Question

Calculate (A⃗ ×B⃗ )⋅C⃗ for the three vectors A⃗ with magnitude A = 5.08 and angle θA = 25.6 ∘ measured in the sense from the +x - axis toward the +y - axis, B⃗ with B = 3.94 and θB = 63.9 ∘, and C⃗ with magnitude C = 6.16 and in the +z - direction. Vectors A⃗ and B⃗ are in the xy-plane. (A⃗ ×B⃗ )⋅C⃗ ( A → × B → ) ⋅ C → = nothing

Answer 1

(A⃗ ×B⃗ )⋅C⃗ = - 76.415

Explanation:

First we need to calculate (A⃗ ×B⃗ ) :

(A⃗ ×B⃗ ) = A.B.sin (α).n

Where A is the magnitude of A⃗

Where B is the magnitude of B⃗

Where α is the angle between A⃗ and B⃗ = 63.9 - 25.6 = 38.3

Finally n is the vector orthogonal to A⃗ and B⃗

n magnitude is 1 and his direction is given by the right hand-rule

so n = ( 0 , 0 , 1 )

(A⃗ ×B⃗ ) = A.B.sin (α).n = 5.08 . 3.94 . sin (38.3) . (0 , 0 , 1 ) = (0,0,12.4)

C⃗ can be written as C.(0,0,-1) because of his +z - direction

C.(0,0,-1) = 6.16.(0,0,-1) = (0,0,-6.16)

(A⃗ ×B⃗ )⋅C⃗ = (0,0,12.4).(0,0,-6.16) = -76.41480787 = -76.415