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Next: The Vector Product Up: Vectors Previous: Vector Area

The Scalar Product

A scalar quantity is invariant under all possible rotational transformations. The individual components of a vector are not scalars because they change under transformation. Can we form a scalar out of some combination of the components of one, or more, vectors? Suppose that we were to define the ``ampersand'' product,
{\bf a}\,\&\,{\bf b} = a_x \,b_y + a_y \,b_z + a_z \,b_x = {\rm scalar~number},
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for general vectors ${\bf a}$ and ${\bf b}$. Is ${\bf a}\,\&\,{\bf b}$ invariant under transformation, as must be the case if it is a scalar number? Let us consider an example. Suppose that ${\bf a} = (1,\,0,\,0)$ and ${\bf b} = (0,\,1,\,0)$. It is easily seen that ${\bf a}\,\&\,{\bf b}= 1$. Let us now rotate the basis through $45^\circ$ about the $z$-axis. In the new basis, ${\bf a} = (1/\sqrt{2},\, -1/\sqrt{2},\, 0)$ and ${\bf b} = (1/\sqrt{2},\,
1/\sqrt{2},\, 0)$, giving ${\bf a}\,\&\,{\bf b} = 1/2$. Clearly, ${\bf a}\,\&\,{\bf b}$ is not invariant under rotational transformation, so the above definition is a bad one.

Consider, now, the dot product or scalar product,

{\bf a} \cdot {\bf b} = a_x \,b_x + a_y \,b_y + a_z \,b_z = {\rm scalar~number}.
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Let us rotate the basis though $\theta$ degrees about the $z$-axis. According to Eqs. (10)-(12), in the new basis ${\bf a} \cdot {\bf b}$ takes the form
$\displaystyle {\bf a} \cdot {\bf b}$ $\textstyle =$ $\displaystyle (a_x\, \cos\theta+a_y\,\sin\theta)\,(b_x\,\cos\theta + b_y\,\sin\theta)$  
    $\displaystyle +(-a_x\,\sin\theta + a_y\,\cos\theta)\,(-b_x\,\sin \theta + b_y\,\cos\theta)
+a_z\, b_z$  
  $\textstyle =$ $\displaystyle a_x\, b_x + a_y\, b_y + a_z\, b_z.$ (18)

Thus, ${\bf a} \cdot {\bf b}$ is invariant under rotation about the $z$-axis. It can easily be shown that it is also invariant under rotation about the $x$- and $y$-axes. Clearly, ${\bf a} \cdot {\bf b}$ is a true scalar, so the above definition is a good one. Incidentally, ${\bf a} \cdot {\bf b}$ is the only simple combination of the components of two vectors which transforms like a scalar. It is easily shown that the dot product is commutative and distributive:
$\displaystyle {\bf a} \cdot {\bf b}$ $\textstyle =$ $\displaystyle {\bf b} \cdot {\bf a},$  
$\displaystyle {\bf a}\cdot({\bf b}+{\bf c})$ $\textstyle =$ $\displaystyle {\bf a} \cdot {\bf b} + {\bf a}\cdot {\bf c}.$ (19)

The associative property is meaningless for the dot product, because we cannot have $({\bf a}\cdot{\bf b}) \cdot{\bf c}$, since ${\bf a} \cdot {\bf b}$ is scalar.

We have shown that the dot product ${\bf a} \cdot {\bf b}$ is coordinate independent. But what is the physical significance of this? Consider the special case where ${\bf a} = {\bf b}$. Clearly,

{\bf a} \cdot {\bf b} = a_x^{~2}+a_y^{~2} + a_z^{~2} = {\rm Length}~(OP)^2,
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if ${\bf a}$ is the position vector of $P$ relative to the origin $O$. So, the invariance of ${\bf a} \cdot {\bf a}$ is equivalent to the invariance of the length, or magnitude, of vector ${\bf a}$ under transformation. The length of vector ${\bf a}$ is usually denoted $\vert a\vert$ (``the modulus of $a$'') or sometimes just $a$, so
{\bf a} \cdot {\bf a} = \vert a\vert^2 = a^2.
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Figure 5: A vector triangle.
\epsfysize =1.75in
Let us now investigate the general case. The length squared of $AB$ in Fig. 5 is
({\bf b} - {\bf a} ) \cdot ({\bf b} - {\bf a} ) = \vert a\vert^2 + \vert b\vert^2 - 2\,{\bf a} \cdot
{\bf b}.
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However, according to the ``cosine rule'' of trigonometry,
(AB)^2 = (OA)^2 + (OB)^2 - 2 \,(OA)\,(OB)\,\cos\theta,
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where $(AB)$ denotes the length of side $AB$. It follows that
{\bf a} \cdot {\bf b} = \vert a\vert\, \vert b\vert\, \cos\theta.
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Clearly, the invariance of ${\bf a} \cdot {\bf b}$ under transformation is equivalent to the invariance of the angle subtended between the two vectors. Note that if ${\bf a} \cdot {\bf b} =0$ then either $\vert a\vert=0$, $\vert b\vert=0$, or the vectors $\bf a$ and $\bf b$ are mutually perpendicular. The angle $\theta$ subtended between two vectors can easily be obtained from the dot product: i.e.,
\cos\theta = \frac{{\bf a} \cdot {\bf b}}{\vert a\vert \,\vert b\vert }.
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Note that $a_x=a\,\cos\theta_x$, etc., where $\theta_x$ is the angle subtended between vector ${\bf a}$ and the $x$-axis.

The work $W$ performed by a constant force $\bf F$ which moves an object through a displacement $\bf r$ is the product of the magnitude of $\bf F$ times the displacement in the direction of $\bf F$. So, if the angle subtended between $\bf F$ and $\bf r$ is $\theta$ then

W = \vert F\vert \,(\vert r\vert\,\cos\theta) = {\bf F}\cdot {\bf r}.
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next up previous
Next: The Vector Product Up: Vectors Previous: Vector Area
Richard Fitzpatrick 2007-07-14