#" A vector perpendicular to: " veca "& "vecb" can be found by finding the cross product" vecaxxvecb#
#"the corresponding unit vector is found by " hatvecu=vecu/|vecu|#
#veca=<4,-3,1>, vecb=<2,3,-1>#
#vecaxxvecb=|(veci,vecj,veck),(4,-3,1),(2,3,-1)|#
expanding by Row 1
#vecaxxvecb=veci|(-3,color(white)-1),(color(white)(-)3,-1)|-vecj|(4,1),(2,-1)|+veck|(4,-3),(2,3)|#
#vecaxxvecb=veci(3-3)+vecj(-4-2)+veck(12- -6)#
#vecaxxvecb=-6vecj+18veck#
since we are only wanting the direction we can simplify teh vector by removing any common factors.
so a perpendicular vector in this direction can be written
#vec n=-vecj+3veck=> |vecn|=sqrt(1^2+3^2)=sqrt10#
so a unit vector will be:
#hatvecn=1/sqrt10(-vecj+3veck)=sqrt10/10(-vecj+3veck)#
