Find the cross product between two vectors then normalize it

Find the cross product between the two vectors u and v, then normalize it.
i. u = <1, 3, 5>, v = <3, 5, 7>
ii. u = <-1, 3, -5>, v = <10, 2, -3>

What does it mean when it ask's "then normalize it"? Isn't it already normalized when you get the cross product?

normalized vector with
length 1

lets say (3,0,4)
to normalize

you need to divide
by sqrt(3^2+0^2+4^2)=5

normalized vector will be

(3/5, 0,4/5)

as for your
vectors
cross product
i. u = <1, 3, 5>, v = <3, 5, 7>

calculated as determinant of

I j k
1, 3, 5
3, 5, 7

(3*7-5*5)i-(1*7-5*3)j+(1*5-3*3)k
=-4i+8j-4k

cross product is <-4,8,-4>

to normalize find

sqrt((-4)^2+8^2+(-4)^2)
4sqrt6

and <-4/4sqrt6,8/4sqrt6,-4/4sqrt6>
=<-1/sqrt6,2/sqrt6,-1/sqrt6)

ii. u = <-1, 3, -5>, v = <10, 2, -3>
cross product of u = <-1, 3, -5>, v = <10, 2, -3>

=<1,-53, -32>

to find normalized
vector
calculate
sqrt(1^2+(-53)^2+(-32)^2)=3sqrt(426)

answer
=<1/3sqrt(426),-53/3sqrt(426), -32/3sqrt(426)>

1)
u X v =<1, 3, 5> X <3, 5, 7>

=<-4,8,-4>

When it says to normalize, find a unit vector in that direction.
To find unit vector divide the vector by its magnitude.

uXv/|uXv|= <-4,8,-4>/4 * sqrt(6)

=<-1/sqrt(6) , 2* sqrt(6), -1/sqrt(6)>

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Similarly for second part uXv/|uXv|= <1,-53,-32>/sqrt(3834)