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.