# Math Help - Vectors

1. ## Vectors

If |x|=13, |y|=17 and |x+y|=45, find |x-y|

x and y are both vectors

(-1109)^0.5

i want to know if this is correct

2. Originally Posted by skeske1234
If |x|=13, |y|=17 and |x+y|=45, find |x-y|
x and y are both vectors
This is an impossible problem. See the red above.
$45 = \left\| {x + y} \right\| \leqslant \left\| x \right\| + \left\| y \right\| = 13 + 17$.
Do you see any thing wrong with that?

3. x and y are bigger than x-y .......
so they have a typo....
what should i write as an answer then?

4. Originally Posted by skeske1234
x and y are bigger than x-y .......
so they have a typo....
what should i write as an answer then?
I would put that there is no answer possible.
I suspect they meant $25 = \left\| {x + y} \right\|$

5. Assuming that it was really |x+y|= 25, then, by the "parallelogram rule" for vector addtion, x+ y is the one diagonal of a parallelogram having sides x and y and x- y is the other diagonal. You can find the angle in the triangle with sides x, y, and x+y by using the cosine law on a triangle with sides of length 25, 13, and 17, then use the cosine law on a triangle with sides x, y, and x- y, and the same angle, to find the length of x-y.

Or you could use the fact that $|x+y|= \sqrt{(x+y)\cdot(x-y)}$ $= \sqrt{x\cdot x+ 2x\cdot y+ y\cdot y}= 25$ to argue that $x\cdot x+ 2 x\cdot y+ y\cdot y= 25^2= 625$. You can easily find $x\cdot x$ and $y\cdot y$ and then find $2x\cdot y$. Now use the fact that $(x-y)\cdot(x-y)= x\cdot x- 2 x\cdot y+ y\cdot y$ to find |x-y|.

In fact, do it both ways, as a check and really shock your teacher!

6. Originally Posted by HallsofIvy
Assuming that it was really |x+y|= 25, then, by the "parallelogram rule" for vector addtion, x+ y is the one diagonal of a parallelogram having sides x and y and x- y is the other diagonal. You can find the angle in the triangle with sides x, y, and x+y by using the cosine law on a triangle with sides of length 25, 13, and 17, then use the cosine law on a triangle with sides x, y, and x- y, and the same angle, to find the length of x-y.
Or you could use the fact that $|x+y|= \sqrt{(x+y)\cdot(x-y)}$ $= \sqrt{x\cdot x+ 2x\cdot y+ y\cdot y}= 25$ to argue that $x\cdot x+ 2 x\cdot y+ y\cdot y= 25^2= 625$. You can easily find $x\cdot x$ and $y\cdot y$ and then find $2x\cdot y$. Now use the fact that $(x-y)\cdot(x-y)= x\cdot x- 2 x\cdot y+ y\cdot y$ to find |x-y|.
On a much simpler level. It is well known that:
$\left\| {x + y} \right\|^2 + \left\| {x - y} \right\|^2 = 2\left\| x \right\|^2 + 2\left\| y \right\|^2$.