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Unformatted text preview: Lecture III
Vector Algebra in Cartesian Coordinates Let us construct a Cartesian coordinates system in E3 . First we choose a
point O, called the origin. Then we chose three mutually perpendicular rays
starting from O. These rays are called the positive x axis, positive y axis, and
positive z axis. Consider the lines containing these rays. For any of these lines,
every point on it can be identiﬁed with a real number: if the point is on the ray,
the real number is the distance to O, if it’s not on the ray, the number is the
distance to O times −1. Let us denote these three lines by X, Y , and Z . Let P
be a point in space. Consider the projectionpoints of P on X, Y , and Z . These
points give the Cartesian coordinates of P , denoted xP , yP , and zP . Any triplet
of real numbers forms the cordinatesfor some p oint P . Diﬀerent points have
diﬀerent coordinates.
The three unit vectors in the directions of the positive x, y , and z axes are
ˆ
�
i j,
customarily denoted by ˆ, ˆ and k . Let A be a vector in E3 and let P be the point
�
�
such that OP = A
. Let (a1 , a2 , a3 ) be the coordinates of P . Consider the vectors
ˆ
�
�
A1 = a1ˆ, A2 = a2 ˆ, and A3 = a3 k . by vector addition and multiplication with
i�
j
scalars, one obtains the following expression:
ˆ
�
�
�
�
A = A1 + A2 + A3 = a1ˆ + a2 ˆ + a3 k
i
j
�
Then a1 , a2 , and a3 are called the scalar components of A, and a1ˆ a2 ˆ and
i, j,
ˆ are called the vector components of A. By using the ˆ ˆ k unit vectors, we
ˆ
�
i, j,
a3 k
obtain coordinate formulas for the four basic vector operations:
1. Multiplication by a scalar
ˆ
�
cA = (ca1 )ˆ + (ca2 )ˆ + (ca3 )k,
i
j
for any scalar c. 1 2. Addition of vectors ˆ
ˆ
�
�
If A = a1ˆ + a2 ˆ + a3 k and B = b1ˆ + b2 ˆ + b3 k , then
i
j
i
j
ˆ
��
A + B = (a1 + b1 )ˆ + (a2 + b2 )ˆ + (a3 + b3 )k.
i
j
3. Dot product
ˆ
ˆ
�
�
If A = a1ˆ + a2 ˆ + a3 k and B = b1ˆ + b2 ˆ + b3 k , considering that ˆ · ˆ =
i
j
i
j
ii
ˆˆ
ˆ ˆi jˆ ˆj
ˆ · ˆ = k · k = 1 and ˆ · ˆ = ˆ · ˆ = ˆ · k = k · ˆ = ˆ · k = k · ˆ = 0, we get that
jj
ij ji i ��
A . B = a1 b1 + a2 b2 + a3 b3 .
4. Cross product
ˆ
ˆ.
�
�
Let A = a1ˆ + a2 ˆ + a3 k and B = b1ˆ + b2 ˆ + b3 k To compute the cross
i
j
i
j
product, we use the following equalities: ˆ × ˆ = ˆ × ˆ = k × k = 0,
iijjˆˆ and ˆ j i, ˆ i
ˆ × ˆ = −ˆ × ˆ = k, ˆ × k = −k × ˆ = ˆ k × ˆ = −ˆ × k = ˆ
ij
j i ˆj ˆ
i ˆ j.
We get the following formula:
ˆ
��
A × B = (a2 b3 − a3 b2 )ˆ + (a3 b1 − a1 b3 )ˆ + (a1 b2 − a2 b1 )k.
i
j
Using determinants one can easily remember the formula, since:
�
�
�ˆ ˆ k�
ˆ�
j
�i
�
�
��
A × B = � a1 a2 a3 �
�
�
�
�
� b1 b2 b3 �
�
Considering that ˆ · ˆ = ˆ · ˆ = k · k = 1, we get that A can be expressed as
ii jj ˆˆ �ii
�ˆˆ
�
�jj
A = (A.ˆ)ˆ + (A.ˆ)ˆ + (A. k )k,
which is known as the frame identity. Triple products
1. Scalar triple product 2 ˆ�
ˆ
ˆ
�
�
Let A = a1ˆ + a2 ˆ + a3 k , B = b1ˆ + b2 ˆ + b3 k ,and C = c1ˆ + c2 ˆ + c3 k .
i
j
i
j
i
j
���
The triple product (A × B ) . C is called a scalar triple product, since it is
a scalar quantity. Its value is given by:
�
�a
�1
�
��
� × B ) . C = A . (B × C ) = � b1
��
�
(A
�
�
� c1 a2
b2
c2 �
a3 �
�
�
b3 � .
�
�
c3 � ���
���
Hence we can simply write (A×B ) . C as [A, B , C ] without specifying the
positions for the cross and dot signs. We will use the following equalities
in computing quadruple products:
���
���
���
���
���
���
[A, B, C ] = [B, C, A] = [C, A, B ] = −[A, C, B ] = −[C, B, A] = −[B, A, C ].
2. Vector triple product
The cross product is not associative so we will give two formulas: ��
�
� ��
���
(A × B ) × C = (C . A)B −(C . B )A, and ���
���
�
�
�
A × (B × C ) = (A . C )B − (A . B )C Quadruple products
1. Scalar quadruple product
���
The expression (A × B ).(C × D) is called a quadruple scalar product, and
by applying the formulas for triple products, we get the value: ���
�
�� ��
�� ��
(A × B ).(C × D) = (C .A)(B .D) − (C .B )(A .D
)
2. Vector quadruple product
�
�
�
�
The expression (A × B ) × (C × D) is called a quadruple vector product,
and by applying the formulas for triple products, we get the value:
��
�
�
� � ��
�� ��
(A × B ) × (C × D) = [C, D, A]B − [C, D, B ]A 3 ...
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 Two '04
 Duorg
 Math, Algebra

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