3D Vectors Problem - Find Cross Product

[TheSerpent]

Homework Statement

Given |\vec{a}| = 8, |\vec{b}| = 9 and the angle between vector \vec{a} and \vec{b} is 48° find the cross product, \vec{a} X \vec{b}.

Homework Equations

Let θ = angle between \vec{a} and \vec{b}.

\vec{a} . \vec{b} = ( \vec{x}1 * \vec{x}2 ) + ( \vec{y}1 * \vec{y}2 ) + ( \vec{z}1 * \vec{z}2 ) = |\vec{a}| * |\vec{b}| * Cosθ

|\vec{a} X \vec{b}| = |\vec{a}| * |\vec{b}| * Sinθ

\vec{a} X \vec{b} = [ ( \vec{y}1 * \vec{z}2 ) - ( \vec{y}2 * \vec{z}1 ) , ( \vec{z}1 * \vec{x}2 ) - ( \vec{z}2 * \vec{x}1 ) , ( \vec{x}1 * \vec{y}2 ) - ( \vec{x}2 * \vec{y}1 ) ]

The Attempt at a Solution

Honestly have no idea how to work this out, the only thing I thought of was assuming the coordinates of one of the vectors. Such as \vec{a} = [0,8,0]. With that use it to solve for the coordinates of \vec{b} with the dot product formula then find the cross product between \vec{a} and \vec{b}. Probably not the right way to do the question though, there might be a formula or method I am not aware.

 

[HallsofIvy]

The data given is sufficient to find the length of \vec{a}\times\vec{b} but not to find \vec{a}\times\vec{b}. To see that just imagine rotating the vectors \vec{a} and vec{b} while maintaining the same lengths and angle between them. Clearly the resultant vector will shift direction as you do that.