What Is The Dot Product Used For

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Fundamental Questions

Why are dot products used for?
What does information technology mean when the dot product of two vectors is nada?
How practice you utilize a dot production to discover the angle between two vectors?
What does it mean when the scalar component of the project \(\|\proj_{\vec{A}}\vec{B}\|\) is negative?

Unlike ordinary algebra where there is but one mode to multiply numbers, there are two singled-out vector multiplication operations. The first is called the dot product or scalar product considering the upshot is a scalar value, and the 2nd is called the cross product or vector product and has a vector result. The dot product will be discussed in this department and the cross product in the next.

For two vectors \(\vec{A}= \langle A_x, A_y, A_z \rangle\) and \(\vec{B} = \langle B_x, B_y, B_z \rangle,\) the dot production multiplication is computed past summing the products of the components.

\begin{equation} \vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z \text{.}\tag{2.7.1} \end{equation}

An alternate, equivalent method to compute the dot product is

\begin{equation} \vec{A} \cdot \vec{B} = | \vec{A} | | \vec{B} |\cos \theta = A\ B\ \cos \theta \tag{2.7.2} \terminate{equation}

where \(\theta\) in the equation is the angle betwixt betwixt the two vectors and \(| \vec{A} |\) and \(| \vec{B} |\) are the magnitudes of \(\vec{A}\) and \(\vec{B}\text{.}\)

We can conclude from this equation that the dot production of 2 perpendicular vectors is zero, because \(\cos \ang{xc} = 0\text{,}\) and that the dot product of two parallel vectors is the product of their magnitudes.

When dotting unit vectors which take a magnitude of one, the dot products of a unit vector with itself is one and the dot product 2 perpendicular unit vectors is zero, so for \(\ihat\text{,}\) \(\jhat\text{,}\) and \(\khat\) we accept

\begin{align*} \ihat \cdot \ihat \amp = 1 \amp \jhat \cdot \ihat \amp = 0 \amp \khat \cdot \ihat \amp = 0\\ \ihat \cdot \jhat \amp = 0 \amp \jhat \cdot \jhat \amp = 1 \amp \khat \cdot \jhat \amp = 0\\ \ihat \cdot \khat \amp = 0 \amp \jhat \cdot \khat \amp = 0 \amp \khat \cdot \khat \amp = 1 \terminate{marshal*}

Dot products are commutative, associative and distributive:

Commutative. The order does non matter.

\begin{equation} \vec{A}\cdot\vec{B} = \vec{B}\cdot\vec{A}\tag{2.7.3} \end{equation}
Associative. It does non matter whether you multiply a scalar value \(C\) by the final dot product, or either of the individual vectors, yous volition yet get the same respond.

\begin{equation} C\left( \vec{A}\cdot\vec{B} \right ) = C\ \vec{A}\cdot\vec{B}= \vec{A}\cdot C\ \vec{B}\tag{2.7.4} \end{equation}
Distributive. If you are dotting one vector \(\vec{A}\) with the sum of ii more \((\vec{B}+\vec{C})\text{,}\) you can either add \(\vec{B}+\vec{C}\) first, or dot \(\vec{A}\) by both and add the final value.

\begin{equation} \vec{A}\cdot\left( \vec{B} + \vec{C} \right ) = \vec{A}\cdot\vec{B}+ \vec{A}\cdot \vec{C}\tag{two.7.5} \cease{equation}

Dot products are a particularly useful tool which can be used to compute the magnitude of a vector, determine the angle between 2 vectors, and find the rectangular component or projection of a vector in a specified direction. These applications will be discussed in the post-obit sections.

Subsection 2.seven.1 Magnitude of a Vector

Dot products tin can be used to notice vector magnitudes. When a vector is dotted with itself using (2.vii.one), the issue is the square of the magnitude of the vector. Past the Pythagorean theorem

\begin{equation} |\vec{A}| = \sqrt{\vec{A} \cdot \vec{A}}\text{.}\tag{2.vii.6} \end{equation}

The proof is trivial. Consider vector \(\vec{A} = \langle A_x, A_y \rangle\text{.}\)

\begin{align*} \vec{A} \cdot \vec{A} \amp = A_x A_x + A_y A_y = A_x^2 + A_y^2 \\ \sqrt{\vec{A} \cdot \vec{A}} \amp = \sqrt{A_x^2 + A_y^ii} = A = | \vec{A} |\text{.} \end{align*}

The results are similar for three-dimensional vectors.

