What is the cross product of #<3 ,1 ,-6 ># and #<7 ,3 ,2 >#?

1 Answer
Dec 26, 2016

Answer:

The cross product is #=〈20,-48,2〉#

Explanation:

The cross product of 2 vectors, #〈a,b,c〉# and #d,e,f〉#

is given by the determinant

#| (hati,hatj,hatk), (a,b,c), (d,e,f) | #

#= hati| (b,c), (e,f) | - hatj| (a,c), (d,f) |+hatk | (a,b), (d,e) | #

and # | (a,b), (c,d) |=ad-bc#

Here, the 2 vectors are #〈3,1,-6〉# and #〈7,3,2〉#

And the cross product is

#| (hati,hatj,hatk), (3,1,-6), (7,3,2) | #

#=hati| (1,-6), (3,2) | - hatj| (3,-6), (7,2) |+hatk | (3,1), (7,3) | #

#=hati(2+18)-hati(6+42)+hatk(9-7)#

#=〈20,-48,2〉#

Verification, by doing the dot product

#〈20,-48,2〉.〈3,1,-6〉=60-48-12=0#

#〈20,-48,2〉.〈7,3,2〉=140-144+4=0#

Therefore, the vector is perpendicular to the other 2 vectors