Lesson Introduction to dot-product

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<H2>Introduction to dot-product</H2><Table>
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<BLOCKQUOTE>Definition

Let &nbsp;u&nbsp; and &nbsp;v&nbsp; be two vectors in a plane.

Then the &nbsp;dot-product&nbsp; (or the &nbsp;scalar product) &nbsp;of the vectors &nbsp;u&nbsp; and &nbsp;v&nbsp; is the real number &nbsp;|u|*|v|*{{{cos(alpha)}}},

&nbsp;where &nbsp;|u|&nbsp; and &nbsp;|v|&nbsp; are the magnitudes &nbsp;(the lengths)&nbsp; of the vectors &nbsp;u&nbsp; and &nbsp;v, &nbsp;and &nbsp;{{{alpha}}}&nbsp; is the angle between the vectors.

The dot-product of vectors &nbsp;|u|&nbsp; and &nbsp;|v|&nbsp; is designated as &nbsp;(u,v)&nbsp; often: &nbsp;&nbsp;(u,v) = |u|*|v|*{{{cos(alpha)}}}.</BLOCKQUOTE>

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locate ( 1.2,  3.1, u), 
            locate ( 2.7,  1.6, v),

arc(1.0, 1.0, 1.2, 1.2, 295, 342),
            locate (1.4, 1.9, alpha)
)}}}

&nbsp;&nbsp;Figure 1. &nbsp;To the definition of the 
dot-product &nbsp;(u,v)&nbsp; of vectors &nbsp;u&nbsp; and &nbsp;v

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Example 1

Calculate the dot-product of the two vectors &nbsp;u&nbsp; and &nbsp;v, &nbsp;if their magnitudes are &nbsp;u = 12, &nbsp;v = 5&nbsp; and the angle &nbsp;{{{alpha}}}&nbsp; between the vectors is &nbsp;{{{alpha}}} = 60�.

Solution

You have &nbsp;(u,v) = |u|*|v|*{{{cos(alpha)}}} = 12*5*cos(60�) = 60*{{{(1/2)}}} = 30.

Example 2

Find the dot-product of the two vectors &nbsp;u&nbsp; and &nbsp;v, &nbsp;if their magnitudes are &nbsp;u = 12, &nbsp;v = 5&nbsp; and the angle &nbsp;{{{alpha}}}&nbsp; between the vectors is &nbsp;{{{alpha}}} = 120�.

Solution

You have &nbsp;(u,v) = |u|*|v|*{{{cos(alpha)}}} = 12*5*cos(120�) = 60*{{{(-1/2)}}} = -30.

Note two important and useful properties of dot-product.<BLOCKQUOTE><TABLE BORDER=2>
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&nbsp;&nbsp;&nbsp;&nbsp;Two vectors &nbsp;u&nbsp; and &nbsp;v&nbsp; in a plane are perpendicular if and only if their dot-product is zero: (u,v) = 0.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(1)&nbsp;&nbsp;&nbsp;&nbsp;

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</TABLE></BLOCKQUOTE>

Indeed, &nbsp;in this case the angle &nbsp;{{{alpha}}}&nbsp; between the vectors is &nbsp;90�, &nbsp;and &nbsp;{{{cos(alpha)}}} = 0.

<BLOCKQUOTE><TABLE BORDER=2>
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&nbsp;&nbsp;&nbsp;&nbsp;For any vector &nbsp;u&nbsp; dot-product of the vector by itself is equal to the square of the length &nbsp;(of the magnitude)&nbsp; of this vector: &nbsp;(u,u) = {{{abs(u)^2}}}.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(2)&nbsp;&nbsp;&nbsp;&nbsp;

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</TABLE></BLOCKQUOTE>

Indeed, in this case the angle &nbsp;{{{alpha}}}&nbsp; between the vectors is &nbsp;0�, &nbsp;and &nbsp;{{{cos(alpha)}}} = 1.

Now it is good time to ask about dot product - to which purposes does it serve for, &nbsp;and where and how is it used?

Dot-product has many applications in &nbsp;Physics, &nbsp;especially in &nbsp;Mechanics, &nbsp;which is the part of &nbsp;Physics, &nbsp;and even in &nbsp;Geometry&nbsp; itself. &nbsp;Below are some examples.

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1. &nbsp;Imagine the body on the horizontal ground surface or on the horizontal floor, &nbsp;which&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

moves in the horizontal direction &nbsp;(Figure 2).

