Multiplication of vectors
Vectors - What Are They? gives an introduction to the subject.
There are two useful definitions of multiplication of vectors, in one the product is a scalar and in the other the product is a vector. There is no operation of division of vectors. In some school syllabuses you will meet scalar products but not vector products but we discuss both types of multiplication of vectors in this article to give a more rounded picture of the basics of the subject
Scalar Multiplication
The scalar product of vectors ${\bf u} = (u_1, u_2, u_3)$ and ${\bf v}=(v_1, v_2, v_3)$ is a scalar defined to be
The Cosine Rule in Euclidean Geometry can be proved without the use of scalar products. Using the Cosine Rule for the triangle $\Delta OPQ$ where $\angle POQ = \theta$ we get:
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Expanding $|{\bf u-v}|^2$ we get
Some texts use the formula (6) to define the angle between two vectors, that is
Note that the product of a row vector and a column vector is defined in terms of the scalar product and this is consistent with matrix multiplication.
Vector Multiplication
The vector product of two vectors ${\bf b}$ and ${\bf c}$, written ${\bf b}\times {\bf c}$ (and sometimes called the cross product ), is the vector
From this definition we can see that ${\bf b}\times {\bf c}=-{\bf c}\times {\bf b}$ so this operation is not commutative. If ${\bf i, j, k}$ are unit vectors along the axes then, from this definition:
Now we prove that the two definitions of vector multiplication are equivalent. The diagram shows the directions of the vectors ${\bf b}$, ${\bf c}$ and ${\bf b}\times {\bf c}$ which form a 'right handed set'.
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You may wish to finish reading here and it is indeed more
important to appreciate that there are two definitions of a vector
product, which can be shown to be equivalent, than it is
mechanically to work through the details of the proof.
Theorem The vector
product of two vectors ${\bf b}$ and ${\bf c}$ is a vector ${\bf
b}\times {\bf c}$ with the following properties:
(i) ${\bf b}\times {\bf c}$ has
magnitude $|{\bf b}||{\bf c}|\sin \theta$ where $\theta$ is the
angle between the directions of ${\bf b}$ and ${\bf c}$;
(ii) ${\bf b}\times {\bf c}$ is
perpendicular to ${\bf b}$ and ${\bf c}$ with direction such that
the vectors ${\bf b}$, ${\bf c}$ and ${\bf b}\times {\bf c}$ form a
right handed set as in the diagram so that ${\bf b}\times {\bf c}$
and ${\bf c}\times {\bf b}$ are in opposite directions.
Proof of part (i)
Consider the area of the parallelogram with sides given by the
vectors ${\bf b}$ and ${\bf c}$ and angle $\theta$ between these
sides. The area of this parallelogram is $|{\bf b}||{\bf c}|\sin
\theta$. The vector ${\bf b}$ can be decomposed into a vector
$k{\bf c}$ of magnitude $|{\bf b}|\cos \theta$ in the direction of
${\bf c}$ and ${\bf b}-k{\bf c}$ of magnitude $|{\bf b}|\sin
\theta$ perpendicular to ${\bf c}$ where $k=|{\bf b}|\cos
\theta/|{\bf c}|=({\bf b.c})/|{\bf c}|^2$.
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So the area of the parallelogram is: In
order to work out the modulus on the right hand side we take the
scalar product of the vector with itself.
Proof of part (ii) To
show that ${\bf b}$ and ${\bf b}\times {\bf c}$ are perpendicular
we show that the scalar product is zero: and similarly the scalar product of ${\bf c}$ and ${\bf
b}\times {\bf c}$ is zero so these vectors are perpendicular.