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MATH 241 (Professor Green) : Homework 2 Solutions
Section 11.
Problem 8: We create the points P, Q and R, and compute the dot product of PQ and PR. P=[2,1];Q=[1,4];R=[-3,2]; dot(Q-P,R-P) ans = 8 Since the dot product is not 0, the vectors are not perpendicular. Problem 14: P=[1,2,3];Q=[-1,0,1];R=[1,1,0]; PQ=Q-P;PR=R-P; projPQPR=(dot(PQ,PR)/dot(PQ,PQ))PQ* projPQPR = -1.3333 -1.3333 -1. Problem 18: We will illustrate two different methods on the two parts of this problem. a) P=[2,3,4];Q=[3,5,5];R=[1,3,11]; veclength(P-Q)^ veclength(P-R)^ veclength(Q-R)^ ans =
ans =
ans = 44 It follows that PQR is a right triangle with the right angle at Q. b) P=[0,-1,2];Q=[1,4,-2];R=[5,0,1]; dot(Q-P,R-P) dot(P-Q,R-Q) dot(P-R,Q-R) ans = 14 ans = 28 ans = 13
Since none of the three dot products is zero, none of the three angles is a right angle. Problem 30: We place the base of the parallelepiped in xy-plane with vertices on the coordinate axes. Since the base is a square with area 1, each side has length 1, which places the vertices at
( c c , and the center at (0,0,c/2). The triangle
formed by the center and the vertices on the positive x and y axes is isoceles with a sixty degree angle between the two equal sides; consequently, it is equilateral. It follows that the line segment connecting the center with any vertex has length 1, and that
c
and (^) c 2.
Section 11.
Problem 2: a=[1,1,1];b=[1,0,-1];c=[1,1,-1]; cross(a,b) dot(c,cross(a,b)) ans = -1 2 - ans = 2 Problem 4: a=[3,4,12];b=[3,4,-12];c=[1/8,-1/12,1/16]; cross(a,b) dot(c,cross(a,b)) ans = -96 72 0 ans =
Problem 14: axb=axc if and only if ax(b-c)=0, which in turn is true if and only if b-c is a multiple of a. This makes it clear how to construct an example. For instance, a= i , b= j , c= i + j.
Section 11.
Problem 12: A=[1,7,5];B=[3,2,-1];C=[2,-2,5];D=[-2,8,17]; cross(B-A,D-C) ans = 0 0 0 This shows that the two lines are parallel or equal. cross(B-A,D-A)
P =
[ x, y, z] planfun = -9x-39+3y+15z So an equation for the plane is -9x+3y+15z=39 or -3x+y+5z=13. Problem 8: From the symmetric equations, we can see that the plane contains the points (1,-1,5) and (- 3,4,0), and also that the vector [3,2,4] is parallel to the plane. This allows us to say P0=[1,-1,5];P1=[-3,4,0]; v=[3,2,4]; N=cross(P1-P0,v) planefun=realdot(N,P-P0) N = 30 1 - planefun = 30x+86+y-23z So that an equation for the plane is 30x+y-23z=-86. Problem 12: a) We begin by simultaneously solving the symmetric equations for the line and the equation for the plane. [x0,y0,z0]=solve('(x+1)/2=(y+3)/3','(x+1)/2=-z','3x-2y+4z=-1')** x0 = 1 y0 = 0 z0 =
This gives us P0=(1,0,-1) b) From the symmetric equations for the line, we can read of the normal [2,3,-1] for the plane in question, giving us the equation 2x+3y-z=3. c) We can also read of [3,-2,4] as a normal to the plane, giving us the symmetric equations (x-1)/3=-y/2=(z+1)/4. Problem 24: From the equation for the plane, we can read off the point (1,-3,0), and the normal [2,2,-1]. We provide a different solution from the one in class: P0=[1,-3,0];N=[2,2,-1];P=[x,y,z]; distsq=realdot(N,P-P0)^2/dot(N,N) distsq = 1/9(2x+4+2*y-z)^ This is an expression for the square of the distance from an arbitrary point to the plane. We set it equal to 9 and solve simultaneously with the symmetric equations for the line.
[x1,y1,z1]=solve('(x-1)/2=(y+1)/3','(y+1)/3=(z+5)/7',distsq-9) x1 = [ -11] [ 1] y1 = [ -19] [ -1] z1 = [ -47] [ -5] This gives the two points (-11,-19,-47) and (1,-1,-5). Problem 26. b. and d. are parallel since their normals are parallel, but the equations are not equivalent. a and c are the same plane, which is not perpendicular to b and d. Problem 30. P0=[-2,1,4];P1=[0,3,1];v=[2,-4,6]; N=cross(P1-P0,v) planefun=realdot(N,P-P0) N = 0 -18 - planefun = -18y+66-12z so that the desired equation is -18y-12z=-66 or, 3y+2z=11.