Let some force &nbsp;F&nbsp; be applied to the body as shown in the &nbsp;Figure 2.

Let the force be directed in some angle &nbsp;{{{alpha}}}&nbsp; to the horizontal plane. &nbsp;Then the body

is subject to the given force, &nbsp;to the weight directed vertically down, &nbsp;to the reaction

force from the floor to the body directed vertically up, &nbsp;and to the friction force

acting to the body along the ground surface or along the floor and directed oppositely

to the body move.

Let all these forces be in equilibrium providing the body uniform movement along the 
surface.

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Figure 2. &nbsp;The body on the horizontal floor
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This is a typical mechanical model or a system. &nbsp;You need to know the horizontal and the vertical components &nbsp;Fx&nbsp; and &nbsp;Fy&nbsp; of the applied force &nbsp;F&nbsp; to analyze the system.

These components are the scalar products &nbsp;(dot-products)&nbsp; of the vector &nbspF&nbsp; and the unit vectors along the selected directions &nbsp;(horizontal and vertical directions in this case):

Fx = (&nbsp;F,Ix) = |F|*{{{cos(alpha)}}} &nbsp; and &nbsp;Fy = (&nbsp;F,Iy) = |F|*{{{cos(pi/2-alpha)}}} = |F|*{{{sin(alpha)}}}, &nbsp;where &nbsp;Ix = (1,0)&nbsp; and &nbsp;Iy = (0,1)&nbsp; are mentioned unit vectors along the &nbsp;x- and &nbsp;y- coordinate axes.

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2. &nbsp;Consider a body which can move &nbsp;(or is forced to move)&nbsp; along a straight line&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

as shown in the &nbsp;Figure 3. &nbsp;You may think about a barge or a ship in the Panama

channel which is pulled by a track moving aside the channel.

Let the force &nbsp;F&nbsp; be applied to the body and compels it &nbsp;(possibly in combination

with other forces or constrains)&nbsp; to move along this line.

The work done by this force when the body makes a displacement &nbsp;s&nbsp; along the

straight line is the scalar product of the vector &nbsp;F&nbsp; and the displacement vector &nbsp;s:

W = (F,s). &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(3)

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            line (10.0, -0.3, 10.3,  0.0), 
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red(line ( 9.0, 0.0, 11.0,  1.0)),
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line (10.5,  1.3, 11.5,  1.3), 
            line (10.5,  0.7, 11.5,  0.7),
            line (10.5,  1.3, 10.5,  0.7), 
            line (11.5,  1.3, 11.5,  0.7),

line ( 2.0,  -0.6,   2.0,  -1.4),
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line (2.0,  -1.3,  5.0,  -1.3),
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line (9.0,  -1.3, 8.7,  -1.2),
            line (9.0,  -1.3, 8.7,  -1.4)

)}}}

Figure 3. &nbsp;The work done by a force &nbsp;(a ship in the 
Panama channel is pulled by a track moving aside the 
channel, &nbsp;plane view atop)
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3. &nbsp;If the body subjected to the acting force &nbsp;F&nbsp; moves along the straight or curved&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

line and has the velocity presented by the vector &nbsp;u&nbsp; in each time moment &nbsp;(Figure 4),

then the power &nbsp;P&nbsp; (in the physical sense)&nbsp; developed by the force &nbsp;F&nbsp; is equal in each

time moment to the scalar product of the vectors &nbsp;F&nbsp; and &nbsp;u:

P = (F,u). &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(4)

It does not matter whether the vectors &nbsp;F&nbsp; and &nbsp;u&nbsp; are co-directed or make some angle.

The force vector &nbsp;F&nbsp; may vary in the magnitude and/or in the direction, &nbsp;as well as

the velocity vector. The formula &nbsp;(4)&nbsp; for the power works in all cases.

If one is required to calculate the work done by the force on the curved path, &nbsp;then

the integral of the power &nbsp;(4)&nbsp; is calculated from the initial to the final time moment.

</TD>
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         locate (1.5, 5.0, u),

line(3.00, 3.2, 3.0,  1.0),
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            line(3.00, 1.0,  2.80, 1.4),
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            line(-2.00, 2.7, -2.85, 2.80),
         locate (-1.80, 3.0, F)
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Figure 4. &nbsp;The power developed by a force. 
The instant force vector &nbsp;F&nbsp; and the instant velocity 
vector &nbsp;u&nbsp; are presented. &nbsp;The orbital movement 
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5. &nbsp;In Geometry, &nbsp;dot-product is used often to prove perpendicularity of vectors. &nbsp;In this case the property &nbsp;(1)&nbsp; of dot-product is used.

