# vectors a and b form the angle of 60 degrees and at the same time is a = 3 and b=12.find |a+b| and |a-b|.

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You need to evaluate `bar a + bar b` and `bar a - bar b` , hence, since you do not know the vectors` bar a` and `bar b` , but you do know the lengths of vectors and the angle between them, you may evaluate its scalar product, such that:

`bar a*bar b = |bar a|*|bar b|*cos(hat(bar a, bar b))`

`bar a*bar b =3*12*cos 60^o => bar a*bar b = (3*12)/2 => bar a*bar b = 18`

Raising to square `bar a + bar b` yields:

`(bar a + bar b)^2 = (bar a)^2 + 2bar a*bar b + (bar b)^2`

`(bar a)^2 = |bar a|^2*cos 0^o => (bar a)^2 = 3^2 => (bar a)^2 = 9`

`(bar b)^2 = |bar b|^2*cos 0^o => (bar b)^2 = 12^2 => (bar b)^2 = 144`

`(bar a + bar b)^2 = 9 + 2*18 + 144 => (bar a + bar b)^2 = 189 => |bar a + bar b| = sqrt 189`

Raising to square `bar a- bar b` yields:

`(bar a - bar b)^2 = (bar a)^2- 2bar a*bar b + (bar b)^2`

`(bar a - bar b)^2 =9 - 36 + 144 => |bar a - bar b| = sqrt 117`

**Hence, evaluating `|bar a + bar b|` and `|bar a - bar b|` , under the given conditions, yields **`|bar a + bar b| = sqrt 189, |bar a - bar b| = sqrt 117.`