Finding Vector Components: Magnitude 15, Angle 315

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Homework Help Overview

The discussion revolves around finding the components of a vector with a given magnitude of 15 and an angle of 315 degrees, which is situated in a two-dimensional coordinate system.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the use of trigonometry to determine the x and y components of the vector. There are questions about how to properly visualize the angle and its implications for the vector's position in the coordinate system.

Discussion Status

Several participants have provided guidance on using trigonometric functions to find the vector components, as well as suggestions for drawing diagrams to aid understanding. There is an ongoing exploration of how to represent the angle and its measurement from the positive x-axis.

Contextual Notes

Some participants note the importance of specifying the reference point for the angle measurement, as well as the quadrant in which the vector lies based on the angle provided.

robfrech
How do you find the components of this vector:

Magnitude 15, angle 315

Thanks!
 
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How do you find the components of this vector:

Magnitude 15, angle 315

Thanks!
 
Your question seems to imply a 2-d vector.

x=15cos(315)
y=15sin(315)
 
Welcome to PF!

robfrech said:
How do you find the components of this vector:

Magnitude 15, angle 315

Thanks!

Hi robfrech! Welcome to PF! :smile:

Draw a diagram, including the x and y axes.

Then use trigonometry to find the x and y coordinates … taking care to use a minus sign where appropriate! :smile:
 
Draw a picture and consider the resulting triangle. You know the length of the hypotenuse, and two of the angles. Use trigonometry to find the other sides.

- Warren
 
How do you draw a triangle with a 315 degree angle?
 
Well, first you should have specified where the angle is measured from!

The convention is that the angle is measured from the positive x-axis so I will assume that. Now 315= 360- 45 so you have a a point in the 4th quadrant with x-coordinate positive and y-coordinate negative. The triangle itself has hypotenuse 15, angle 45 degrees. x= 15 cos(45), y= -sin(45). Since cos(315)= cos(45) and sin(315)= -sin(45), that is exactly what mathman said originally.
 
Draw two crossed lines, representing the x and y axes. Draw a circle centered on the origin. Count 315 degrees around the circle, counterclockwise, starting from the point where the circle meets the positive x-axis. The point is in the lower-right quadrant. Draw two lines: one from the origin to the point, the other from the x-axis to the point. The second line should make a right angle with the x-axis.

You've now created a triangle. You know one angle and two sides. You can solve for the others.

- Warren
 

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