# 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

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

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 **dot product** with

(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

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

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

Since both results were

**Hence,** **is perpendicular to** **and is also perpendicular to**