Question 1084407: If I've: (U-3V)=(1;5) and (-U+V)=(-5;3)
What's the angle between U and V?? Thank you VERY MUCH !!
Found 2 solutions by jim_thompson5910, rothauserc:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Let U and V be two vectors of the form
U = (a,b)
V = (c,d)
where a,b,c,d are scalars and are in the set of real numbers.
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Using those definitions, we can say
U - 3V = 1*U + (-3)*V
U - 3V = 1*(a,b) + (-3)*(c,d)
U - 3V = (a,b) + (-3c,-3d)
U - 3V = (a-3c,b-3d)
Since U-3V also equals (1,5), which is given, this means
(a-3c,b-3d) = (1,5)
further breaking down to
a-3c = 1
b-3d = 5
Let's refer to these equations as (1) and (2).
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Similarly,
-U + V = -1*U + 1*V
-U + V = -1*(a,b) + 1*(c,d)
-U + V = (-a,-b) + (c,d)
-U + V = (-a+c,-b+d)
Since -U + V also equals (-5,3), this means
(-a+c,-b+d) = (-5,3)
further breaking down into
-a+c = -5
-b+d = 3
Let's refer to these equations as (3) and (4).
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Group equation (1) and (3) together to get this system
a-3c = 1
-a+c = -5
Adding those equations together leads to -2c = -4 so c = 2
If c = 2, then...
a-3c = 1
a-3(2) = 1
a-6 = 1
a-6+6 = 1+6
a = 7
So far we know that a = 7 and c = 2
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Now group equation (2) and (4) together and solve the system
b-3d = 5
-b+d = 3
Adding those equations leads to -2d = 8 so d = -4
If d = -4, then,
-b+d = 3
-b+(-4) = 3
-b-4 = 3
-b-4+4 = 3+4
-b = 7
b = -7
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To wrap up everything so far, we found that
a = 7
b = -7
c = 2
d = -4
making
U = (a,b) and V = (c,d)
turn into
U = (7,-7) and V = (2,-4)
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We have the coordinates of U and V, so we can now compute the angle theta between them.
First we'll need the dot product
U dot V = a*c + b*d
U dot V = 7*2 + (-7)*(-4)
U dot V = 14 + 28
U dot V = 42
Now we need the length of each vector
Let's find the length of vector U
|U| = sqrt(U dot U)
|U| = sqrt(a*a + b*b)
|U| = sqrt(a^2 + b^2)
|U| = sqrt(7^2 + (-7)^2)
|U| = sqrt(49 + 49)
|U| = sqrt(49*2)
|U| = sqrt(49)*sqrt(2)
|U| = 7*sqrt(2)
And we also need the length of vector V
|V| = sqrt(V dot V)
|V| = sqrt(c*c + d*d)
|V| = sqrt(c^2 + d^2)
|V| = sqrt(2^2 + (-4)^2)
|V| = sqrt(4 + 16)
|V| = sqrt(20)
|V| = sqrt(4*5)
|V| = sqrt(4)*sqrt(5)
|V| = 2*sqrt(5)
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It's a lot of work, but we're at the home stretch. Plug the dot product result, and the vector lengths, into the formula below to find  (greek letter theta) which is the angle between the two vectors U and V
 Be sure to be in degree mode
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Final Answer: 18.4349488229 degrees
The answer is approximate. Round it however you need to.
Side Note: The answer has been confirmed with GeoGebra (free graphing software) as shown below
The vectors "check1" and "check2" are defined to be
check1 = U - 3V
check2 = -U + V
so that part is confirmed as well.
Answer by rothauserc(4718)  (Show Source):
You can put this solution on YOUR website! 1) (U-3V) = (1;5)
2) (-U+V) = (-5;3)
:
solve equation 1 for U
:
U = (1;5) + 3V
:
substitute in equation 2
:
-[(1;5) + 3V] + V= (-5;3)
:
(-1;-5) -3V + V = (-5;3)
:
-2V = (-4;8)
:
V = (2;-4)
:
U = (1;5) + 3(2;-4) = (7;-7)
:
U is (7,-7) and V is (2,-4)
:
we want the angle(theta) between vectors U and V
:
cos theta = (U . V) / |U| |V|, where . means dot product
:
U . V = 7*2 + -7*-4 = 42
|U| = square root(49 + 49) = square root(98) = 7 * square root(2)
|V| = square root(4 + 16) = square root(20) = 2 * square root(5)
:
cos theta = 42 / 14 * square root(2) * square root(5) = 3 / square root(10) = 0.9487
:
inverse cos 0.9487 = 18.4 degrees
:
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The angle between vectors U and V is 18.4 degrees
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