Example 2.vii.1 . Find Vector Magnitude using the Dot Product.

Find the magnitude of vector \(\vec{F}\) with components \(F_x = \N{30}\text{,}\) \(F_y=\North{-40}\) and \(F_z = \N{l}\)

Respond .

\begin{equation*} F = |\vec{F}| = \N{lxx.vii} \end{equation*}

Solution .

\begin{marshal*} \vec{F} \amp = \langle \N{30}, \N{-xl}, \N{50} \rangle\\ \\ \vec{F} \cdot \vec{F} \amp = F_x^ii + F_y^ii + F_z^2\\ \amp = (\Due north{30})^two +(\N{-40})^2 + (\Due north{fifty})^ii\\ \amp = \N{5000}^2\\ \\ F \amp = |\vec{F}| = \sqrt{\vec{F} \cdot \vec{F}}\\ \amp = \sqrt{\North{5000}^2}\\ \amp = \N{70.7} \cease{align*}

Subsection 2.7.2 Angle between Two Vectors

A second awarding of the dot product is to find the angle betwixt 2 vectors. Equation (2.7.two) provides the procedure.

\begin{align} \vec{A} \cdot \vec{B} \amp = | \vec{A} | | \vec{B} |\cos \theta \notag\\ \cos \theta \amp = \frac{\vec{A} \cdot \vec{B}}{ | \vec{A} | | \vec{B} |}\tag{two.7.7} \end{align}

Example two.7.2 . Bending between Orthogonal Unit of measurement Vectors.

Detect the bending between \(\ihat= \langle one,0,0 \rangle\) and \(\jhat =\langle 0,ane,0 \rangle\text{.}\)

Reply .

\begin{equation*} \theta= \ang{90} \end{equation*}

Solution .

\begin{marshal*} \cos \theta \amp = \frac{\ihat \cdot \jhat}{ | \ihat | | \jhat |}\\ \amp = \frac{(ane)(0) + (0)(1) + (0)(0)}{(1)(i)}\\ \amp = 0 \\ \\ \theta \amp = \cos^{-one}(0) \\ \amp = \ang{90} \end{align*}

This shows that \(\ihat\) and \(\jhat\) are perpendicular to each other.

Case 2.vii.3 . Bending between Ii Vectors.

Observe the angle between \(\vec{F} = \langle \N{100}, \Due north{200}, \North{-fifty} \rangle \)and \(\vec{G} = \langle \N{-75}, \N{150}, \N{-forty} \rangle\text{.}\)

Answer .

\begin{equation*} \theta= \ang{51.seven} \stop{equation*}

Solution .

\begin{align*} \cos \theta \amp = \frac{\vec{F} \cdot \vec{One thousand}}{ | \vec{F} | | \vec{Grand} |}\\ \amp = \frac{ F_x G_x + F_y G_y + F_z G_z }{\sqrt{F_x^2 + F_y^2 + F_z^ii}\sqrt{G_x^2 + G_y^2 + G_z^2}}\\ \amp = \frac{(100)(-75) + (200)(150) + (-50)(-twoscore)} {\sqrt{100^two + 200^2 + (-50)^two} \sqrt{(-75)^ii + 150^2 + (-forty)^ii}}\\ \amp = \frac{24500}{(229.ane)(172.4)}\\ \amp = 0.620\\ \\ \theta \amp = \cos^{-1}(0.620) \\ \amp = \ang{51.7} \finish{marshal*}

Subsection 2.seven.3 Vector Projection

The dot product is used to find the projection of one vector onto another. You can think of a project of \(\vec{B}\) on \(\vec{A}\) as a vector the length of the shadow of \(\vec{B}\) on the line of action of \(\vec{A}\) when the sun is straight above \(\vec{A}\text{.}\) More precisely, the projection of \(\vec{B}\) onto \(\vec{A}\) produces the rectangular component of \(\vec{B}\) in the direction parallel to \(\vec{A}\text{.}\) This is 1 side of a rectangle aligned with \(\vec{A}\text{,}\) having \(\vec{B}\) as its diagonal.

