What is the angle between #<5,1,4 ># and #< 3,1,9 >#?

1 Answer
Dec 16, 2016

# 32.7 °#(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=<<5,1,4>># and #vec v=<<3,1,9>>#

The modulus is given by;

# |vec u| = |<<5,1,4>>| = sqrt(5^2+1^2+4^2)=sqrt(25+1+16)=sqrt(42) #
# |vec v| = |<<3,1,9>>| = sqrt(3^2+1^2+9^2)=sqrt(9+1+81)=sqrt(91) #

And the scaler product is:

# vec u * vec v = <<5,1,4>> * <<3,1,9>>#
# \ \ \ \ \ \ \ \ \ \ = (5)(3) + (1)(1) +(4)(9) #
# \ \ \ \ \ \ \ \ \ \ = 15+1+36#
# \ \ \ \ \ \ \ \ \ \ = 52#

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

# 52 = sqrt(42) * sqrt(91) * cos theta #
# :. cos theta = (52)/(sqrt(42) * sqrt(91))#
# :. cos theta = 0.841120 ... #
# :. theta = 32.741412 °#
# :. theta = 32.7 °#(3sf)