Given the vectors u=a*i+2*j and v=3*i+(a-5)*j, |u+v|=5square root2, find the real number a.

First, we'll add the vectors u and
v:

u + v =
a*i+2*j+3*i+(a-5)*j

We'll factorize by i and by
j:

u + v = (a+3)*i + (2 + a -
5)*j

u + v = (a+3)*i + (a -
3)*j

The absolute value of the resultant vector
is:

|u+v| = sqrt[(a+3)^2 +
(a-3)^2]

We'll expand the
binomials:

|u+v| = sqrt(a^2 + 6a + 9 + a^2 - 6a +
9)

We'll eliminate like terms inside
brackets:

|u+v| = sqrt(2a^2 +
18)

But, from enunciation, we'll have |u+v| =
5sqrt2

5sqrt2 = sqrt(2a^2 +
18)

We'll raise to square both
sides:

50 = 2a^2 + 18

We'll
use symmetrical property:

2a^2 + 18 =
50

2a^2 = 50 - 18

2a^2 =
32

a^2 = 16

a1 =
+sqrt16

a1 = 4 and a2 =
-4

The requested real values of "a" are: {-4
; 4}.