How do you normalize #(2i - 3 j +k)#?

1 Answer
ali ergin
Feb 9, 2016

#bar x=1/sqrt 14#
#bar y=-3/sqrt 14#
#bar z=1/sqrt 14#

Explanation:

#a_x=2#
#a_y=-3#
#a_z=1#
#|\a|=a_x*a_x+a_y*a_y+a_z*a_z#
#|\a|=sqrt(2*2+(-3)*(-3)+1*1)#
#|\a|=sqrt (4+9+1)#
#|\a|=sqrt 14#
#bar x=2/|\a|#
#bar y=-3/|\a|#
#bar z=1/|\a|#
#bar x=1/sqrt 14#
#bar y=-3/sqrt 14#
#bar z=1/sqrt 14#

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