How do you find the cross product #<0,1,2>times<1,1,4># and verify that the resulting vectors are perpendicular to the given vectors?

1 Answer
Jul 7, 2016

To take the cross product, for some

#veca = << a_1,a_2,a_3 >>#

and

#vecb = << b_1,b_2,b_3 >>,#

you can use this formula:

#\mathbf(veca xx vecb = << a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 >>)#

If you notice, the subscripts are cyclic. They go #23, 31, 12#, and you subtract the "mirror-image" subscripts.

So, the cross product is:

#color(blue)(veca xx vecb)#

#= << 1*4 - 2*1, 2*1 - 0*4, 0*1 - 1*1 >>#

#= color(blue)(<< 2,2,-1 >>)#

An easy way to verify the perpendicularity of #vecaxxvecb# is by performing the dot product with #veca# and then with #vecb#. If they are perpendicular, and neither of them is #vec0# (which would defeat the purpose), then the dot product should be #0#.

(It doesn't matter which direction you do it either, because dot products are commutative.)

#color(blue)((vecaxxvecb)cdotveca) = vecacdot(vecaxxvecb)#

#= << 2,2,-1 >>cdot<< 0,1,2 >>#

#= 2*0 + 2*1 + (-1)*2#

#= color(blue)(0)#

#color(blue)((vecaxxvecb)cdotvecb) = vecbcdot(vecaxxvecb)#

#= << 2,2,-1 >>cdot<< 1,1,4 >>#

#= 2*1 + 2*1 + (-1)*4#

#= color(blue)(0)#

And to further prove this, here is another formula you should know (modified to substitute #vecaxxvecb# for one of the vectors):

#\mathbf((vecaxxvecb)cdotveca = ||(vecaxxvecb)||cdot||veca||costheta)#
#\mathbf((vecaxxvecb)cdotvecb = ||(vecaxxvecb)||cdot||vecb||costheta)#

Since both results were #0#, and since neither vector was #vec0# (hence their norms are nonzero), it must be that #costheta = 0#. Therefore, #theta = 90^@# or #270^@#, and either way, CW or CCW, that indicates perpendicularity.

Hence, #\mathbf(vecaxxvecb)# is perpendicular to #\mathbf(veca)# and is also perpendicular to #\mathbf(vecb)#.