SOLUTION: The question is
Show that there are no vectors u and v such that the magnitude of u=1, the magbnitude of v=2, and the (dot product) u*v=3.
Hope this makes sense. It is easier to
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Question 98494: The question is
Show that there are no vectors u and v such that the magnitude of u=1, the magbnitude of v=2, and the (dot product) u*v=3.
Hope this makes sense. It is easier to understand with the symbols that they use.
I wasn't even sure how to start this problem. All I could get was 1*2 which doesn't =3 but that seems too easy.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
I'm not sure how far you are in your course, but to find the angle between any two vectors, you would use this formula:
=\frac{\v{u}\cdot\v{v}}{\left|\v{u}\right|\left|\v{v}\right|})
where 
is the magnitude (ie length) of vector 
and 
is the magnitude (ie length) of vector 
So let's assume that 
is true. That means we can replace 
with 3
=\frac{3}{\left|\v{a}\right|\left|\v{b}\right|})
Now plug in the given magnitudes 
and 
=\frac{3}{1*2})
Now multiply the values in the denominator
=\frac{3}{2})
So this equation is implying that there is some value of 
that will make true. However, there is no such value. Remember, the range of cosine is \le1)
. Since equals 1.5 (which is greater than 1), it is outside the range of cosine. So if you try to take arccosine of both sides, you will not get a real answer for
So that means there are no two vectors that have magnitudes of 1 and 2 while having a dot product of 3.
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