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

2 Answers
Apr 8, 2018

Answer:

The vector is #=〈27,-36,-8〉#

Explanation:

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#

Apr 8, 2018

Answer:

#<27, -36, -8>#

Explanation:

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>#