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

$ˆ - - \to A B = (\frac{\sqrt{6}}{6}) (- \to i + \to j - 2 \to k)$ $hat(vec(AB))=(sqrt6/6)(-veci+vecj-2veck)$

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

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