Vectors X, Y, and Z?

Question

Find the magnitude and direction(s) of the following vector.

C=-2i -3j +4k (2 angles are needed)

Details Please

Answer 1

| C | = √(4 + 9 + 16) = √29
cos α = ( - 2 ) / √29
cos β = ( - 3 ) / √29
cos γ = (4) / √29

Answer 2

magnitude = the length of the vector <-2, -3, 4>
|c| = sqrt((-2)² + (-3)² + (4)²) = sqrt(29)

cos α = x/|c| = -2/sqrt(29)
α = arccos(-2/sqrt(29)) = 111.80°
cos β = y/|c| = -3/sqrt(29)
β = arccos(-3/sqrt(29)) = 123.85°
cos γ = z/|c| = 4/sqrt(29)
γ = arccos(4/sqrt(29)) = 42.03°