How do you find the cross product and state whether the resulting vectors are perpendicular to the given vectors #<-1,1,0>times<2,1,3>#?

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Answer 1 · 2017-01-04T13:33:57

#vecaxxvecb=+3hatveci+3hatvecj-3hatveck#

The vector cross product of two vectors #veca" "#&#" "vecb" "#

is defined as

#vecaxxvecb=|veca||vecb|sinthetahatvecn#

where #theta" "#is the angle between the vectors & #" "hatvecn" "isa unit vector mutually perpendicular to

#veca" "#&#" "vecb" "#

If the vectors are in component form, in particular:

#veca=((a_1),(a_2),(a_3))#and #vecb=((b_1),(b_2),(b_3))#

we can evaluate the cross product by the use of a determinate

#vecaxxvecb=|(hatveci,hatvecj,hatveck),(a_1,a_2, a_3),(b_1,b_2,b_3)|#

In this case we have

#veca=((-1),(1),(0))##" "vecb=((2),(1),(3))#

#vecaxxvecb=|(hatveci,hatvecj,hatveck),(-1,1,0),(2,1,3)|#

expanding the determinant as usual

#vecaxxvecb=+hatveci|(1,0),(1,3)|-hatvecj|(-1,0),(2,3)|+hatveck|(-1,1),(2,1)|#

#vecaxxvecb=+3hatveci+3hatvecj-3hatveck#

By definition this is perpendicular tot eh original two vectors. A quick check using the dot product will confirm this.

#veca.((3),(3),(-3))=((-1),(1),(0)).((3),(3),(-3))#

#=-3+3+0=0" "#perpendicular

#vecb.((3),(3),(-3))=((2),(1),(3)).((3),(3),(-3))#

#=6+3-3=0" "#perpendicular