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The Scalar, or Dot Product

It is also possible form several ``multiplication-like'' products of two (or more) vectors. We can take two vectors and make a scalar, another vector, or a ``bivector'' (tensor). Some of these might be regular version of the objects, some might be ``pseudo'' versions that we will come to understand. However, we have to be careful not to get swept off of our feet by the dazzling array of possibilities right at the beginning.

We will therefore start with the arguably simplest form of vector multiplication: the scalar or dot product: $ C = \vA \cdot
\vB$ . Note that the dot product turns two vectors into a scalar. It is also often called an inner product, although the latter is somewhat more general than the dot product in a Euclidean (e.g. Cartesian) space.

The dot product is commutative:

$\displaystyle \vA \cdot \vB = \vB \cdot \vA $

It is distributive:

$\displaystyle \vA \cdot (\vB + \vC) = \vA \cdot \vB + \vA \cdot \vC $

It can be evaluated two (important) ways:

$\displaystyle C = \vA \cdot \vB = A B \cos(\theta) = A_xB_x + A_yB_y + A_zB_z $

where $ A$ and $ B$ are the scalar magnitudes of the vectors $ \vA$ and $ \vB$ respectively3.1 and $ \theta$ is the angle in between them:

AB-theta-vectors.eps

From the first of these forms, we see that the dot product can be thought of as the magnitude of the vector $ \vA$ times the magnitude of the component of the vector $ \vB$ in the same direction as $ \vA$ , indicated as $ B_\parallel$ in the figure above. Indeed:

$\displaystyle \vA \cdot \vB = A B_\parallel = A_\parallel B $

(The latter the magnitude of $ \vB$ times the component of $ \vA$ parallel to $ \vB$ .)

The second follows from the following multiplication table of unit vectors, which can be thought of as defining the dot product and the unit vectors of ``orthonormal coordinates'' simultaneously:

$\displaystyle \hx\cdot\hx = \hy\cdot\hy = \hz\cdot\hz = 1 $

$\displaystyle \hx\cdot\hy = \hy\cdot\hz = \hz\cdot\hx = 0 $

(plus the commutated forms of the last row, e.g. $ \hy\cdot\hx = 0$ as well).

Two vectors that are perpendicular (orthogonal) have a dot product of zero and vice-versa. If and only if (written henceforth as ``iff'') $ \vA \cdot \vB = 0$ then $ \vA \perp \vB$ . We might say that $ \vA$ is normal to, perpendicular to, at right angles to, or orthogonal to $ \vB$ . All of these mean the same thing.



Subsections
next up previous contents
Next: The Law of Cosines Up: Vectors and Vector Products Previous: Scalars and Vectors   Contents
Robert G. Brown 2017-07-11