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

1 Answer
Aug 14, 2017

Answer:

#hatv = -sqrt(2/3)hati - 1/(sqrt6)hatj - 1/(sqrt6)hatk#

Explanation:

Normalization of a vector is the process of finding a unit vector in the same direction of the vector in question.

The equation for the normalization of a vector (which I'll call #vecv#) is given by

#(vecv)/(||vecv||)#

where #||vecv||# is the magnitude of vector #vecv#.

The magnitude of #vecv# is

#||vecv|| = sqrt((-2)^2 + (-1)^2 + (-1)^2) = color(red)(ul(sqrt6#

Thus, the unit vector (#hatv#) will be

#hatv = (-2)/(color(red)(sqrt6))hati - 1/(color(red)(sqrt6))hatj - 1/(color(red)(sqrt6))hatk#

#color(blue)(ulbar(|stackrel(" ")(" "hatv = -sqrt(2/3)hati - 1/(sqrt6)hatj - 1/(sqrt6)hatk" ")|)#