# If a and b are unit vectors where (3a-2b) dot product (a+2b) = 4, find the acute angle between a and b.

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The dot product of (3a - 2b) and (a + 2b) is 4

a dot b = |a||b|cos x, where x is the angle between the two vectors.

(3a - 2b) dot (a+2b) = 4

=> 3a dot a + 3a dot 2b - 2b dot a - 2b dot 2b = 4

=> 3|a|^2 + 6a dot b - 2 b dot a - 4|b|^2 = 4

as a and b are unit vectors |a| = |b| = 1

=> 3 - 4 + 4*(a dot b) = 4

=> 4*(a dot b) = 5

=> a dot b = 5/4

But as a and b are unit vectors a dot b < 1.

**It is not possible for (3a - 2b) dot (a+2b) to be equal to 4.**

The question itself is wrong somewhere:

4 = (3a - 2b) (a +2 b) = 3 a a + 6 a b - 2 b a - 4b b

= 3 + 4 a b -4 = -1+ 4 a b

a b = 5/4 but | a b| <= |a| |b| = 1, so we have

5/4 <= 1, a contradictory. The question has some mistakes.