Vector Joining Two Points

[nycmath]

See attachment.

Show your work.

[Eigenvalue]

A) To determine the vector joining the 2 points, subtract the coordinates of A from the coordinates of B
AB=((-1-2),(-2-3),(-4-0))=(-3,-5,-4)


B) To find the magnitude of AB, find the distance of (-3,-5,-4) to the origin
d=[tex]\sqrt{(-3)²+(-4)²+(-5)²}[/tex]
d=[tex]\sqrt{9+16+25}[/tex]
d=[tex]\sqrt{50}[/tex]
d=5[tex]\sqrt{2}[/tex]

C) To find the unit vector, divide each component by the magnitude
unit vector for AB=([tex]\frac{-3}{5 \sqrt{2} }[/tex], [tex]\frac{-4}{5 \sqrt{2} }[/tex], [tex]\frac{-5}{5 \sqrt{2} }[/tex])
Simplify and rationalize the denominator
unit vector for AB=([tex]\frac{-3 \sqrt{2} }{10}[/tex], [tex]\frac{-4 \sqrt{2} }{10}[/tex], [tex]\frac{- \sqrt{2} }{2})[/tex]

[nycmath]

A) To determine the vector joining the 2 points, subtract the coordinates of A from the coordinates of B
AB=((-1-2),(-2-3),(-4-0))=(-3,-5,-4)


B) To find the magnitude of AB, find the distance of (-3,-5,-4) to the origin
d=[tex]\sqrt{(-3)²+(-4)²+(-5)²}[/tex]
d=[tex]\sqrt{9+16+25}[/tex]
d=[tex]\sqrt{50}[/tex]
d=5[tex]\sqrt{2}[/tex]

C) To find the unit vector, divide each component by the magnitude
unit vector for AB=([tex]\frac{-3}{5 \sqrt{2} }[/tex], [tex]\frac{-4}{5 \sqrt{2} }[/tex], [tex]\frac{-5}{5 \sqrt{2} }[/tex])
Simplify and rationalize the denominator
unit vector for AB=([tex]\frac{-3 \sqrt{2} }{10}[/tex], [tex]\frac{-4 \sqrt{2} }{10}[/tex], [tex]\frac{- \sqrt{2} }{2})[/tex]

You are on a roll. Very cool.