What is the cross product of #[1, 3, 4]# and #[2, -5, 8] #?

1 Answer
Mar 8, 2017

Answer:

The vector is #=〈44,0,-11〉#

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=〈1,3,4〉# and #vecb=〈2,-5,8〉#

Therefore,

#| (veci,vecj,veck), (1,3,4), (2,-5,8) | #

#=veci| (3,4), (-5,8) | -vecj| (1,4), (2,8) | +veck| (1,3), (2,-5) | #

#=veci(44)-vecj(0)+veck(-11)#

#=〈44,0,-11〉=vecc#

Verification by doing 2 dot products

#veca.vecc#

#=〈1,3,4>.〈44,0,-11〉=44-44=0#

#vecb.vecc#

#=〈2,-5,8〉.〈44,0,-11〉=88-88=0#

So,

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