- #1

TheSerpent

- 12

- 0

## Homework Statement

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

## Homework Equations

Let θ = angle between [itex]\vec{a}[/itex] and [itex]\vec{b}[/itex].

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

|[itex]\vec{a}[/itex] X [itex]\vec{b}[/itex]| = |[itex]\vec{a}[/itex]| * |[itex]\vec{b}[/itex]| * Sinθ

[itex]\vec{a}[/itex] X [itex]\vec{b}[/itex] = [ ( [itex]\vec{y}[/itex]1 * [itex]\vec{z}[/itex]2 ) - ( [itex]\vec{y}[/itex]2 * [itex]\vec{z}[/itex]1 ) , ( [itex]\vec{z}[/itex]1 * [itex]\vec{x}[/itex]2 ) - ( [itex]\vec{z}[/itex]2 * [itex]\vec{x}[/itex]1 ) , ( [itex]\vec{x}[/itex]1 * [itex]\vec{y}[/itex]2 ) - ( [itex]\vec{x}[/itex]2 * [itex]\vec{y}[/itex]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 [itex]\vec{a}[/itex] = [0,8,0]. With that use it to solve for the coordinates of [itex]\vec{b}[/itex] with the dot product formula then find the cross product between [itex]\vec{a}[/itex] and [itex]\vec{b}[/itex]. Probably not the right way to do the question though, there might be a formula or method I am not aware.