Question 98494
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:


*[Tex \LARGE \cos(\theta)=\frac{\v{u}\cdot\v{v}}{\left|\v{u}\right|\left|\v{v}\right|}] where   *[Tex \Large \left|\v{u}\right|] is the magnitude (ie length) of vector *[Tex \Large \v{u}] and *[Tex \Large \left|\v{v}\right|] is the magnitude (ie length) of vector *[Tex \Large \v{v}]





So let's assume that *[Tex \Large \v{u}\cdot\v{v}=3] is true. That means we can replace *[Tex \Large \v{u}\cdot\v{v}] with 3



*[Tex \LARGE \cos(\theta)=\frac{3}{\left|\v{a}\right|\left|\v{b}\right|}] 



Now plug in the given magnitudes *[Tex \Large \left|\v{u}\right|=1] and *[Tex \Large \left|\v{v}\right|=2]



*[Tex \LARGE \cos(\theta)=\frac{3}{1*2}] 


Now multiply the values in the denominator 


*[Tex \LARGE \cos(\theta)=\frac{3}{2}] 



So this equation is implying that there is some value of *[Tex \LARGE \theta] that will make *[Tex \LARGE \cos(\theta)=\frac{3}{2}] true. However, there is no such value. Remember, the range of cosine is *[Tex \LARGE -1\le cos(\theta)\le1]. Since {{{3/2}}} 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 *[Tex \LARGE \theta]



So that means there are no two vectors that have magnitudes of 1 and 2 while having a dot product of 3.