find the sine of the acute angle between -6i + 3j + 2k and the x axis, leaving your answer in exact form

Answer 1

x axis vector = 1 i + 0 j + 0 k
vector product = 1 * sqrt(36+9+4) sin A = 7 sin A
i j k
1 0 0
-6 3 2

= -2 j + 3 k
---> sqrt (4+ 9) = sqrt13
so sin A = (1/7) sqrt 13

Answer 2

or, using the dot product,
-6 = 7 cosθ
cosθ = -6/7
so sinθ = √13/7