To find the magnitude of 3A + 2B, we need to add 3A and 2B.

The angle between A and B is given as 120 degrees. The component of 3A in the direction of 2B is given by -3A* cos 30.

So we have 2B - (3/2)*A*sqrt 3

As A...

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To find the magnitude of 3A + 2B, we need to add 3A and 2B.

The angle between A and B is given as 120 degrees. The component of 3A in the direction of 2B is given by -3A* cos 30.

So we have 2B - (3/2)*A*sqrt 3

As A and B are unit vectors, they have a magnitude equal to 1.

=> 2 - (3/2)*sqrt 3

The component of 3A perpendicular to 2B is 3A* sin 30.

=> 3/2

The magnitude of the resultant of 2 - (3/2)*sqrt 3 and 3/2 is

sqrt [ ( 2 - (3/2)*sqrt 3)^2 + (3/2)^2]

=> sqrt[ 4 + (9/4)*3 - 6*sqrt 3 + 9/4]

=> sqrt [ 4 + 27/4 + 9/4 - 6*sqrt 3]

=> sqrt [ 13 - 6*sqrt 3]

**The required magnitude of 3A + 2B is sqrt [13 - 6*sqrt 3]**