anonymous
  • anonymous
Let v1 = (2, -6) and v2 = (-4, 7). Compute the unit vectors in the direction of |v1| and |v2|. And can anyone double check if this graph is right? Draw and label v1, v2, and v1+v2. https://gyazo.com/ca330d1301b8dd28e0cdfa3e72f6443c
Mathematics
katieb
  • katieb
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Alex_Mattucci
  • Alex_Mattucci
Your graph is looking great!
Alex_Mattucci
  • Alex_Mattucci
is that all?
anonymous
  • anonymous
Compute the unit vectors in the direction of |v1| and |v2|. What exactly is this question trying to find?

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SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \frac{\vec{V_1} }{\left|\left| \vec{V_1}\right|\right|} }\)
SolomonZelman
  • SolomonZelman
This is the unit vector (with magnitude 1) in direction of \(\vec{V_1}\)
anonymous
  • anonymous
How would you plug in the vector v1 into this equation to find a value?
SolomonZelman
  • SolomonZelman
Note: V\(_1\) with two bars on each side, means "magnitude of V\(_1\).
SolomonZelman
  • SolomonZelman
You want units vecotrs with the same directions as \(\vec{V_1}\) and \(\vec{V_2}\), right?
SolomonZelman
  • SolomonZelman
|dw:1442448091015:dw|
SolomonZelman
  • SolomonZelman
that means that if you take each component and divide by this magnitude, you get a unit vector in same direction.
SolomonZelman
  • SolomonZelman
\(\left|\vec{V_1}\right|=\sqrt{(-2)^2+(6)^2{\color{white}{\large|}}}\)
SolomonZelman
  • SolomonZelman
I mixed it up, 2, and -6. For magnitude that doesn't matter though. you still get 2√10
SolomonZelman
  • SolomonZelman
|dw:1442448470003:dw|
SolomonZelman
  • SolomonZelman
Now, you need to simplify this and you are done with the unit vector for V\(\large _1\)
anonymous
  • anonymous
Oh, so the value just ends up being the x value of the vector/ magnitude and the y value/magnitude?
SolomonZelman
  • SolomonZelman
yes
SolomonZelman
  • SolomonZelman
And same for 3 D vector: < x-component/magnitude, y-component/magnitude, z-component > (and same for any N-dimensional vector (only physics stops at 3 D))
SolomonZelman
  • SolomonZelman
Then you need to find the magnitude of \(\vec{V_2}\) and divide each of the \(\vec{V_2}\)'s components by this magnitude. And lets recall that: \(\vec{V_2}=<4,-7>\)
SolomonZelman
  • SolomonZelman
So the rest is yours:)
anonymous
  • anonymous
So magnitude of v2 ends up being √65, then the unit vector would be 4/√65 and -7/√65?
SolomonZelman
  • SolomonZelman
let me see: √{4²+(-7)²}=√{16+49}=√65 Yes, then it would be: < 4/√65 , -7/√65 >
SolomonZelman
  • SolomonZelman
You can use a taylor polynomial approximation of some nth degree near a=64, of f(x)=√x. (If you want.... :D)
SolomonZelman
  • SolomonZelman
Imagine you didn't have a calculator, then you would need one ... jk
anonymous
  • anonymous
Oh jeez, I have no idea about 3d vectors or taylor polynomials at all....But thanks so much for all the help!
SolomonZelman
  • SolomonZelman
3D is just more complicated to draw, that is all. And tailor polynomial just comes from standard theorem of calculus and integration of parts, starting from: \(\displaystyle \int_{0}^{x}f'(t)dt\)
SolomonZelman
  • SolomonZelman
You will learn it pretty soon I am sure.... as for now, if you don't have any questions regarding your problem, then good luck:)

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