Question

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`

Explanation:

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`