# Dot Products and Vector Projections

In this article we
will introduce a kind of vector multiplication in space-2 and space-3. The
properties of the multiplication arithmetic will be determined and some of the
applications will be given.

Suppose

**u**and**v**are two nonzero vectors in space-2 and space-3, and suppose that these vectors have been located so that the starting point coincides. What we mean by the angle between**u**and**v**, is the angle ÆŸ which is determined by**u**and**v**which satisfies**0****≤****ÆŸ****≤Ï€**, and is shown in the following image,
Definition

If

**u**and**v**are vectors in space-2 or space-3 and ÆŸ is the angle between**u**and**v**, then the product dot (dot product) or product in Euclidis (*Euclidean Inner Product*)**u.v**is defined as,
Example 1.

Determine the product of
vector

**u**= (0, 0, 1) and vector**v**= (0, 2, 2) with the angle between**u**and**v**is 45^{0}.
Solution:

Theorem 1.

Suppose

**u**and**v**are vectors in Space 2 or space 3,**v**.**v**= |**v**|^{2}; that is, |**v**|= (**v**.**v**)^{1/2}- if
**u**and**v**are nonzero vectors and ÆŸ is the angle between the two vectors, then

ÆŸ taper if and only if
u.v> 0

ÆŸ blunt if and only if u.v <0

ÆŸ = Ï€ / 2 if and only if u.v = 0

Proof:

Because the angle ÆŸ between

**v**and**v**is 0, we get it**v.v**= |v| |v| cos ÆŸ = |v|

^{2}cos ÆŸ = |v|

^{2}

^{}

Because |u| > 0, |v|
> 0, and u.v = |u| |v| cos ÆŸ, then

**u.v**has an equal sign like cos ÆŸ. Because ÆŸ satisfies ÆŸ ≤ 0 ≤ Ï€, then the angle ÆŸ is acute if and only if cos ÆŸ> 0; and ÆŸ blunt if and only if cos ÆŸ <0; and ÆŸ = Ï€ / 2 if and only if cos ÆŸ = 0.
Example 2.

If u = (1, -2, 3), v = (-3, 4, 2), and w = (3, 6, 3), then

u.v = (1)(-3) + (-2)(4) +
(3)(2) = -5

v.w = (-3)(3) + (4)(6) +
(2)(3) = 21

u.w = (1)(3) + (-2)(6) + (3)(3)
= 0

So,

**u**and**v**form blunt angles,**v**and**w**form sharp angles, and**u**and**w**are perpendicular to each other.
A perpendicular vector is
also called an orthogonal vector. In the explanation of Theorem 1 (2), two non-zero vectors are perpendicular if and only if
the result of the point is zero. If we agree that

**u**and**v**are rather perpendicular then one or both of these vectors must be**0**, because we can declare without exception that both vectors**u**and**v**will be orthogonal if and only if**u****.****v**is 0. To determine that u and v are orthogonal vector then we can write u _|_ v.### Row Vector Equation

Suppose i = (1, 0) and j =
(0, 1) and note that these vectors are perpendicular and that the length is
equal to one. These vectors

**i**and**j**are called row vectors, because each vector**u**= (u*, u*_{1}*) can be expressed singly with*_{2}**i**and**j**, namely:**u**= (u

*, u*

_{1}*) = u*

_{2}*(1, 0) + u*

_{1}*(0, 1) = u*

_{2}*i + u*

_{1}*j*

_{2}
the geometric meaning of
the relationship can be seen in Figure 3.

Example 3.

If

**u**are arrows from P (2, 1) and Q (-3, 7), write**u**as**u***i +*_{1}**u***j*_{2}
Solution:

We slide the arrow, so
that its base coincides with the origin (Figure 4). This can be done by
reducing the starting point component from the endpoint. Then obtained (-3-2,
7-(-1)=(-5, 8).

So that,

**u**= -5

**i**+ 8

**j**

Example 4.

Determine the angle of ABC
with A = (4, 3), B = (1, -1) and C = (6, -4).

Solution:

Based on Figure 5 we get,

Then the ABC angle is:

The following theorem will
describe most of the important properties of the results of that point. The
results of this point will be useful in calculations that include vectors.

Theorem 2.

If

**u**,**v****,**and**w**are vectors in space-2 or space-3 and**k**is scalar, then**u.v**=**y.u**commutative**u.(v + w) = u.v + u.w**(distributive)**k(u.v) = (k u).v = u.(k v)****v.v**> 0 jika**v**≠ 0 dan**v.v**= 0 jika**v**= 0

Proof:

We will prove (c) for
vectors in Space-3 and
leave the rest of the evidence as training for you.

Suppose

**u**= (u*, u*_{1}*, u*_{2}*) and*_{3}**v**= (v*, v*_{1}*, v*_{2}*)*_{3}
Then,

k (u.v) = k ( u

*v*_{1}*+ u*_{1}*v*_{2}*+ u*_{2}*v*_{3}*)*_{3}
k (u.v) = (k u

*) v*_{1}*+ (k u*_{1}*) v*_{2}*+ (k u*_{2}*) v*_{3}*)*_{3}
k (u.v) = (k u).v

As well,

K(u.v) = u. (k v)

In many applications this
is interesting enough to "decipher" the vector

**u**into the sum of two terms, the one equal to the vector**a**not zero while the other is perpendicular to**a**. If**u**and**a**are placed in such a way that the starting point will occupy point**Q**, we can describe the vector**u**as follows (Figure 6), decrease the perpendicular line from top**u**to the current through**a**, and form the vector w_{1}from Q to the line perpendicular to it. The next form will be different**w**

*= u – w*

_{2}

_{1}

_{}

_{}
As shown in Figure 6,
vector w

_{1}is parallel to**a**, vector w_{2}is perpendicular to**a**, and
W

*+ w*_{1 }*= w*_{2}*+ (u – w*_{1}*) = u*_{1}
We call the vector w

_{1}an orthogonal projection**u**in**a**or sometimes we call the vector component**u**along**a**. We declare this
Proj

_{a}**u**

The vector w

_{2}we call the vector component**u**is orthogonal to a. Because w_{2}=**u**- w1, we can write this vector in notation as
W

*= u – proj*_{2}_{a}u
The following theorem
provides a formula for calculating proj

_{a}u and u – proj_{a}u vectors
Theorem 3.

If

**u**and**a**are vectors in space -2 or in space -3 and if**a**≠ 0, then
proof:

Suppose that w

*= proj*_{1}_{a}u and w*= u - proj*_{2}_{a}u. Because w_{1}is parallel to**a**, we must multiply**a**scalar, so we can write w_{1}= k**a**.
So,

**u = w**= k

*+ w*_{1}_{2}**a**+

**w**

_{2}

_{}
By taking the results of
the points from both sides with

**a**, namely:**u**.

**a**= (k

**a**+ w

*) .*

_{2}**a**= k ||

**a**||

^{2}+ w

*.*

_{2}**a**

But w

_{2}.**a**= 0 because w_{2}is perpendicular to**a**, so
Because proy

_{a}u = w1 = k**a,**we get it
Example 5.

Suppose

**u**= (2, -1, 3) and**a**= (4, -1, 2). Find the component vector**u**along**a**and the component vector**u**which is orthogonal to**a**.
Solution:

**u . a**= (2)(4) + (-1)(-1) + (3)(2) = 15

|a|

^{2}= 4^{2}+ (-1)^{2}+ 2^{2}= 21
So, the component of vector

**u**along**a**is
And the component vector

**u**which is orthogonal to**a**isSUBSCRIBE TO OUR NEWSLETTER

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