the vectors with one component equal 1 and all others components
equal zero. For example, in R3, the standard unit vectors are i: (1,0,0), j: (0,1,0), k: (0,0,1) Examples: F09. W." snows unit utctots A2E
(1,0550), (o,\,o,o) , (o,o,l,o), (0,093) 8I e1 63 6H
1'11 2 (1/27 _1/27 1'01“: 8‘Q+éi +42 = 4, So 61 'K A ou‘w W602. A .  J..."
2. V = (1,—3,2) “Chis VH :{E u (S M A 01m Wae’wum V ‘5. 3. w = (6, —3,5) = Q? 4315?.
See page 147 in text for the determinant form of cross product using the standard unit vectors, and how to get the direction of the cross
product from the “right—hand rule”. Dr. Doom Lin Alg MATHl850/2050 Sections 3.45 Page 4 of 7 Theorem 3.4.3: Area of a Parallelogram. If u and V are
vectors in R3, then Hu >< vH is equal to the area of the parallelogram
determined by u and V. This gives a geometric interpretation for the cross product. See page
147 and 148 in text. Am, [Lam],
_‘.J;[”“’ aw use (minder“Stag /
/ / /
6' / 49' Sir See Theorem 3.4.4 on page 150 of text for a geometric interpretation
of determinants for vectors in R2 and R3. ,/ / I};
<\ f / _> _s A
‘ ‘4” / \lo‘klw '(w‘ull
.s
V 7‘ x N_}/ L‘ V g
/ r’ W\ wt d}
/ I, k\ “1' u"
/ V\ V1. v‘ Dr. Doom Lin Alg MATH1850/2050 Sections 3.45 Page 5 of 7 Changing coordinate system. The cross product u >< V does not depend on the coordinate system
you choose to write u and V in. Example: Z Dr. Doom Lin Alg MATHl850/2050 Sections 3.4,5 Page 6 of 7 Some stuff from Section 3.5. Lines and Planes in
3Space Planes in 3Space. A line in R2 can be speciﬁed by giving its
slope and one of its points. Similarly, one can specify a plane in R3 by
giving its inclination and specifying one of it’s points. A convenient
method for describing the inclination of a plane is to specify a nonzero
vector, called a normal, that is perpendicular to the plane. Suppose we want to ﬁnd the equation of the plane passing through the
point X0 = (500, yo, ZQ) and having the nonzero vector r1 = (a, b, c) as
normal. Then, the plane that we want consists of the points (50, y, z)
such that the vector (50 — :60, y — yo, z — ZQ) is orthogonal to the
normal vector r1. That is, the plane can be described as: n°($_$07y_y07Z_Z0):O "‘*°'l°“‘\°’cz°
R:(QO¢;C) (anh'zJ is A 9011!! 135165 PUMP. ll a(a: — 560) + My — yo) + C(Z — zo) = O [email protected]= 0
We call this the pointnormal form of the equation of a plane. Examples: x. 7. 2.
1. Find an equation of the plane passing through the point (—1, l, 2) with normal vector r1 = (l, l, —1)
Q lo C 2. Find an equation of the plane that contains the points (1, 1, 1), (2, —1, 1), (1, 3, 0) @ usme Qxl‘o ﬁz+ol=o , x4 La z,+ol——o , due.» —\ + \ —1.*al~‘0 .
MS: l x*§':12 *Q'Eo
@ TA“ 1‘: (23">\)"Llal,l) = (‘3Z)°)lg\ a): C\)\>l) “(‘3320) =E So all=2. 'MCE vux‘xjv. \ k $(_1 4 4') So Qx+loia+Cz+al=o 7
t —L 0 anon *2x—«3—Zz+ol=0 41:,
° '7' ‘ PLoeiA (I,\,\) ) 41:45
S" ’Zx ’21 kg :0
Dr. Doom Lin Alg MATHl850/2050 Sections 3.45 Page 7 of 7 See Example 2 page 158 of text book for the geometric interpretation of solutions to systems of equations (we already discussed this when
we did Chapter 1). Lines in 3Space. Suppose that l is the line in R3 through the
point X0 = (:60, yo, zo) and parallel to a nonzero vector V = (a, b, c).
The line I consists of those points (56, y, z) for which (a: — 560, y —
yo, z — zo) is parallel to V. That is7 for which there is a scalar t such
that (a: — :60, y — yo, z — ZQ) = tv 2 (ta, 1519, 750) If we write this out in parametric form7 we have (x ‘5 Z ) x=x0+ta, y=y0+tb, z=z0+tc
(1‘1‘1)=(x.,\1.,2.)+ 1: Cake)
for —oo < t < 00. These are called the parametric equations for the line I. Examples: X0 V. 20
1. Find parametric equations of the line containing the point (—1, 2, 3) and having as direction vector (1 = (l, O, 2)
q, ‘o c p
2. FindCequation of a plane containing the point (1, 2, —l) and par—
allel to the 2 lines given by: [1: (:6,y,z) = (1,—1,1)+t(1,0,2) linen) lg: (cc,y,z) = (0,1,2) + 8(3, 1,1) 358,50
><= ‘o‘d‘o‘ X=~\+k CD ‘3’?“ 5° {pa 0L wapﬁ(—\,1.1>)+k0,o,2). z= 294k 2:3«115 So USE. 2{i‘tc \l 3 k I o 7— : (‘1)g)
3 l \ a b QX’VDtaTCZﬁcl‘o 2x*S‘3*7‘*°L=O Ye“? (M Yak}; (
—24(0—\+oL=O {5° 49*“7 AUS’. i'Zx—kSaVZ7c'ol ...
View
Full Document
 Spring '09
 MihaiBeligan
 Linear Algebra, Algebra

Click to edit the document details