Linear algebra normalising a vector?

1. Oct 29, 2009

1. The problem statement, all variables and given/known data

Let the vector V = (x1,y1,z1)

x1' = x1/sqrt(x1^2 + y1^2 + z1^2)
y1' = x1/sqrt(x1^2 + y1^2 + z1^2)
z1' = x1/sqrt(x1^2 + y1^2 + z1^2)

What do you call x1', y1' and z1' in mathematics terms?

is x1' the norm of x1 and y1' the norm of y1?

Thanx

2. Relevant equations

3. The attempt at a solution

2. Oct 29, 2009

HallsofIvy

Staff Emeritus
No. x1', y1', and z1' are the x, y, and z components, respectively, of the normalized vector- the unit vector parallel to V.

3. Oct 31, 2009

LCKurtz

If $\vec V = \langle a,b,c \rangle$ is a vector and
$$|\vec V| = \sqrt{a^2 + b^2 + c^2}$$
is its length, or norm, then the vector
$$\hat V = \frac 1 {|\vec V|}\ \vec V = \langle \frac a {\sqrt{a^2 + b^2 + c^2}},\frac b {\sqrt{a^2 + b^2 + c^2}},\frac c {\sqrt{a^2 + b^2 + c^2}}\rangle$$
is called a unit vector. The components of this unit vector are sometimes called the direction cosines of $\vec V$.

4. Oct 31, 2009

HallsofIvy

Staff Emeritus
I assume you mean
y1' = y1/sqrt(x1^2 + y1^2 + z1^2)
z1' = z1/sqrt(x1^2 + y1^2 + z1^2)
rather than having "x1" as every numerator.