Maths Warm-Up Answers


 
1. What is a point? What is a vector? How are they related?
 
2. Find the dot products for the following pairs of vectors:
 
(a) (1,1,1) . (1,1,-1)
(b) (2,5,3) . (2,5,3)
(c) (1,0,0) . (10,1,5)
 
 
 
a = 1x1 + 1x1 + 1x-1 = 1
b = 38
c = 10
 
 
 
3. Find the length of the vector (2,5,3).
 
 
root (2x2 + 5x5 + 3x3) = root(38) = 6.16
 
 
 
 
4. Let i = (1,0,0), j = (0,1,0) and k = (0,0,1). By considering the
meaning of cross product (*), write down the answers to:
i*i, j*j, k*k, i*j, j*i, i*k, k*i, j*k, k*j.
 
 
i*i = 0 0 0
j*j = 0 0 0
k*k = 0 0 0
i*j = 0 0 1
j*i = 0 0 -1
i*k = 0 -1 0
k*i = 0 1 0
j*k = 1 0 0
k*j = -1 0 0
 
 
i*j Method1 (1,0,0)*(0,1,0)
 
x = y1z2 - y2z1 = 0x0 - 1x0 = 0
y = z1x2 - z2x1 = 0x1 - 0x1 = 0
z = x1y2 - x2y1 = 1x1 - 0x0 = 1       = (0,0,1)
 
 
 
i*j Method2 (1,0,0)*(0,1,0)
 
reminder that the determinant of 2x2 matrix:
 
         | a b |
         | c d | = ad - cb
 
consider unit vectors
 
        ux=(1,0,0) uy=(0,1,0) uz=(0,0,1)
 
then the determinant of the following 3x3 matrix
 
         | ux uy uz |    | ux uy uz | 
         | ix iy iz | =  | 1  0  0  | 
         | jx jy jz |    | 0  1  0  | 
 
that can be simplified to 2x2 determinants
 
        det|0 0|ux - det|1 0|uy + det|1 0|uz
           |1 0|        |0 0|        |0 1|
 
        det|0 0| = (0x0 - 1x0) = 0
           |1 0|
        
      - det|1 0| = (1x0 - 1x0) = 0
           |0 0|
 
        det|1 0| = (1x1 - 0x0) = 1
           |0 1|
 
giving the cross product
 
        0ux - 0uy + 1yz = (0,0,1)
 
Note that this is the same as calculated previously!
 
 
 
 
5. Using the result in 4. and the fact that (x,y,z) = x i + y j + z k
find an expression for p*q where p = (x1,y1,z1) and q = (x2,y2,z2).
Hence find the cross products for each of (a), (b), (c) in 2.
 
 
 
(a) -2 2 0
(b) 0 0 0
(c) 0 -5 1
 
 
 
6. The equation of a plane is ax+by+cz = d. (Under what circumstances
will d=0?). 
 
When the plane goes through the origin
 
Find a,b,c,d for each of the cases:
 
(a) The plane is the XY plane.
(b) The plane is the XZ plane.
(c) The plane is the YZ plane.
(d) The plane is parallel to the XY plane and of distance D from the origin.
 
 
Reminder: In the plane equation the a,b,c is the plane norm (the vector perpendicular
to the plane) that gives the precise position of the plane. The d is the translation 
distance of the plane from the origin. The x,y,z is a point in 3D space. So, values 
of x,y,z that satisfy the equation are on the plane.
 
Visualise the plane as the XY axis (using a piece of paper helps). A vector that is 
perpendicular to this plane does not travel in the X or Y direction but will travel 
in the Z direction. Such a vector is thus (x=0,y=0,z=1).
 
(a) a=0  b=0  c=1  d=0   (a+b+c=1 in a norm)
(b) a=0  b=1  c=0  d=0
(c) a=1  b=0  c=0  d=0
(d) a=0  b=0  c=1  d=D
 
 
 
Describe the geometric meaning of the planes:
(a) x=y                Reformulate as x-y=0
(b) x+y = 0
(c) x+y+z = 0
(d) x+y+z = 1.
 
 
Imagine values of x y and z that will satisfy this (a) plane equation: 
(x=+ve, y=+ve, z=any).
 
This corresponds to (a=+ve,b=+ve,c=0) as the plane norm with d=0. From
this we can deduce the position of the plane.
 
