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MA 26100 (13)
Chapter 12&13

# MA261 Chapters 12 & 13 Notes.docx

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School
Department
Mathematics
Course
MA 26100
Professor
David Catlin
Semester
Fall

Description
MA261 Notes Chapter 12: Vectors and the Geometry of Space Section 12.1 Three-Dimensional Coordinate Systems • x,y,z regular coordinates o O(0,0,0) o P(x,y,z) √x +y +z 2 o |OP|= • Example 1: Equation of a sphere with radius 2, centered at (1,-1, 1/2) 2 (x−1 + (y− −1 )+ z− 1 =4 o ( ) 2 2 2 2 7 o x −2x+ y +2y+z −z= 4 Section 12.2 Vectors • • • • • • Example: a=<5,-12>, b= o Find a + b  <5+-1,-12+-4>=<4,-16> o Find 2a+3b  <2*5+2(-1),3(-12)+3(-4)>=<7,-36> • Special Vectors o i=<1,0,0> o j=<0,1,0> o k=<0,0,1> Section 12.3 The Dot Product • • • • Example: a=i+2j-2k, b=4i-3k o Find the angle between the vectors:  a dot b=1(4)+2(0)+(-2)(-3)=10  |a|=sqrt(1+2^2+(-2)^2)=3  |b|=sqrt(4^2+(-3)^2)=5  Theta=cos^(-1)(10/(3*5)=cos^(-1)(2/3) • Two vectors a and b are orthogonal if and only if a dot b = 0 • • Example: Use vectors to decide whether the triangle with vertices P(1,-3,-2), Q(2,0,-4), and R(6,-2,-5) is right-angled. o PQ=<1,3,-2> o PR=<5,1,-3> o PQ dot PR=(1)(5)+(3)(1)+(-2)(-3)=5+3+6=14, doesn’t = 0 o QP = o QR = <4,-2,-1> o QP dot QR = (-1)(4)+(-3)(-2)+(2)(-1)=-4+6-2=0 o So the angle at vertex Q is a right angle Section 12.4 The Cross Product • • The vector a x b is orthogonal to both a and b • • Two nonzero vectors a and b are parallel if and only if a x b = 0 • • The volume of the parallelepiped determined by the vectors a, b, and c is the magnitude of their scalar triple product: V=|a dot (b x c)| Section 12.5 Equations of Lines and Planes r=r 0tv •  a vector equation of L o Each value of parameter t gives the position vector r of a point on L • Two vectors are equal if corresponding components are equal o o Called parametric equations • Example: (a) Find a vector equation and parametric equations for the line that passes through the point (5,1,3) and is parallel to the vector i+4j-2k. (b) Find two other points on the line r0=¿5,1,3>¿ o (a) =5i+j+3k and v=i+4j-2k  r=(5i+j+3k)+t(i+4j-2k)=(5+t)i+(1+4t)j+(3-2t)k o (b) choosing the parameter value t=1 gives x=6,y=5,z=1 so (6,5,1) is a point on the line. t=-1 gives the point (4,-3,5) • Example: Find parametric equations and symmetric equations for the line. The line through (5,1,0) and perpendicular to both i+j and j+k o v=(i+j) x (j+k)=i-j+k o parametric equations: x=5+t, y=1-t, z=t o symmetric equations: x-5 = -(y-1)= z • Symmetric equations o • Example: (a) Find parametric equations and symmetric equations of the line that passes through the
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