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.

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

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