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?

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Answer 1 · 2016-08-01T10:56:35

$102.6^o$

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 $triangle$

$sqrt(84) cos varphi = -2$

$cos varphi = -2/ (2sqrt(21))$

$varphi = 102.6^o$