How do you normalize # (2i + 3j – 7k) #?

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Answer 1 · 2016-02-27T05:03:59

#hat v# = #(2/sqrt62)hat i+(3/sqrt62)hat j-(7/sqrt62)hat k#

Normalizing vector #v=(2i+3j–7k)# means finding a unit vector in the direction given by vector #v#.

For a vector #u=ahat i+bhat j+chat k#, this is represented by #hat u#, and is equal to #u/|u|#,

where #|u|# is absolute value of #u# and is given by #sqrt(a^2+b^2+c^2)#.

Hence in te given case

#hat v=(2i+3j–7k)/sqrt(2^2+3^3+(-7)^2)# or

#(2i+3j–7k)/sqrt62# i.e.

#(2/sqrt62)hat i+(3/sqrt62)hat j-(7/sqrt62)hat k#