Question #871b4

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Answer 1 · 2017-06-15T08:19:38

you are correct

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

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

putting the vectors in to column vectors for ease of formatting

$vec(AB)=-((1),(3),(p))+((-7),((1-p)) , (p))$

$vec(AB)=((-8),(-2-p),(0))$

substitute for $p$

$vec(AB)=((-8),(-2- -8),(0))=((-8),(6),(0))$

$vec(AB)=-8hatveci+6hatvecj$

$|vec(AB)|=sqrt(8^2+6^2)=10$

unit vector for $vec(AB)$

$hatvec(AB)=vec(AB)/|vec(AB)|$

$hatvec(AB)=1/10(-8hatveci+6hatvecj)=-4/5hatveci++3/5hatvecj$

which is what you got.