How do you find the angle between the vectors #u=5i+5j# and #v=-6i+6j#?

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Answer 1 · 2016-11-02T03:19:33

Please see the explanation to understand why we know that the angle between the two vectors is #pi/2#

Given two vectors of the form:

#bara = a_(hati)hati + a_(hatj)hatj#

and

#barb = b_(hati)hati + b_(hatj)hatj#

The dot-product is:

#bara*barb = (a_hati)(b_hati) + (a_hatj)(b_hatj)#

Therefore, the dot-product of the two vectors:

#baru = 5hati + 5hatj# and #barv = -6hati + 6hatj# is

#baru*barv = (5)(-6) + (5)(6) = 0#

Normally, we would use the above dot-product the for the left side of the equation:

#baru*barv = |baru||barv|cos(theta)#

And then compute the magnitudes and, ultimately, solve for #theta#.

But, because the dot-product is zero and we know that the magnitudes cannot be zero, then we know that:

#cos(theta) = 0#

Which means that:

#theta = pi/2#