Sometimes dot-product is used to prove congruency of angles. &nbsp;In this way, &nbsp;the optical (reflective) properties of conic sections - ellipses, &nbsp;parabolas and hyperbolas, &nbsp;are proved in the lessons &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections/Optical-property-of-an-ellipse.lesson>Optical property of an ellipse</A>, &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections/Optical-property-of-a-parabola.lesson>Optical property of a parabola</A> &nbsp;and &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections/Optical-property-of-a-hyperbola.lesson>Optical property of a hyperbola</A> &nbsp;under the topic &nbsp;Conic sections&nbsp; of the section &nbsp;Algebra-II&nbsp; in this site.

The advantage of this approach is in replacing a geometric proof by an algebraic one. &nbsp;You will find examples of solved problems in lessons that follow under the current topic.

My lessons on Dot-product in this site are

&nbsp;&nbsp;&nbsp;&nbsp;- Introduction to dot-product&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(this lesson)

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Vectors/Formula-for-Dot-product-of-vectors-in-a-plane-via-the-vectors-components.lesson>Formula for Dot-product of vectors in a plane via the vectors components</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Vectors/Dot-product-of-vectors-in-a-plane-and-the-angle-between-two-vectors.lesson>Dot-product of vectors in a coordinate plane and the angle between two vectors</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Vectors/Perpendicular-vectors-in-a-coordinate-plane.lesson>Perpendicular vectors in a coordinate plane</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Vectors/Solved-problems-on-Dot-product-of-vectors-and-the-angle-between-two-vectors.lesson>Solved problems on Dot-product of vectors and the angle between two vectors</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Vectors/Properties-of-Dot-product-of-vectors-in-a-coordinate-plane.lesson>Properties of Dot-product of vectors in a coordinate plane</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Vectors/The-formula-for-the-angle-between-two-vectors-and-the-formula-of-cosines-of-the-difference-of-two-angles.lesson>The formula for the angle between two vectors and the formula for cosines of the difference of two angles</A>

There are short lessons of the &nbsp;"HOW TO . . . "&nbsp; type on Dot-product:

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/HOW-TO-find-the-length-of-the-vector-in-a-coordinate-plane.lesson>HOW TO find the length of the vector in a coordinate plane</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/HOW-TO-find-dot-product-of-two-vectors-in-a-plane.lesson>HOW TO find dot-product of two vectors in a plane</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/HOW-TO-find-scalar-product-of-two-vectors-in-a-coordinate-plane.lesson>HOW TO find scalar product of two vectors in a coordinate plane</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/HOW-TO-find-the-angle-between-two-vectors-in-a-coordinate-plane.lesson>HOW TO find the angle between two vectors in a coordinate plane</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/HOW-TO-prove-that-two-vectors-in-a-coordinate-plane-are-perpendicular.lesson>HOW TO prove that two vectors in a coordinate plane are perpendicular</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/HOW-TO-prove-that-a-triangle-in-a-coordinate-plane-is-a-right-triangle.lesson>HOW TO prove that a triangle in a coordinate plane is a right triangle</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/HOW-TO-prove-that-a-quadrilateral-in-a-coordinate-plane-is-a-parallelogram.lesson>HOW TO check if a quadrilateral in a coordinate plane is a parallelogram</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/HOW-TO-check-if-a-quadrilateral-in-a-coordinate-plane-is-a-rectangle.lesson>HOW TO check if a quadrilateral in a coordinate plane is a rectangle</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/HOW-TO-check-if-a-quadrilateral-in-a-coordinate-plane-is-a-rhombus.lesson>HOW TO check if a quadrilateral in a coordinate plane is a rhombus</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/HOW-TO-check-if-a-quadrilateral-in-a-coordinate-plane-is-a-square.lesson>HOW TO check if a quadrilateral in a coordinate plane is a square</A>

For the full list of my lessons on dot-product with short annotations see the file &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Vectors/REVIEW-of-the-lessons-on-Dot-product.lesson>OVERVIEW of lessons on Dot-product</A>.

Use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A>&nbsp; to navigate over all topics and lessons of the online textbook &nbsp;ALGEBRA-II.