What is the cross product of `<4 , 5 ,-9 >` and `<4, 3 ,0 >`?

The cross product of 2 vectors is calculated with the determinant

`| (veci,vecj,veck), (d,e,f), (g,h,i) |`

where `veca=〈d,e,f〉` and `vecb=〈g,h,i〉` are the 2 vectors

Here, we have `veca=〈4,5,-9〉` and `vecb=〈4,3,0〉`

Therefore,

`| (veci,vecj,veck), (4,5,-9), (4,3,0) |`

`=veci| (5,-9), (3,0) | -vecj| (4,-9), (4,0) | +veck| (4,5), (4,3) |`

`=veci((5)*(0)-(-9)*(3))-vecj((4)*(0)-(4)*(-9))+veck((4)*(3)-(4)*(5))`

`=〈27,-36,-8〉=vecc`

Verification by doing 2 dot products

`〈27,-36,-8〉.〈4,5,-9〉=(27)*(4)+(-36)*(5)+(-8)*(-9)=0`

`〈27,-36,-8〉.〈4,3,0〉=(27)*(4)+(-36)*(3)+(-8)*(0)=0`

So,

`vecc` is perpendicular to `veca` and `vecb`

The easiest way to compute is to make a matrix whose determinant is the cross product.

`M = [ (hat i, hat j , hat k), (v_(1x), v_(1y), v_(1z)), (v_(2x), v_(2y), v_(2z)) ]`

For our vectors we have:

`=> v_1 = <4,5,-9>`
`=> v_2 = <4,3,0>`

Hence, we have:

`M =[ (hat i, hat j , hat k), (4, 5, -9), (4, 3, 0) ]`

Now we just need to take the determinant of this matrix to get the cross product.

`v_1 ox v_2 = "det"(M)`

`= (5*0-(-9)*3)hat i + ((-9)*4-(4)*0)hat j+ (4*3-(5)*4)hat k`

`= (27) hat i + (-36)hat j + (-8)hat k`

`= 27hat i -36hat j - 8hat k`

Hence:

`v_1 ox v_2 = <27, -36, -8>`