Math 225: Calculus III     Solution 2

Initialization

<< Calculus`Math225`

i  =  {1,  0,  0};  j  =  {0,  1,  0};  
    k  =  {0,  0,  1};

Problem 1

Let  a = i - 2j + 3k and  b = -i + 4j + k. Find the following.
a)   Norm[a], Norm[b]
b)  a.b,  b.a
c)  Cross[a,b], Cross[b,a]
d)  The angle between a and b, and the projection of b on a (Pr[b,a]).

Solution

a  =  i  -  2  j  +  3  k

b = -i + 4j + k

Norm[a]

N[%]

Norm[b]

N[%]

a.b

b.a

Cross[a, b]

Cross[b, a]

The angle between a and b is

ArcCos[a.b/(Norm[a] Norm[b])]

N[%]

The projection of b on a is given by:

Pr[b, a]

Problem 2

a)  Find the area of the parallelogram determined by
     a = i + j - 3 k and b = -2 i + j + k
  using a cross product.
b)  Find the volume of the parallelopiped determined by a, b, and
     c = i - 2 j + k
     using a triple products (or a determinant, Det[]).

Solution

a  =  i  +  j  -  3 k

b  =  -2 i  +  j  +  k

Area is given by:

Norm[Cross[a, b]]

N[%]

c  =  i  -  2 j  +  k

Abs[c.Cross[a, b]]

or:

Abs[Det[{a, b, c}]]

Problem 3

a)  Find the distance from the point p = {1,-3,2} to the plane
    5x + y - 4z == 1.
b)  Find the distance from the point q = {2,1,3} to the line
    x == 1 - t, y == 1 + t, z == 3t.

Solution

a) The normal vector is

n  =  5  i  +  j  -  4  k

A point on the plane is

p0  =  {0, 1, 0};

The distance is given by

p  =  {1, -3, 2};

d  =  Abs[n.(p - p0)]/Norm[n]

N[%]

b) The formula for the distance uses the direction vector v of the line and a point p on the line.
The direction of the line is given by the coefficients of t:

v  =  {-1, 1, 3};

A point on the line is found by plugging in t->0:

p  =  {1, 1, 0};

The given point is:

q  =  {2, 1, 3};

Norm[Cross[q - p, v]]/Norm[v]

N[%]

Problem 4

a)  Find an equation for the plane through the points p = {1,-1,2},
    q = {3,0,1}, and r = {-1,1,1}.
b)  Find the parametric equations for the line in which the planes x - y + z == 1
    and x + y + 3z == -2 intersect.

Solution

a) Use the cross product of two vectors parallel to the plane to get a vector perpendicular.

p  =  {1, -1, 2}; q  =  {3, 0, 1}; r  =  {-1, 1, 1};

v1  =  q - p

v2  =  r - p

n  =  Cross[v1, v2]

The equation of the plane is

n.{x, y, z}  ==  n.p

b) To find the intersection we solve the equations simultaneously, say, solve for x and y in terms of z. We then get a parameterization of the line by substituting z == t.

soln  =  
      Solve[{x  -  y  +  z  ==  1,  
          x  +  y  +  3  z  ==  -2}, {x, y}]

The parametric equations of the line are:
    x == -1/2 - 2t
    y == -3/2 - t
    z == t

Another way to solve this problem is to notice that the normal vectors of the two planes are both perpendicular to the line, so their cross product will be parallel to the line, i.e. give the direction vector of  the line.

n1  =  {1, -1, 1};  n2  =  {1, 1, 3};

v  =  Cross[n1, n2]

Two get a point p on the line, find any simultaneous solution to the equations. For example, set z->0 in the above solution to get the point

soln /. z->0

Set z->0 to get the point

p = {-1/2,-3/2,0};

The parametric equations of the line are then

p + t v

or x == -1/2 - 4t, y == -1/2 - 2t, and z == 2t. This looks different then the previous solution, but it does represent the same line. The direction vector from these equations is {-2,-1,1} which is parallel to the direction vector {-4,-2,2} from the cross product and the point {-1/2,-3/2,0} is on both lines (t==0 in both solutions).

Problem 5

a)  Find the point in which the line x == 2 - 2t, y == 1 - t, z == 3 + 5t meets the plane 3x + 2y + 6z == 4.
b)  Find the point in which the three planes x - y + z == 1, x + y - z == 2 and
-x + y + z == 3 intersect.

Solution

a) Solve the equations!

Solve[{x  ==  2 - 2  t,  y  ==  1 - t,  z  ==  3 + 5  t, 
        3  x  +  2  y  +  6  z  ==  4}]

The point is {4,2,-2}.

b)

Solve[{x  -  y  +  z  ==  1,  x  +  y  -  z  ==  2,  
        -x  +  y  +  z  ==  3}]

the point is {3/2, 5/2, 2}.