If |x| = 11, |y| = 23, and |x-y| = 30, find |x+y|.
Textbook answer: 20
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 !!!
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
Somehow you can tell one of us does physics and the other does math :)
Thank you so much :)))
Well, the textbook answer is 20, but let me give you a clown answer just for a good laugh.
Now, let's see. If |x| = 11 and |y| = 23, then we know both x and y are either 11 or -11 in absolute value.
So, if we have |x-y| = 30, that means the difference between x and y is 30. But we also know that x and y can only be 11 or -11.
Now, let's do some clown math! If x = 11 and y = -11, then the sum of x and y is 0.
But wait, let's not stop there! If x = -11 and y = 11, then the sum of x and y is 0 again.
So, according to my very clowny calculations, |x+y| can either be 0 or a big ol' mystery. But hey, let's go with the textbook answer of 20 because that's what we're supposed to do.
To find the value of |x+y|, we can use the given information about |x|, |y|, and |x-y|.
We know that the absolute value of a number is always positive. Therefore, |x| = 11 means that x can be either 11 or -11. Similarly, |y| = 23 means that y can be either 23 or -23.
Now, let's use the third piece of information, |x-y| = 30, to determine the possible values of x and y. When taking the absolute value of the difference between two numbers, it gives the distance between them.
If x and y are positive numbers, then |x-y| is the same as x-y. So, if x=11 and y=23, we have |11-23| = |-12| = 12, which does not equal 30. Therefore, x and y cannot both be positive numbers.
If x and y are negative numbers, then |x-y| is the same as -(x-y), since the result would be negative. So, if x=-11 and y=-23, we have |-11-(-23)| = |-11+23| = |12| = 12, which does not equal 30. Therefore, x and y cannot both be negative numbers.
This means that one of x or y is positive, and the other is negative. We need to find the combination that satisfies |x-y| = 30. Let's consider the case where x is positive and y is negative.
If x is positive (x=11) and y is negative (y=-23), then |x-y| becomes |11-(-23)| = |11+23| = |34|. However, we are given that |34| = 30, which is not true.
Now let's consider the case where x is negative and y is positive.
If x is negative (x=-11) and y is positive (y=23), then |x-y| becomes |-11-23| = |-11+23| = |12| = 12. This satisfies the given condition |x-y| = 30.
Now we have x=-11 and y=23. To find |x+y|, we substitute these values into the expression: |(-11)+23| = |12| = 12.
Therefore, |x+y| = 12, not 20 as stated in the textbook answer.