lecture5compl-09[1]

lecture5compl-09[1] - Math 18.02(Spring 2009 Lecture 5...

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Unformatted text preview: Math 18.02 (Spring 2009): Lecture 5 Planes. Parametric equations of curves and lines February 12 Reading Material: From Simmons: 17.1 and 17.2. Last time: Square Systems. Word problem. How many solutions? Equations of planes Today: Planes. Parametric equations of curves and lines 2 Equations of Planes In this section we recall the analytic description of a plane through a point P0 = (mmyo, :0) and perpendicular to a vector N = (a, b, c). This plane is described by all points P : (any, 2) such that _, _. the vector P0P = (.13 a moyy — yo, 2 — 20) is perpendicular to the given vector N: Since —-—> N- 0P:o(\$—\$U)+b(y—yol+C(3*ZU)=0! it follows that the (analytic) equation for our plane is all—Hiol-l'bly—yol +007 —50) =0- (2-1) Exercise 1. Find a plane containing P1 = (1,2,3),P2 = (1,4,4) and P3 = (0,2,6). - —> w—> . —. w ——3 Metllod: Find a vector N normal to two vectors P1132 and P1133 by computing N : P1132 x P1P3. Use N and any of P1, P2, P3 to write the appropriate equation like in (2.1). Now the details: P1: (1,2,3) 13—91132: (0,211) P2: (1,4,4) _, P3: (0,2,5) ea: (4,0,3) 3 j l: P1Pg><P1P3= 0 2 1 = 6ie1j+2t=(6,—1,2)=N. —1 0 3 To write equation (2.1) we take the vector A? we just computed and the point P2 (1,4,4) for example1 and we get 6(3—1)—1(y—4)+2(z—4)=0 that can also be written as 61' — y + 2:: 2 10. Let’s now check if 131,132 and P3 belong to this plane: P1 : 6 — 2 + 6 z 10 P2 : 6 — 4 + 8 : 10 P3:O—2+12=10. Exercise 2. Given the two planes 3:1; + 4y = 10 <— note a plane in 3D parallel to the z-aa‘is 2a: — y + 22 = 5 find the angle between them. 1Here one can use either one of the other two given points, the equation for the plane at the end will be the same! [0 Method: Consider the general situation of two planes 0.13: -l- (313} + C121 = d1 H.221: + bay + C22 : d2 with angle 3 between them: _-, —+ Observe that 6 is the same as the angle between the two normals N1 and N 2 to each plane re~ spectively. We know that the coefﬁcients in the equation of a plane represent the coordinates of a normal vector to the plane itself. Hence TV: —} N2 (‘11: bllcl) (a21b2102) Now recall that from the deﬁnition of dot product-one can deduce that JV N cos 6 = 1 2 (2-2) |N1| - |N2| and the problem is solved. Let‘s go back to Exercise 2. From the two planes 33; + 4y = 10 2.7; — y + 22 = we obtain the corresponding normals N1=(3,4,U) and N2 = (2,—1,2), moreover N1 - N2 = 6 — 4 = 2 and |N1l = 5 and “V21 : 3 Finally applying (T?) we obtain 9 9 = "1 i _ cos (15) 3 Parametric equations of curves and lines In Math 18.01 you described curves as 1. either the graph of a function y = f(1'} [explicit] A 3 a J : X > X 2. or as the points on the plane such that F(.1',y) = 0 [implicit] A fl These two ways were okay in 2D but net in 3D! In fact the graph of a two variable function 2 = f(.1', y) gives a smface not a. curve: 2‘: Xe-laa {3 In 3D in order to describe a curve we need the so called Parametric Equations Deﬁnition. A curve 7 in 3D is deﬁned by the points (egg, 2) such, that 1‘ = f6) 11 = 90) z = h(t) with the parameter 7? ranging in a certain interval [(1,1)]. The functions f (t), g(t) anar Mt) are, called parametric equations. 2D example: Consider the curve 7 described by the parametric equations 1' : cost y = sin 1‘, for t in [0, 4n]. 5 L5, (550 Wl lid-1L! 3D example: Consider the curve 7 described by the parametric equations Equations for lines in 3D Typical problem: Given a point P0 and a vector 17, ﬁnd the line through P0 and parallel to 17. In comdinates: Pu: {3:0, 0,50) )and V:( (,a b c are iven. y g V/ A generic point P: (3 y,z) is on the line if P0P is parallel to I7 or in other words --—1 _. POP = 'tV for any scalar L. In coordinates this is written as :1: — 3; = to, U LU — 'Uu = 125 z " = tc or equivalently it = 1(1 + w 1! = Iiu + H) a + to zinc] these are called the parametric equations of a line through Pu = (3:9, y“: :0) and parallel to V = (a.,b,c). Exercise 3. Find the line through the two points PU = (1I 2,3) and P] = (0.1.0). Method: Let‘s consider the general case Pu : ((11,012, 013) and P1 : (bl,bg.b3). Observe that the vector ———-—’ P0131 = (9-1 — (111(k) — bzﬁa "- 113) is by construction parallel (actually on top of it!) to the line we are looking for. Then one can use the parametric equations above and obtain 1' = a1 +£(o1— ()1) y: as +3012 - (’2) z = (1.3 + t(a.3 — b3). Going back to Exercise 3 we then have the parametric equations I=l—t _=2—t 5:3k3t. Handy Fact: If you want to describe only the segment of line from P0 to P1 then you restrict the parameter in the interval [0, 1]. T Study Guide 1. Answer the foitoioing questions: I! You have now two ways to describe a line in 3D: as the set of solutions to a system of two linear equations ( intersection of two planes) or as a curve given by three very special parametric equations as described above. Can you see a connection between the two? a When a curve is given certainly the parametric equations are essentiat. But it is also very important to describe the range of the parameter. Can you give an example of this fact? ...
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