What is the cross product of #[2, 5, 4]# and #[1, -4, 0] #?

2 Answers
Mar 4, 2017

Answer:

#[16,4,-13].#

Explanation:

#[2,5,4]xx[1,-4,0]=|(i,j,k),(2,5,4),(1,-4,0)|,#

#=16i+4j-13k,#

#=[16,4,-13].#

Mar 4, 2017

Answer:

The vector is #=〈16,4,-13〉#

Explanation:

The vector perpendicular to 2 vectors is calculated with the determinant (cross product)

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

where #〈d,e,f〉# and #〈g,h,i〉# are the 2 vectors

Here, we have #veca=〈2,5,4〉# and #vecb=〈1,-4,0〉#

Therefore,

#| (veci,vecj,veck), (2,5,4), (1,-4,0) | #

#=veci| (5,4), (-4,0) | -vecj| (2,4), (1,0) | +veck| (2,5), (1,-4) | #

#=veci(16)-vecj(-4)+veck(-13)#

#=〈16,4,-13〉=vecc#

Verification by doing 2 dot products

#veca.vecc#

#=〈2,5,4>.〈16,4,-13〉=32+20-52=0#

#vecb.vecc#

#=〈1,-4,0〉.〈16,4,-13〉=16-16+0=0#

So,

#vecc# is perpendicular to #veca# and #vecb#