Find the angle between the two vectors [5,2] and [-2,5]?

possible answers are


a. zero degrees


b. 30 degrees


c. 60 degrees


d. 90 degrees

Find the angle between the two vectors [5,2] and [-2,5]?
Hi,





The correct answer is D, 90°.





This angle is found from the formula:





scalar product of 2 vectors


------------------------------------- = cos θ


product of magnitudes





The scalar product of [5,2] and [-2,5] is found by





X1*X2 + Y1*Y2 = 5(-2) + 2(5) = -10 + 10 = 0





The magnitude of the 2 vectors is found by √(x² + y²) which is


..________.....________


√(5² + (-2)²) + √(-2)² + 5²





Substituting these in the formula for the angle, it becomes:





scalar product of 2 vectors


------------------------------------- = cos θ


product of magnitudes





0


--------------- = cos θ


√50 * √50





0


---- = 0 = cos θ


50





If cos θ = 0, then arc-cos(0) = 90. so θ = 90°.





I hope that helps. :-)
Reply:cos @ = (a1b1 + a2b2) / (|A||B|)


So here, that would be arccos 0 or 90 degrees.
Reply:Use dot product.


a * b = |a||b|cosx





a * b


= [5,2] * [-2,5]


= -10 + 10 = 0





cosx = (a * b) / (|a||b|)


= 0 / (|a||b|)


= 0


x = cos^-1(0)


= 90 degrees OR 0 degrees





BUT since [5,2] and [-2,5] are not scalar multiples of each other the answer is just 90 degrees so


Answer: d.
Reply:Use the dot product.





Gasp! - dont tell me that you dont know what dot product is.
Reply:cosθ = dot product/product of magnitudes





cosθ = 0/whatever


cosθ = 0


θ = 90 [d]