How do you find a unit vector which is parallel to the vector which points from (4,13)(19,8)?

1 Answer
Mar 3, 2018

Answer:

#sqrt10/10((3),(-1))=sqrt10/10(3hati-hatj)#

Explanation:

call the first coordinate #A#, and the second #B#

then

#vec(OA)=((4),(13))#

#vec(OB)=((19),(8))#

#now

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

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

#vec(AB)=-((4),(13))+((19),(8))#

#vec(AB)=((15),(-5))#

now

#|vec(AB)|=sqrt(15^2+5^2#

#|vec(AB)|=sqrt250=5sqrt10#

a unit vector parallel to# " "vec(AB)#

is given by

#1/(5sqrt10)((15),(-5))#

#sqrt10/50((15),(-5))#

cancelling

#sqrt10/10((3),(-1))=sqrt10/10(3hati-hatj)#