# If |x| = 11, |y| = 23, and |x-y| = 30, find |x+y|.

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1. Somehow you can tell one of us does physics and the other does math :)

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2. x = a i + b j
y = c i + d j

a^2 + b^2 = 11^2 = 121
c^2 + d^2 = 23^2 = 529

x-y = (a-c)i + (b-d)j
(a-c)^2 + (b-d)^2 = 30^2 = 900 = a^2 - 2ac + c^2 + b^2 -2bd + d^2
so 900 = 121+529 -2(ac+bd)
so 250 = -2(ac+bd)
now
x+y = (a+c)i + (b+d)j
|x+y|^2 = a^2+2ac +c^2 +b^2+2bd+d^2 = 121+529 +2(ac+bd)
but we know 2(ac+bd) = -250
so
|x+y|^2 = 121+529-250 = 400
so in the end
|x+y| = 20
Whew !!!

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3. Thank you so much :)))

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4. or ...
make a sketch of a triangle with sides 11, 23 and 30
side 11 would be vector x, 23 would be vector -y and 30 is the resultant x-y
by the cosine law: 30^2 = 11^2 + 23^2 - 2(11)(23)cosØ
cosØ = -250/506

reflect vector -y and draw in vector y. Complete the new triangle with a resultant of x+y
|x+y|^2 = 11^2 + 23^2 - 2(11)(23)cos (180-Ø) , but cos(180-x) = -cosx = 250/506
= 650 - 2(11)(23)(250/506)
= 650 - 250
= 400
|x+y| = 20

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