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 Ɵ which is determined by u and v which satisfies 0≤Ɵ≤π, and is shown in the following image,

Definition

If u and v are vectors in space-2 or space-3 and Ɵ 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 45⁰.

Solution:

Theorem 1.

Suppose u and v are vectors in Space 2 or space 3,

1. v.v = |v|²; that is, |v|= (v.v)^1/2
2. if u and v are nonzero vectors and Ɵ is the angle between the two vectors, then

           Ɵ taper if and only if u.v> 0

           Ɵ blunt if and only if u.v <0

           Ɵ = π / 2 if and only if u.v = 0

Proof:

Because the angle Ɵ between v and v is 0, we get it

v.v = |v| |v| cos Ɵ = |v|² cos Ɵ = |v|²

Because |u| > 0, |v| > 0, and u.v = |u| |v| cos Ɵ, then u.v has an equal sign like cos Ɵ. Because Ɵ satisfies Ɵ ≤ 0 ≤ π, then the angle Ɵ is acute if and only if cos Ɵ> 0; and Ɵ blunt if and only if cos Ɵ <0; and Ɵ = π / 2 if and only if cos Ɵ = 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 = (u₁, u₂)  can be expressed singly with i and j, namely:

u = (u₁, u₂) = u₁(1, 0) + u₂(0, 1) = u₁i  + u₂j

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 u₁i  + u₂j

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 = -5i + 8j

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

1. u.v = y.u commutative
2. u.(v + w) = u.v + u.w (distributive)
3. k(u.v) = (k u).v = u.(k v)
4. 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 = (u₁, u₂, u₃) and v = (v₁, v₂, v₃)

Then,

k (u.v) = k ( u₁ v₁ + u₂ v₂ + u₃ v₃)

k (u.v) = (k u₁) v₁ + (k u₂) v₂ + (k u₃) v₃)

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 w₁ from Q to the line perpendicular to it. The next form will be different

w₂ = u – w₁

As shown in Figure 6, vector w₁ is parallel to a, vector w₂ is perpendicular to a, and

W₁ + w₂ = w₁ + (u – w₁) = u

We call the vector w₁ an orthogonal projection u in a or sometimes we call the vector component u along a. We declare this

Proj _a u

The vector w₂ we call the vector component u is orthogonal to a. Because w₂ = u - w1, we can write this vector in notation as

W₂ = u – proj _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 w₁ = proj _a u and w₂ = u - proj _a u. Because w₁ is parallel to a, we must multiply a scalar, so we can write w₁ = ka.

So,

u = w₁ + w₂ = ka + w₂

By taking the results of the points from both sides with a, namely:

u . a = (ka + w₂) .  a = k ||a||² + w₂ . a

But w₂. a = 0 because w₂ is perpendicular to a, so

Because proy _a u = w1 = ka, 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 = 4² + (-1)² + 2² = 21

So, the component of vector u along a is

And the component vector u which is orthogonal to a is

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