Linear algebra normalising a vector?

[Chadlee88]

Homework Statement

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

Homework Equations

The Attempt at a Solution

 

[HallsofIvy]

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

 

[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 $\vec V$.

 

[HallsofIvy]

Homework Statement

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)

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.

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

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

Thanx

Homework Equations

The Attempt at a Solution