Given two points (4i, -2j, 3k) and (i, j, -3k), how do you find a unit vector parallel to the line segment AB?

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Answer 1 · 2016-12-20T10:04:36

#hat(vec(AB))=(sqrt6/6)(-veci+vecj-2veck)#

assuming #vec(OA)=((4),(-2),(3))" "#&#" "vec(OB)=((1),(1),(-3))#

#vec(AB)=vec(AO)+vec(OB)#

#vec(AB)=-vec(OA)+vec(OB)#

#vec(AB)=-((4),(-2),(3))+((1),(1),(-3))#

#vec(AB)=((-4+1),(2+1),(-3-3))#

#vec(AB)=((-3),(3),(-6))#

unit vector parralel to #" "vec(AB)#

#hat(vec(AB))=vec(AB)/|vec(AB)|#

#|vec(AB)|=sqrt(3^2+3^2+6^2)=3sqrt6#

#hat(vec(AB))=(1/(3sqrt6))((-3),(3),(-6))#

#hat(vec(AB))=(1/(3sqrt6))(-3veci+3vecj-6veck)#

#hat(vec(AB))=(sqrt6/18)(-3veci+3vecj-6veck)#

cancelling a factor of #3# we have

#hat(vec(AB))=(sqrt6/6)(-veci+vecj-2veck)#