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 Depdendent Variable

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 Dependent Variable

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# Vectors

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. and if there is only one way to do so for any given vector (i.e. the vector determines the coefficients ). 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 vector 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 .

One has

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 AB Theorem 43.3. In a parallelogram OACB the sum of the squares of the lengths of the two diagonals equals the sum of the squares of the lengths of all four sides. Proof. The squared lengths of the diagonals are 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
implies

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

so that

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

whence

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 ยง 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 description 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 not. 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 parallel to
• 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 line .

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.

 43.5 Problem. Find a defining equation for the line which goes through A(1, 1) and is perpendicular 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. With we then get the following equation for If you choose the normal instead, you get Both equations and are equivalent