Given u = (1-m)*i + 1+m)*j and v=(1+2m)*i+4j determine the vectors u and v whether u is perpendicular to v.

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By definition, two vectors are perpendicular if the value of their dot product is zero.

u*v = 0 (1)

[(1-m)*i+(1+m)*j]*[(1+2m)*i+4*j] = (1-m)*(1+2m) + 4*(1+m) (2)

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

(1-m)*(1+2m) + 4*(1+m) = 0

We'll remove the brackets:

1 + 2m - m - 2m^2 + 4 + 4m = 0

We'll combine like terms:

-2m^2 + 5m + 5 = 0

We'll multiply by -1:

2m^2 - 5m - 5 = 0

We'll apply quadratic formula:

m1 = [5+sqrt(25+40)]/4

m1 = (5+sqrt65)/4

m2 = (5-sqrt65)/4

**The values of the parameter "m", for the vector u to be perpendicular to the vector v, are: {(5-sqrt65)/4 ; (5+sqrt65)/4}.**

We have vectors u and v perpendicular to each other.

u = (1-m)*i + 1+m)*j and v=(1+2m)*i+4j

If the two are perpendicular the cross vector is 0.

The cross vector is given by

(1 - m)*(1 + 2m) + (1 + m)*4 = 0

=> 1 - m + 2m - 2m^2 + 4 + 4m = 0

=> 5 + 5m - 2m^2 = 0

=> 2m^2 - 5m - 5 = 0

m = 5/4 + sqrt( 25 + 40) / 4

=> 5/4 + sqrt 65 / 4

=> 5/4 + (sqrt 65)/4

m = 5/4 - (sqrt 65)/4

**The values of m can be m = 5/4 + (sqrt 65)/4 and m = 5/4 - (sqrt 65)/4**

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