(a) a=1  b=-1  c=0   d=0
(b) a=1  b=1   c=0   d=0
(c) a=1  b=1   c=1   d=0
(d) a=1  b=1   c=1   d=1
 
 
 
 
 
 
7. Find the equation of the plane that passes through the following
three points: (1,0,0), (0,1,0), (0,0,1).
 
 
Simply visualise the three points. The plane that goes through them
has a norm that seems to point straight away from (or at, according
to in which direction you're looking at) the origin. The direction of 
this norm vector is clearly something like (1,1,1) meaning a plane
equation of x+y+z=1
 
 
 
 
8. In general given the points (x1,y1,z1), (x2,y2,z2), (x3,y3,z3) find
the equation of the plane which passes through these points, by using
the plane equation, ax+by+cz = d.
 
 
Let p1,p2,p3 be our three points in the plane. We can construct two vectors
that go between the points that definately exist on the plane. To compute the
plane equation we simply want to find the norm of these two vectors (which will
give us the norm of the plane as well!). 
 
So, the two vectors, (using the points from the previous example)
 
v1 = p2 - p1 = (-1,1,0)               (point 1 to point 2)
v2 = p3 - p1 = (-1,0,1)        (point 1 to point 3)
 
both lie in the plane. If we take their cross product we find our normal vector:
 
n = v1 x v2 = (1,1,1)
 
and our displacement from the origin:
 
d = n.p1 = (1,1,1).(1,0,0) = 1
 
so the equation of the plane is x+y+z=1
 
 
 
 
 
 
 
9. Let p0, p1, p2 be three distinct points in 3D. Then consider the
cross product n = (p2-p0)*(p1-p0). What does this vector mean
geometrically? 
 
 
It is the norm to the plane on which the three points lie.
 
 
Let p=(x,y,z) be any point on the plane formed by the
three points. What is the geometric relationship between the vector
(p-p0) and the vector n? 
 
 
They are orthogonal.
 
 
Hence show that the equation of the plane
must be the dot product: n.(p-p0) = 0. 
 
 
The equation for the plane consists of all vectors r satisfying (r-a).n=0 where
a is a known point on the plane and n is the plane norm.
 
 
 
 
By explicitly evaluating the
expressions, show that this is the same plane equation as you have
found in question 8.
 
10. Using the argument in 9., show that n must be normal to the plane,
and that in question 8 the normal to the plane is (a,b,c). What are
the normals to the planes given in question 6 (note that there are at
least two answers in each, case, give both)?
 
 
 
 
 
11. Evaluate the following product of a row vector and a matrix:
 
   (1, 2, 3, 4) | 1 2  3 4 |
               | 2 1  0 3 |
               | 2 0  1 3 |
               | 1 2 -1 1 |
 
 
 
1x1 + 2x2 + 3x2 + 4x1  = 15
1x2 + 2x1 + 3x0 + 4x2  = 12
1x3 + 2x0 + 3x1 + 4x-1 = 2
1x4 + 2x3 + 3x3 + 4x1  = 23
 
= (15, 12, 2, 23)
 
 
 
12. Evaluate the following product of two matrices:
 
 | 1 2  3 4 | * | 2 1 3  4 |
 | 2 1  0 3 |   | 1 2 2  0 |
 | 2 0  1 3 |   | 1 1 1  1 |
 | 1 2 -1 1 |   | 2 3 1 -1 |
 
 
 
eg: 
at pos 1,1 we have 1x2 + 2x1 + 3x1  + 4x2  = 15
at pos 4,4 we have 1x4 + 2x0 + -1x1 + 1x-1 = 2
 
so:
 
 | 15 20 14 3 |
 | 11 13 11 5 |
 | 11 12 10 6 |
 |  5  7  7 2 |
 
 
13. What is the transpose of the matrix in question 11?
 
 
 | 1  2  3  4 |         | 1  2  2  1 |
 | 2  1  0  3 |         | 2  1  0  2 |
 | 2  0  1  3 | becomes | 3  0  1 -1 |
 | 1  2 -1  1 |         | 4  3  3  1 |
 
obtained by switching its rows and columns
 
 
 
 
14. What is the inverse of a matrix? How can you define it?
 
 
The inverse is written A^-1  (A to the power -1). The definition is:
 
A . A^-1 = A^-1 . A = I
 
where I is the identity matrix:
 
 | 1 0 0 0 |
 | 0 1 0 0 |
 | 0 0 1 0 |
 | 0 0 0 1 |
 
If A has an inverse then A is called a nonsingular matrix. An inverse only
exists if and only if: det(A) != 0. We compute the inverse of a matrix using 
the Gauss-Jordan elimination. But we're not going to do that here:-)
 
 
 
15. Suppose that A is a square matrix, and its inverse and transpose
are equal.  (Such a matrix is called orthogonal). Show that, for any
angle q, the following matrix is orthogonal:
 
  cos(t) sin(t)
 -sin(t) cos(t)
 
Suppose further that x and b are vectors of the appropriate dimension
and that xA=b.  How would you find x? In the example, suppose that b =
(1,2) find x.