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Vectors: The Scalar Product
Figure 1: {\bf a} \cdot{\bf b} = ab\cos\theta.
If the vectors a and b have magnitudes a and b respectively, and if the angle
between them is \theta, then the scalar product of a and b is defined
to be
{\bf a} \cdot{\bf b} = ab\cos\theta.
The scalar product is quite clearly commutative:
{\bf a} \cdot{\bf b} = {\bf b}\cdot{\bf a}.
It's less easy to show that it's also distributive over addition:
{\bf a} \cdot({\bf b}+{\bf c})={\bf a}\cdot{\bf b}+{\bf a}\cdot{\bf c}.
However, this is indeed the case, and it's why we call it a "product" at
all.
Note that
\begin{array}{ccccccc}
{\bf i}\cdot{\bf i}&=&{\bf j}\cdot{\bf j}&=&{\bf k}\cdot{\bf k}&=&1,\\
{\bf i}\cdot{\bf j}&=&{\bf j}\cdot{\bf k}&=&{\bf k}\cdot{\bf i}&=&0.
\end{array}
From these results, together with the distributivity property, it's easy
to show that the scalar product of
a = a_1{\bf i}+a_2{\bf j}+a_3{\bf k}
and
b = b_1{\bf i}+b_2{\bf j}+b_3{\bf k}
is
a \cdot{\bf b}=a_1b_1+a_2b_2+a_3b_3.
This yields an easy method for calculating the angle
between two vectors given in component form. For example, the angle \theta
between the vectors
a = 9{\bf i} - 2{\bf j} -6{\bf k}
and
b = {\bf i} - 2{\bf j} + 2{\bf k}
is calculated as follows. We know that
a\cdot{\bf b}=ab\cos\theta.
That is,
1\times9+(-2)\times(-2)+(-6)\times2=\sqrt{9^2+2^2+6^2}\times\sqrt{1^2+2^2+2^2}\cos\theta,
or simply
1=33\cos\theta.
it follows that \theta = \cos^{-1}(1/33) \approx 1.54 radians.
Figure 2: a\cos\theta. , the length of the projection of a in the direction of b
Note that the dot product of a with a unit vector \hat {\bf b} is
a \cos \theta, the length of the projection of a in the direction
of b.