Vector field calculations- dumb question

Since \vec{u} = \vec{\hat{x}} and \vec{v} = \vec{\hat{y}}, the magnitudes are 1 and the dot product is 0, so the resulting equation is:\cos(\theta) = \frac{0}{1 \cdot 1} = 0Therefore, the angle between the two vectors is 90 degrees. In summary, if the two vectors have a relationship of barXx-barYy and their magnitudes are calculated using the equation sqrt(x2+y2), then the angle between them
  • #1
bookdad
2
0
Given 2 vectors, say x and y with a relationship of barXx-barYy (barX =1,0,0 and barY=0,1,0) calculations for the vector field in x and y plane are:
Magnitude = sqrt(x2+y2)
tails of vectors begin at input points (IE: if I choose 1,0 or 0,-1 etc) and the magnitudes are calcluated from these values with the equation above.
Question: how to find the angle of the vectors? If I use ones and zeros and the BarX dot BarYCos theta how to get theta? also barXdot BarY is zero making the whole thing zero?
 
Physics news on Phys.org
  • #2
bookdad said:
Given 2 vectors, say x and y with a relationship of barXx-barYy (barX =1,0,0 and barY=0,1,0) calculations for the vector field in x and y plane are:
Magnitude = sqrt(x2+y2)
tails of vectors begin at input points (IE: if I choose 1,0 or 0,-1 etc) and the magnitudes are calcluated from these values with the equation above.
Question: how to find the angle of the vectors? If I use ones and zeros and the BarX dot BarYCos theta how to get theta? also barXdot BarY is zero making the whole thing zero?

Does this reference help?

http://www.algebralab.org/lessons/lesson.aspx?file=Trigonometry_TrigVectorDotProd.xml

.
 
  • #3
this gives the formula cos theta = barV dot bar U / the magnitudes of vu. as mentioned before this returns a value of zero whether the y component or the x component is zero. so this means each vector with a zero Y component has a 90 degree vector and this cannot be.
 
  • #4
bookdad said:
this gives the formula cos theta = barV dot bar U / the magnitudes of vu. as mentioned before this returns a value of zero whether the y component or the x component is zero. so this means each vector with a zero Y component has a 90 degree vector and this cannot be.
Since the two vectors are obviously perpendicular, their dot product is zero, and hence the angle between them is 90 degrees.

One way that the dot product of two vectors is defined is:
[tex]\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}| \cos(\theta)[/tex]

You can solve this equation for cos(theta) like so:
[tex]\cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}| }[/tex]
 

Related to Vector field calculations- dumb question

1. What is a vector field calculation?

A vector field calculation is a mathematical process used to determine the magnitude and direction of a vector at a given point in space. This is useful in many scientific fields, including physics, engineering, and meteorology.

2. How is a vector field calculated?

A vector field is calculated by first defining the vector at each point in space and then using mathematical operations to determine the magnitude and direction of the vector. This can be done using equations, computer algorithms, or physical measurements.

3. What are some real-world applications of vector field calculations?

Vector field calculations have many practical applications, such as predicting the movement of fluids in engineering, modeling weather patterns in meteorology, and analyzing the force and direction of magnetic fields in physics.

4. Are there different types of vector fields?

Yes, there are several types of vector fields, including conservative and non-conservative fields, gradient fields, and curl fields. Each type has different properties and is used for different purposes in vector field calculations.

5. What are some common tools and techniques used in vector field calculations?

In addition to mathematical equations and computer algorithms, scientists may use specialized software, physical models, and experimental data to perform vector field calculations. Visualization tools, such as vector field plots, are also commonly used to analyze and interpret the results.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
21
Views
7K
Replies
3
Views
1K
Replies
2
Views
1K
Replies
39
Views
3K
  • Advanced Physics Homework Help
2
Replies
44
Views
3K
Replies
8
Views
879
  • Calculus and Beyond Homework Help
Replies
5
Views
824
  • Classical Physics
Replies
6
Views
2K
  • Precalculus Mathematics Homework Help
Replies
6
Views
8K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top