m114-ex10.html

CHAPTER 10 - Vectors and Analytic Geometry in Space

Section 10.1, page 794

Problem 30

> eqn:=x^2-6*x*y(x)+8*y(x)^2-2*x-1=0;

> slope:=solve(diff(eqn,x),diff(y(x),x));

slope := (x-3*y(x)-1)/(3*x-8*y(x))

> subs(x=1,y(1)=1,slope);

So the slope of the graph is 3/5 at the point. A vector in this direction is [1,3/5], but to make it a unit vector, we divide by the length:

> len:=sqrt(1^2+(3/5)^2);

len := 1/5*34^(1/2)

> tanvec:=simplify([1/len,3/5/len]);

tanvec := [5/34*34^(1/2), 3/34*34^(1/2)]

> opptanvec:=-tanvec;

opptanvec := [-5/34*34^(1/2), -3/34*34^(1/2)]

> normvec:=[tanvec[2],-tanvec[1]];

normvec := [3/34*34^(1/2), -5/34*34^(1/2)]

> oppnormvec:=-normvec;

oppnormvec := [-3/34*34^(1/2), 5/34*34^(1/2)]

> oppnormvec[1]^2+oppnormvec[2]^2;

(Just checking.)

Section 10.2, page 804

Problem 53

> P1:=[1,4,5]; P2:=[4,-2,7];

> dotp:=(V,W)->V[1]*W[1]+V[2]*W[2]+V[3]*W[3];

> P1-P2;

> dist:=sqrt(dotp(P1-P2,P1-P2));

> direction:=(P2-P1)/dist;

direction := [3/7, -6/7, 2/7]

> midpt:=(P1+P2)/2;

midpt := [5/2, 1, 6]

Section 10.3, page 812

Problem 57

> y1:=3/2-x^2; y2:=x^2;

y1 := 3/2-x^2

> xvals:=solve(y1=y2,x);

xvals := -1/2*3^(1/2), 1/2*3^(1/2)

At the first point:

> tanvec1:=[1,subs(x=xvals[1],diff(y1,x))];

> tanvec2:=[1,subs(x=xvals[1],diff(y2,x))];

We'll use the "dotp" program defined as above to get the angle between these vectors:

> dotp:=(V,W)->V[1]*W[1]+V[2]*W[2];

> angle1:=arccos(dotp(tanvec1,tanvec2)/sqrt(dotp(tanvec1,tanvec1)*dotp(tanvec2,tanvec2)));

angle1 := 2/3*Pi

Now for the second point:

> tanvec1:=[1,subs(x=xvals[2],diff(y1,x))];

> tanvec2:=[1,subs(x=xvals[2],diff(y2,x))];

Of course, since these are the same vectors in the opposite order, the angle is the same.

Section 10.4, page 820

Problem 37

We'll assume that the vertices are in fact those of a parallelogram. So we'll take two vectors AB and AD. We'll put them in three dimensions so we can take cross products.

> AB:=[2,0,0]-[-1,2,0]; AD:=[4,3,0]-[-1,2,0];

> cprod:=(V,W)->[V[2]*W[3]-V[3]*W[2],V[3]*W[1]-V[1]*W[3],V[1]*W[2]-V[2]*W[1]];

> cprod(AB,AD);

So the area of the parallelogram is 13.

Section 10.5, page 827

Problem 27

We'll write the lines parametrically as follows:

> line1:=[2*t+1,3*t+2,4*t+3]; line2:=[s+2,2*s+4,-4*s-1];

We want to solve for the values of s and t where the lines meet - you can only solve two equations for two unknowns, so we'll just use two of the coordinates (and then check the third -- remember, in general, lines in 3-space don't meet):

> sol:=solve({line1[1]=line2[1],line1[2]=line2[2]},{s,t});