42.2. A Basis of Vectors (in general)*
The vectors are called the standard basis vectors. They are an example of what is called a
“basis”. Here is the definition in the case of space vectors:
Definition 42.1. A triple of space vectors
is a basis if every space vector
can be written as a linear
combination of , i.e.
For plane vectors the definition of a basis is almost the same, except
that a basis consists of two
vectors rather than three:
Definition 42.2. A pair of plane vectors
is a basis if every plane
can be written as a linear
combination of , i.e. , and if there is only one way to do so for any given vector (i.e. the
vector determines the coefficients ).
|43. Dot Product
|Definition 43.1. The “inner product” or “dot product” of two vectors is given by
Note that the dot-product of two vectors is a number!
The dot product of two plane vectors is (predictably) defined by
An important property of the dot product is its relation with the length of a vector:
43.1. Algebraic properties of the dot product
The dot product satisfies the following rules,
which hold for all vectors
and any real number t.
43.2 Example. Simplify .
43.2. The diagonals of a parallelogram
Here is an example of how you can use the algebra of the dot product to
prove something in geometry.
Suppose you have a parallelogram one of whose vertices is the origin. Label the vertices, starting at
the origin and going around counterclockwise, O, A, C and B. Let . One has
These vectors correspond to the diagonals OC and
Adding both these equations you get
The squared lengths of the sides are
Together these also add up to .
43.3. The dot product and the angle between two vectors
Here is the most important interpretation of the dot product:
Theorem 43.4. If the angle between two vectors and is θ, then one has
Proof. We need the law of cosines from high-school
trigonometry. Recall that for a triangle OAB with
angle θ at the point O, and with sides OA and OB of lengths a and b, the length c of the opposing side
AB is given by
In trigonometry this is proved by dropping a perpendicular line from B onto the side OA. The triangle
Figure 20. Proof of the law of cosines
OAB gets divided into two right triangles, one of which has AB as hypotenuse. Pythagoras then
After simplification you get (60).
To prove the theorem you let O be the origin, and then observe that the length of the side AB is the
length of the vector . Here , and hence
Compare this with (60), keeping in mind that
and : you
are led to conclude that
, and thus .
43.4. Orthogonal projection of one vector onto another
The following construction comes up very often. Let be a given vector. Then for any other
vector there is a number λ such that
where . In other words,
you can write any vector
as the sum of one vector parallel to
and another vector orthogonal to . The two vectors and are called the parallel and orthogonal
components of the vector (with respect to), and sometimes the following notation is used
There are moderately simple formulas for
and ), but it is better
to remember the following derivation
of these formulas.
Assume that the vectors and vx are given. Then we look for a number λ such that is
perpendicular to . Recall that if and only if
Expand the dot product and you get this equation for λ
To compute the parallel and orthogonal components of
you first compute λ according to (61),
which tells you that the parallel coponent is given by
The orthogonal component is then “the rest,” i.e. by definition , so
43.5. Defining equations of lines
In § 41 we saw how to generate points on a line given two points on that line by means of a
“parametrization.” I.e. given points A and B on the line the point whose position vector is
will be on for any value of the “parameter” t
In this section we will use the dot-product to give a different
of lines in the plane (and planes in three dimensional space.) We will
derive an equation for a line. Rather than generating points on the line
this equation tells us if any given point X in the plane is on the line or
Here is the derivation of the equation of a line in the plane. To produce
the equation you need two ingredients:
1. One particular point on the line (let’s call this point A, and write for
its position vector),
2. a normal vector for the line, i.e. a nonzero vector which is perpendicular
to the line.
Now let X be any point in the plane, and consider the line segment AX.
Is X on ?
• Clearly, X will be on the line if and only if AX is
• Since is perpendicular to , the segment AX and the line will be parallel if and only if
• holds if and only if
So in the end we see that X lies on the line if and only if the following vector equation is satisfied:
This equation is called a defining equation for the
Any given line has many defining equations. Just by changing the length of the normal you get a
different equation, which still describes the same line.
Find a defining equation for the line which goes through A(1, 1) and
to the line segment AB where B is the point (3,−1).
Solution. We already know a point on the line, namely A, but we still need a normal vector. The
line is required to be perpendicular to AB, so is a normal vector:
Of course any multiple of is also a normal vector, for instance
is a normal vector.
If you choose the normal instead, you get
Both equations and are equivalent