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Project 1
1. Vector Calculus page 77 problem 1
Let P0 , P1 , . . . , Pk−1 be the vertices of a regular polygon having k sides. Let O be the center of the polygon.
k−1 Show that i=0 OP
i = 0.
Remember that adding vectors can be expressed geometrically by placing the tail of one addend at the head
of the other. The vector from the tail of the first vector to the tail of the last is the vector sum. We obtain the
desired result by adding OP0 + OP1 + . . . + OPk−1 .
Note that the smaller angle between any two of these vectors is given by
π(k−2)
π − 2π
.
k =
k
2π
k .
Thus the supplementary angle is
By definition each of the vectors has the same magnitude (length). Thus our object is a k sided object with
. These are the properties of a regular polygon with k-sides. Thus
equal length sides each at an angle of π(k−2)
k
k−1 geometrically i=0 OPi is the sum of vectors in a cycle along the sides of a regular polygon. Thus the head of
OPk−1 is on the tail of OP0 which means that the sum is 0.
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2. Vector Calculus page 77 problem 9
Suppose people make a set of n statements with a 1 if they agree and a -1 if they disagree. Each set of answers
forms a vector in Rn . Interpret the dot product of two answer vectors.
Let A = [a1 , a2 , . . . , an ] and B = [b1 , b2 , . . . , bn ] be two answer vectors. A · B = a1 b1 + a2 b2 + . . . + an bn .
Note that if person A and person B agree on question j, then aj bj = 1 · 1 = 1 or aj bj = −1 · −1 = 1. If they
disagree then aj bj = −1 · 1 = −1.
As a result for every question that the answers are the same the sum increases by one and for every question
on which they disagree the sum decreases by one. Thus the dot product ranges from −n (disagreement on all
questions) to n (agreement on all questions). The dot product can be interpreted as the measure of agreement
on the set of statements.
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3. Angles between planes
Find the angles between the following two pairs of planes. 2x + 3y + 5z = 0, 5x − 5y + z = 0 and 2x + 3y + 5z =
0, 2x + 3y + 4z = 0.
We must first determine a consistent means of finding the angle between two intersecting planes. Angles are
formed by two lines. We need to find a representative line from each plane. An arbitrary line will not work.
For example we could pick the line of intersection for one of the planes. Then the second plane has an infinite
number of lines incident with the line of intersection and forming an infinite number of angles with it. Consider
a line in each plane that is orthogonal to the line of intersection and these two lines intersect. To obtain the
angle we will use the vector defining the lines. A plane is determined by two non-parallel vectors. The vector
defining the line of intersection and the vector that we choose orthogonal to it define the plane. Any third
vector orthogonal to both would from a hyperplane (3 space object). Thus there is only one vector orthogonal
to the line of intersection in each plane.
Calculating these orthogonal vectors could be time consuming, so we use a shortcut. The normal vector to a
plane is orthogonal to every vector in the plane. Thus the normal vectors for the planes form the same angle
as two vectors orthogonal to the line of intersection. The normal vectors can be read from the equations of the
planes in standard form which simplifies our work.
Find the angle between 2x + 3y + 5z = 0, 5x − 5y + z = 0.
θ
=
arccos
=
=
=
[2, 3, 5] · [5, −5, 1]
[2, 3, 5] · [5, −5, 1]
10 − 15 + 5
arccos √
2
2
2 + 3 + 52 52 + (−5)2 + 12
0
arccos √ √
38 51
π
.
2
Find the angle between 2x + 3y + 5z = 0, 2x + 3y + 4z = 0.
θ
[2, 3, 5] · [2, 3, 4]
= arccos
[2, 3, 5] · [2, 3, 4]
4 + 9 + 20
√
= arccos √
22 + 32 + 52 22 + 32 + 42
33
= arccos √ √
.
38 29
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