Find Angle Between Vectors A & B: 105.27

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SUMMARY

The discussion focuses on calculating the angle between two vectors, A = -2.00i + 6.00j and B = 2.00i - 3.00j. The initial approach used the tangent function to find angles θ(A) and θ(B), resulting in 71.57 degrees and 33.70 degrees, respectively. However, the correct method involves using the dot product formula, A·B = |A||B|cos(θ), or the cross product formula, |A×B| = |A||B|sin(θ), to accurately determine the angle between the vectors, which is 105.27 degrees.

PREREQUISITES
  • Understanding of vector components in Cartesian coordinates
  • Familiarity with trigonometric functions, specifically tangent and inverse tangent
  • Knowledge of dot product and cross product operations
  • Basic grasp of angle measurement in degrees
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  • Study the dot product and its geometric interpretation in vector mathematics
  • Learn about the cross product and its applications in determining angles between vectors
  • Explore the use of trigonometric identities in vector calculations
  • Practice solving problems involving angles between vectors using both dot and cross product methods
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w0lfed
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Homework Statement


Find the angle between the following vectors?
A = -2.00i^ + 6.00j^ and B = 2.00i^ - 3.00j^ (the ^ means 'i hat')



Homework Equations


Theta = tan^-1 = (opposite/adjacent)


The Attempt at a Solution


i said for the first part of the component that
theta(A) = tan^1 (6/2) = 71.57
and then did the same for the other component
theta(B) = tan^1 (2/3) = 33.70

and the angle between them should be the addition of those two? ie 105.27?
im confused...this doesn't look right
 
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Have you done dot or cross products yet? If you have the easiest way to compute the angle is through the association that

\mathbf{A}\cdot\mathbf{B}=|\mathbf{A}||\mathbf{B}|\cos{\theta}

or with cross products

|\mathbf{A}\times\mathbf{B}|=|\mathbf{A}||\mathbf{B}|\sin{\theta}

where \theta is the angle in between them.
 
nope, I've never seen that before?
 

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