3.6 The scalar or dot product

The scalar product, also called dot product or inner product, of a and b is written as a⋅b, and is defined as

This is clearly a number (scalar) and not a vector. The angle {θ}_{ab} is the angle between the first and second vector, and thus

(One usually suppresses the subscript ab on the angle θ.) We thus see that order does not matter, or more formally, that the dot product is commutative.

Let us look at some special cases

1. a is perpendicular to b. In that case θ = 9{0}^{∘} = π∕2, and the cosine is zero: a⋅b = 0.
2. a is parallel to a, i.e., θ = 0. a⋅a = {a}^{2}. For that reason one sometimes writes {a}^{2} for {a}^{2}. Also

   \class{boxed}{a = \sqrt{a ⋅ a}\quad . }

3. i⋅i = j ⋅j = k ⋅k = 1\quad ,
   i⋅j = j ⋅k = k ⋅i = 0\quad .

   This is a straighforward application of the previous two properties! A set where each vector is orthogonal to all the others is called an orthogonal set of vectors; if the vectors also have unit length, one speaks of an orthonormal set.

It is generally useful to list a few more properties:

1. (ma) ⋅b = (ma)b\mathop{cos}\nolimits θ = m(ab\mathop{cos}\nolimits θ) = ma⋅b. (What is (2a) ⋅ (2b)?)
2. (a⋅b)c is the product of the scalar a⋅b with the vector c. Thus the result has the same direction as c, with magnitude (a⋅b)c.
3. We can divide by a⋅b since it is a scalar! (Conversely, we can not divide by a vector!)
4. a⋅ (b + c) = a⋅b + a⋅c. (Distributive law). This will not be proven here, but can easily be done using component form.

Example 3.6: 

Simplify {(a + b)}^{2}

Solution: 

\begin{eqnarray*}{ (a + b)}^{2}& =& (a + b) ⋅ (a + b) %& \\ & =& (a + b) ⋅a + (a + b) ⋅b %& \\ & =& a⋅a + b⋅a + a⋅b + b⋅b%& \\ & =& {a}^{2} + {b}^{2} + 2a⋅b %& \\ \end{eqnarray*}

3.6.1 Component form of dot product

Let a = {a}_{1}i + {a}_{2}j + {a}_{3}k, b = {b}_{1}i + {b}_{2}j + {b}_{3}k, then

Example 3.7: 

Find a unit vector which is perpendicular to (1,2,−1) and has y-component zero.

Solution: 

This vector has the form a = ({a}_{x},0,{a}_{z}). Must be orthogonal to (1,2,−1), so

({a}_{x},0,{a}_{z}) ⋅ (1,2,−1) = 0,

which leads to

{a}_{x} − {a}_{z} = 0,\qquad {a}_{x} = {a}_{z} = α.

For this to be a unit vector {a}^{2} = 2{α}^{2} = 1, or α = ±1∕\sqrt{2} (we can choose either sign. Explain!). Thus

\class{boxed}{a = ( {1\over \sqrt{2}},0, {1\over \sqrt{2}})\quad . }