What are the vectors AB and CD if AB = (3-2a)*i+(a+1)*j and CD=(2a+1)*i+2*j? AB is perpendicular to CD.

If the given vectors are perpendicular, the value of their
dot product is zero.

AB*CD = 0
(1)

[(3-2a)*i+(a+1)*j]*[(2a+1)*i+2*j] = (3-2a)*(2a+1) +
2*(a+1) (2)

We'll equate (1) and
(2):

(3-2a)*(2a+1) + 2*(a+1) =
0

We'll remove the
brackets:

6a + 3 - 4a^2 - 2a + 2a + 2 =
0

We'll eliminate like
terms:

-4a^2 + 6a + 5 =
0

We'll multiply by -1:

4a^2 -
6a - 5 = 0

We'll apply quadratic
formula:

a1 =
[6+sqrt(36+80)]/8

a1 =
(6+sqrt116)/8

a1 =
(6+2*sqrt29)/8

a1 =
(3+sqrt29)/4

a2 =
(3-sqrt29)/4

The values of the parameter "a",
for the vector AB to be perpendicular to CD, are: {(3-sqrt29)/4 ;
(3+sqrt29)/4}.