What is the cross product of #[4, -4, 4]# and #[-6, 5, 1] #?

2 Answers
Jul 19, 2018

Answer:

\begin{pmatrix}-24&-28&-4\end{pmatrix}

Explanation:

Use the following cross product formula:

#(u1,u2,u3)xx(v1,v2,v3) = (u2v3 − u3v2 , u3v1 − u1v3 , u1v2 − u2v1)#

#(4,-4,4)xx(-6,5,1) = (-4*1 − 4*5 , 4*-6 − 4*1 , 4*5 − -4*-6)#

#=(-24,-28,-4)#

Jul 19, 2018

Answer:

The vector is #= 〈-24,-28,-4〉#

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,-4,4〉# and #vecb=〈-6,5,1〉#

Therefore,

#| (veci,vecj,veck), (4,-4,4), (-6,5,1) | #

#=veci| (-4,4), (5,1) | -vecj| (4,4), (-6,1) | +veck| (4,-4), (-6,5) | #

#=veci((-4)*(1)-(5)*(4))-vecj((4)*(1)-(-6)*(4))+veck((4)*(5)-(-4)*(-6))#

#=〈-24,-28,-4〉=vecc#

Verification by doing 2 dot products

#〈4,-4,4〉.〈-24,-28,-4〉=(4)*(-24)+(-4)*(-28)+(4)*(-4)=0#

#〈-24,-28,-4〉.〈-6,5,1〉=(-24)*(-6)+(-28)*(5)+(-4)*(1)=0#

So,

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