Vectors a and b have scalar product -7.00 and their vector product has magnitude 6.00. what is the angle between these vectors?

Convention: A letter followed by an underscore means a vector: A_ 

NOTE: Angle between vectors is always between 0º and 180º, inclusive. 

A_•B_ = AB cosθ = -7 
|A_x B_| = AB sinθ = 6 

Sum of their squares is: 
(AB cosθ)^2 + (AB sinθ)^2 = -7^2 + 6^2 = 85 
= (AB)^2 ((cosθ)^2 + (sinθ)^2) = (AB)^2 

and since 

A, B = |A_|, |B_| ≥ 0, 
AB = sqrt(85) (and not -10) 

So 

sqrt(85) cosθ = -7; sqrt(85) sinθ = 6 
(cosθ, sinθ) = (-0.76, 0.65) 
θ = acos(-0.76) = 139.46 degrees

Therefore, the angle between the two vectors is 139.46 degrees.

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