How do you use the definition of the scalar product, find the angles between the following pairs of vectors: 2i - 3 j k and - 2i - j - k?

Answer 1

# 102.6^o#

we use the definition of the scalar dot product, namely

#vec a * vec b = abs (vec a) abs(vec b) cos varphi#

doing the LHS first [check the typo in the question, BTW]

#((2), (-3), (1)) * ((-2), (-1), (-1)) = -4 + 3 -1 = -2 qquad triangle#

next the RHS

# |(2), (-3), (1)| |(-2), (-1), (-1)| cos varphi#
#= sqrt(2^2 + (-3)^2 + 1^2) sqrt((-2)^2 + (-1)^2 +(-1)^2) cos varphi#
#= sqrt(14) sqrt(6) cos varphi qquad square#
comparing #square# and #triangle#
# sqrt(84) cos varphi = -2#
# cos varphi = -2/ (2sqrt(21)) #
#varphi = 102.6^o#
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