This is illustrated in Effigy two.7.4, where \(\vec{u}\) is the projection of \(\vec{B}\) onto \(\vec{A}\text{,}\) or alternately \(\vec{u}\) is the rectangular component of \(\vec{B}\) in the direction of \(\vec{A}\text{.}\)

In this text we will utilize the symbols

\(\proj_{\vec{A}}\vec{B}\) to mean the projection of \(\vec{B}\) on \(\vec{A}\text{,}\) a vector quantity,
\(|\proj_{\vec{A}}\vec{B}|\) to mean the magnitude of the projection, a positive or goose egg valued scalar, and
\(\|\proj_{\vec{A}}\vec{B}\|\) to mean the scalar component of the project (the scalar projection), a signed scalar.

Equally we have mentioned before, the magnitude of a vector is its length and is always positive or cypher, while a scalar component is a signed value which can exist positive or negative. When a scalar component is multiplied by a unit of measurement vector the result is a vector in that direction when the scalar component is positive, or \(\ang{180}\) reverse when the scalar component is negative.

Figure two.7.4. Vector projection in ii dimensions.

The interactive shows that the project is the adjacent side of a right triangle with \(\vec{B}\) as the hypotenuse. From the definition of the dot product (two.vii.two) nosotros find that

\begin{equation*} \vec{A}\cdot \vec{B} = A ( B\ \cos \theta) = A\ \|\proj_A B\|\text{,} \end{equation*}

where \(B\ \cos \theta\) is the scalar component of the project. So, the dot production of \(\vec{A}\) and \(\vec{B}\) gives u.s. the project of \(\vec{B}\) onto \(\vec{A}\) times the magnitude of \(\vec{A}\text{.}\) This value will exist positive when \(\theta < \ang{90}\text{,}\) negative when \(\theta > \ang{90}\text{,}\) and cipher when the vectors are perpendicular because of the properties of the cosine function.

So, to find the scalar value of the projection of \(\vec{B}\) onto \(\vec{A}\) we divide by the magnitude of \(\vec{A}\)

\brainstorm{equation} \|\proj_{\vec{A}}\vec{B} \| =\frac{\vec{A}\cdot \vec{B}}{A}=\lid{\vec{A}}\cdot \vec{B}\tag{2.7.8} \cease{equation}

The final simplified course is written in terms of the unit vector in the direction vector \(\hat{\vec{A}}=\dfrac{\vec{A}}{A}\text{.}\)

If you desire the vector projection of \(\vec{B}\) onto \(\vec{A}\text{,}\) every bit opposed to the scalar projection we just found, multiply the scalar projection by the unit vector \(\hat{\vec{A}}\)

\begin{equation} \proj_{\vec{A}}\vec{B} = \|\proj_{\vec{A}}\vec{B} \| \lid{\vec{A}} = \left (\hat{\vec{A}} \cdot \vec{B} \right )\chapeau{\vec{A}}\text{.}\tag{2.7.9} \end{equation}

Similarly, the vector projection of \(\vec{A}\) onto \(\vec{B}\) is

\begin{equation} \proj_{\vec{B}}\vec{A} ={\left (\vec{A} \cdot \hat{\vec{B}} \right )\lid{\vec{B}}}\text{.}\tag{two.7.10} \terminate{equation}

The spatial interpretation of the results the scalar projection \(\|\proj_A B\|\) is

Positive value.
means that \(\vec{A}\) and \(\vec{B}\) are more often than not in the same direction.
Negative value.
ways that \(\vec{A}\) and \(\vec{B}\) are generally in opposite directions.
Cypher.
means that \(\vec{A}\) and \(\vec{B}\) are perpendicular.
Magnitude smaller than \(\vec{B}\).
This is the most common answer. The vectors are neither parallel nor perpendicular.
Magnitude equal to \(\vec{B}\).
\(\vec{A}\) and \(\vec{B}\) signal in the same management, thus 100% of \(\vec{B}\) acts in the direction of \(\vec{A}\text{.}\)
Magnitude larger than \(\vec{B}\).
This answer is impossible. Bank check your algebra; y'all might have forgotten to divide past the magnitude of \(\vec{A}\text{.}\)

Figure 2.7.5. Vector projections in 3 dimensions.

Subsection two.7.4 Perpendicular Components

The final application of dot products is to find the component of one vector perpendicular to some other.

To detect the component of \(\vec{B}\) perpendicular to \(\vec{A}\text{,}\) first find the vector projection of \(\vec{B}\) on \(\vec{A}\text{,}\) then decrease that from \(\vec{B}\text{.}\) What remains is the perpendicular component.

\begin{equation} \vec{B}_\perp = \vec{B} - \proj_{\vec{A}}\vec{B}\tag{2.vii.eleven} \end{equation}