How do you normalize # <0,8,5>#?

1 Answer
May 13, 2018

Answer:

#bb(hat(a))=[(0,(8sqrt(89))/89,(5sqrt(89))/89)]#

Explanation:

A unit vector in the direction of some vector #bba# is given by:

#bb(hat(a))=(bba)/||bba||#

Let:

#bba=[(0,8,5)]#

#||bba||=sqrt((0)^2+(8)^2+(5)^2)=sqrt(89)#

#:.#

#bb(hat(a))=[(0,(8sqrt(89))/89,(5sqrt(89))/89)]#