What is the cross product of #<9,2,8 ># and #<1,3,-4 >#?

1 Answer
Apr 3, 2017

Answer:

The vector is #=〈-32,44,25〉#

Explanation:

The cross product is a vector perpendiculat to 2 other vectors

#| (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=〈9,2,8〉# and #vecb=〈1,3,-4〉#

Therefore,

#| (veci,vecj,veck), (9,2,8), (1,3,-4) | #

#=veci| (2,8), (3,-4) | -vecj| (9,8), (1,-4) | +veck| (9,2), (1,3) | #

#=veci(-2*4-3*8)-vecj(-9*4-8*1)+veck(9*3-2*1)#

#=〈-32,44,25〉=vecc#

Verification by doing 2 dot products

#〈-32,44,25〉.〈9,2,8〉=-32*9+44*2+25*8=0#

#〈-32,44,25〉.〈1,3,-4〉=-32*1+44*3-4*25=0#

So,

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