The next definition we shall make is that of scalar product, whereby two vectors are combined to give a number.
Let A=\left(a_{1}, a_{2}, \ldots, a_{n}\right), B=\left(b_{1}, b_{2}, \ldots, b_{n}\right). We define the scalar product A\boldsymbol{\cdot} B by A \boldsymbol{\cdot} B=a_{1} b_{1}+a_{2} b_{2}+\ldots+a_{n} b_{n}
It is left as an exercise to prove the following:
- A\boldsymbol{\cdot}(B+C)=A\boldsymbol{\cdot} B+A\boldsymbol{\cdot} C
- A\boldsymbol{\cdot} B=B\boldsymbol{\cdot} A
- a(B\boldsymbol{\cdot} C)=(a B)\boldsymbol{\cdot} C=B\boldsymbol{\cdot}(a C)
In particular, we define A^{2}=A\boldsymbol{\cdot} A. Thus A^{2}=a_{1}^{2}+a_{2}^{2}+\ldots+a_{n}^{2}. But if we recall that |A|=\sqrt{a_{1}^{2}+a_{2}^{2}+\ldots+a_{n}^{2}} we have A^{2}=|A|^{2}.
(Note that this equation makes sense, since both sides are numbers. Note also that if A \neq(0,0, \ldots, 0), A^{2}>0.)
Now let us try to express, in terms of the language we have developed, what is to be meant by saying that two vectors are perpendicular. Suppose that A and B of the diagram are perpendicular. In particular, this means that if we lay off a segment of length |B| to the left of their common initial point, the two line segments marked || are of equal length. But if we consider the broken line segment as a vector directed to the left, we observe that this vector is -B, and that the two equal line segments, directed as indicated by the arrows, are A-B and A-(-B)=A+B. Thus a necessary condition for perpendicularity is |A-B|=|A+B|.
Conversely, if this is the case, then A is perpendicular to B in the sense of plane geometry. Squaring both sides of this equation and using the fact that A^{2}=|A|^{2} for any vector A, we have (A-B)^{2}=|A-B|^{2}=|A+B|^{2}=(A+B)^{2} Using the rules developed in the exercises, (A-B)^{2}=A^{2}-2 A\boldsymbol{\cdot} B+B^{2},(A+B)^{2}=A^{2}+2 A\boldsymbol{\cdot} B+B^{2}
Thus A^{2}-2 A\boldsymbol{\cdot} B+B^{2}=A^{2}+2 A\boldsymbol{\cdot} B+B^{2}, or
-2 A \boldsymbol{\cdot} B=2 A \boldsymbol{\cdot} B, \quad \text { i.e., } \quad A\boldsymbol{\cdot} B=0
Conversely, if A\boldsymbol{\cdot} B=0, we see that |A-B|^{2}=|A+B|^{2}. But since both |A-B| and |A+B| are non-negative, this means that |A-B|=|A+B|, or that A and B are perpendicular.
Thus we may regard this result as our definition of perpendicularity in n dimensions:
A is perpendicular to B will mean simply that A\boldsymbol{\cdot} B=0.
Now suppose A and B are non-zero vectors (recalling that the zero vector is (0,0, \ldots, 0) ). Then the quantity |B| A-|A| B is a vector. Therefore the number (|B| A-|A| B)^{2} is either positive or zero, and is zero only if |B| A-|A| B=0 (vector), i.e., only if |B| A=|A| B Now \begin{aligned} (|B| A-|A| B)^{2} &=|B|^{2} A^{2}-2|A||B| A \boldsymbol{\cdot} B+|A|^{2} B^{2} \\ &=|B|^{2}|A|^{2}-2|A||B| A \boldsymbol{\cdot} B+|A|^{2}|B|^{2} \\ &=2|A||B|(|A||B|-A\boldsymbol{\cdot} B) \end{aligned} and this is \geq 0. Since 2|A||B| is positive, we can cancel it and get |A||B|-A B \geq 0, or A \boldsymbol{\cdot} B \leq|A||B| The two are equal only if |B| A=|A| B, i.e., only if A and B have the same direction.
Likewise, (|B| A+|A| B)^{2} \geq 0, and is zero only if |B| A=-|A| B, i.e., only if A and B have opposite directions. But (|B| A+|A| B)^{2}=2|A||B|(|A||B|+A \boldsymbol{\cdot} B) \geq 0 giving |A||B|+A \boldsymbol{\cdot} B \geq 0, \quad \text { or } \quad-|A||B| \leq A \boldsymbol{\cdot} B
Equality holds only if A and B have opposite directions. Combining the two results, we have in general -|A||B| \leq A \boldsymbol{\cdot} B \leq|A||B| and we know exactly when one or the other of the inequalities becomes an equality. This result is known as Schwarz’s inequality.
Dividing the inequality through by |A||B| (which is positive), we have
-1 \leq \frac{A \boldsymbol{\cdot} B}{|A||B|} \leq 1
Therefore the number \dfrac{A \boldsymbol{\cdot} B}{|A||B|} can be the cosine of some angle. If we require that this angle \theta be between 0^{\circ} and 180^{\circ} (or in radians, between 0 and \pi ), then \theta is uniquely determined by giving \cos \theta. Thus we define the angle between A and B to be that angle \theta between 0^{\circ} and 180^{\circ} such that \cos \theta=\frac{A \boldsymbol{\cdot} B}{|A||B|}.
Observe that \theta=0^{\circ} if and only if A and B have the same direction, \theta=180^{\circ} if and only if they have opposite directions. So our terminology is consistent in this respect. Plane trigonometry will provide a more complete justification of our choice of definitions. Consider the triangle shown. The law of cosines tells us that |A-B|^{2}=|A|^{2}+|B|^{2}-2|A||B| \cos \theta
Using what we know about vectors, this means that (A-B)^{2}=A^{2}+B^{2}-2|A||B| \cos \theta or \begin{gathered} A^{2}-2 A\boldsymbol{\cdot} B+B^{2} \\ =A^{2}+B^{2}-2|A||B| \cos \theta \end{gathered} i.e., -2 A \boldsymbol{\cdot} B=-2|A||B| \cos \theta and finally \bbox[5px,border:1px solid black;background-color:#f2f2f2]{A \boldsymbol{\cdot} B=|A||B| \cos \theta} which agrees with our definition of \theta.
