How do you use the definition of the scalar product, find the angles between the following pairs of vectors: A = - 3 i + j and B = 21 j - 3k?

1 Answer
Jan 8, 2017

# 71.8 °#(3sf)

Explanation:

The angle #theta# between two vectors #vec A# and #vec B# is related to the modulus (or magnitude) and scaler (or dot) product of #vec A# and #vec B# by the relationship:

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# vec A * vec B = ||A|| * ||B|| * cos theta #

By convention when we refer to the angle between vectors we choose the acute angle.

So for this problem, let the angle betwen #vecu# and #vecv# be #theta# then:

#vec u=<<3, 1, 0>># and #vec v=<<0, 21, -3>>#

The modulus is given by;

# ||vec u|| = ||<<3, 1, 0>>|| = sqrt(3^2+1^2+0^2)=sqrt(9+1)=sqrt(10) #
# ||vec v|| = ||<<0, 21,-3>>|| = sqrt(0^2+21^2+(-3)^2)=sqrt(441+9)=sqrt(450) #

And the scaler product is:

# vec u * vec v = <<3, 1,0>> * <<0,21,-3>>#
# \ \ \ \ \ \ \ \ \ \ = (3)(0) + (1)(21) + (0)(-3)#
# \ \ \ \ \ \ \ \ \ \ = 21#

And so using # vec A * vec B = ||A|| * ||B|| * cos theta # we have:

# 21 = sqrt(10) * sqrt(450) * cos theta #
# :. cos theta = (21)/sqrt(4500)#
# :. cos theta = (7sqrt(5))/50 #
# :. cos theta = 0.31304 ... #
# :. theta = 71.7568 ... °#
# :. theta = 71.8 °#(3sf)