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

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Answer 1 · 2018-03-03T13:47:37

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

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

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