# 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.

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### 2 Answers

The vectors AB and CD are perpendicular to each other. In this case the dot product is equal to 0, as cos 90 = 0.

The dot product of the vectors AB = (3-2a)*i+(a+1)*j and CD=(2a+1)*i+2*j is also equal to (3 - 2a)*(2a + 1) + (a +1)*2

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

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

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

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

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

=> 3/4 + (sqrt 116) /8

=> 3/4 + (sqrt 29)/4

a2 = 3/4 - (sqrt 29)/4

**The values of a are 3/4 + (sqrt 29)/4 and 3/4 - (sqrt 29)/4**

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}.**