# Exploring Vector Decomposition: Unraveling Data Components and Force Analysis

Topics: Calculus

## Vector decomposition

• Vector decomposition is a technique used in statistical analysis to break down a dataset into its various components. The purpose of vector decomposition is to determine the variance of each component, which allows for a more accurate analysis of the data.
• Vector decomposition can be used to analyze stock prices, economic data and other types of data sets that contain multiple variables. Vector decomposition uses a series of mathematical equations to break down each variable into its most basic components.

The result is a set of equations that shows how each variable affects the entire dataset.

So, this is not merely an exercise in determining the magnitude of dot products by examining a picture. Rather, it is often used in problems dealing with vector quantities. Let me illustrate with a motivating example that is similar to things you already know. Motivation. Suppose that we have a mass on an inclined plane. Gravity is the force that pulls mass toward Earth’s center.

The resultant force vector, known as “g,” is directed straight down. [Graph] So to determine what will happen to the mass, the plane is also pushing on the mass, which it prevents it from falling through the plane. To make correct force diagram for this situation, it is helpful to divide the force of gravity into 3 component that is perpendicular to the plane and one that is tangent to the plane. Here is a sample of what they might look like. So there is a component of the force that is tangent to the plane, and there is also a component of the force that is normal to it.

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Let’s call this red one g tangent, and call this blue one g norm. [Graph] So the total force is decomposed into these different pieces. The gravitational force normal to the plane plus the gravitational force tangent. g->=g->nor=g->tan OK, so if we consider the problem of decomposing a vector into a part that goes in a certain direction–for example, along a tangent line to a plane known as the normal vector–and another component that is perpendicular to that line, then we are asking ourselves what sort of general mathematical problem we might be dealing with here. In general, I have a vector, v, that I would like to decompose into two orthogonal components. One component is going in the direction of vector u and the other is perpendicular to it. [Graph] Let us draw basically identical picture. To visualize the u direction, you draw a line like this. The piece of v going in the u direction looks like this, and then there is a right angle there. There is a piece of vector v that is moving in the u direction. That is the red component of v. A blue component of v is moving perpendicular to it, and we will call that a and b. [Graph] That is an image in symbols: v is equal to a plus b. v->=a->+b-> u And a is in the u direction, and b is perpendicular to the u direction. a-> in u-> dir b-> perp to u-> dir The one here is called the component of v in the u direction. a in u dir – component of vin u dir. And the one here, b, is called the component of v that is perpendicular to u. byerp to u dir – component of v perp. to u The problem is that we are given v and u and must find these components. Given v and u. the task is to find a and b. Given v, u -) Find a, b So this setup probably reminds us of this image, and we will see in the following question that our understanding of what we just fieured out here will help us to do it. Let’s try to find a. What do we know about a? We are familiar that a is in the same direction as u. and that a equals lambda times u. To find lambda, we must first solve for u. a-> same dir. as u-> –> a-> = λu-> We need to know more about a in order to do this; the additional information can be found in the picture that we just worked out. The two images are essentially the same and the thing we called w we also call a. Also, we found that u dot y is the same as u dot W or u dot a. Thus, that’s the second important thing to remind yourself. u dot v is the same as u dot a. u->*v->=u->*a-> And a is lambda times u, so this is u dotted with lambda times u. u->*v->=u->*a->=u->*(λu) Our algebra warm up showed us that the dot product is simply lambda times u times v. This is the way the dot product of vectors works. u->*v->=u->*a->=u->*(λu)=λ(u->*v->) We wanted to find the value of lambda, and we know that v and u are equal. So we can solve for lambda by dividing v dot u by u dot u λ=(u->*v->)/(u->*u->) If you prepare to become a math teacher, there is a lesson in which you must practice writing u’s and v’s so that they can always be distinguished. OK. so lambda represents the function of one variable. a is just lambda times u over u dot u times u. So here’s a. u dot v over u dot u times u. a->=((u->*u->)/(u->*u->))u->

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