Linear algebra with dot product?

In summary, the conversation discusses how to calculate (x dot b)/(|x||b|) when given the equation A(theta)x = b. It is related to theta by using the standard dot-product and magnitude operations for vectors in Rn, with x dot y = x(transpose)y and |x| = sqrt(x(transpose)*x). The matrix A is a rotation matrix with theta indicating the amount of rotation. To calculate the equation, x can be determined using the inverse of A and the vector b, and the inverse rotation is equal to the rotation about the same axis by a negative theta.
  • #1
SpiffyEh
194
0

Homework Statement



Let A(theta)x = b for each theta in S. Calculate,
(x dot b)/(|x||b|)

A =
[cos(theta) -sin(theta)
sin(theta) cos(theta)]

How is this related to theta?
Recall that x dot y and |x| are the standard dot-product and magnitude, respectively, from vector-calculus. These operations hold for vectors in Rn
but now have the following definitions, x dot y = x(transpose)y and |x =sqrt(x(transpose)*x)

Homework Equations





The Attempt at a Solution



I'm not sure how to do this because i don't know what x is or what b is, so I'm confused.
 
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  • #2
x is a vector in R2, and so is b. The matrix A is a rotation matrix, where the parameter theta indicates how much rotation.
 
  • #3
I don't understand how I would go about calculating it without knowing anything besides that
 
  • #4
Pick a value for theta, then pick a few values for x. Now calculate Ax. That should give you an idea of what A does, and how A is related to theta.
 
  • #5
don't I need to actually calculate the dot product equation though? I can't just pick something for x and b and actually have it be correct to the equation.
 
  • #6
The equation is:[tex]A(\theta )x=b[/tex]. Then [tex]x=A^{-1}(\theta )b[/tex], You can calculate what the inverse of A is. Denote [tex]\mathbf{b}=(b_{1},b_{2})[/tex], you can then calculate x in terms of theta and b, from there compute your dot product.
 
  • #7
Nice to know: if [itex]A(\theta)[/itex] is 'rotation about a given axis by angle [itex]\theta[/itex]' then the inverse rotation is just the rotation about the same axis by angle [itex]-\theta[/itex].

That is, [itex]A(\theta)^{-1}= A(-\theta)[/itex].
 

FAQ: Linear algebra with dot product?

What is a dot product in linear algebra?

A dot product, also known as an inner product, is a mathematical operation that takes two vectors and produces a single scalar value. It is a way to measure the similarity between two vectors and is often used in linear algebra to calculate the length of a vector, the angle between two vectors, and to determine whether two vectors are orthogonal.

How is the dot product calculated?

The dot product is calculated by multiplying the corresponding components of two vectors and then adding the products together. For example, the dot product of vectors a and b would be written as a · b = a1b1 + a2b2 + ... + anbn. This can also be written using summation notation as a · b = Σi=1n aibi.

What is the significance of the dot product in linear algebra?

The dot product is an essential tool in linear algebra as it allows for the calculation of important quantities such as vector length, angle, and projection. It is also used in many other areas of mathematics, physics, and engineering, including vector calculus, mechanics, and computer graphics.

How is the dot product related to orthogonality?

Two vectors are considered orthogonal if their dot product is equal to 0. This means that the angle between the two vectors is 90 degrees, and they are perpendicular to each other. This property of the dot product is used to determine whether vectors are orthogonal, which has important applications in geometry and physics.

Can the dot product be extended to higher dimensions?

Yes, the dot product can be extended to higher dimensions. In fact, the dot product is only defined for vectors in three-dimensional space, but it can be generalized to any number of dimensions. This is known as the inner product, and it follows the same principles as the dot product, but with more components and a more complex formula for calculation.

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