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 July 11th, 2017, 11:39 AM #1 Newbie   Joined: Jul 2017 From: US Posts: 7 Thanks: 0 Unit vectors Given points on the Cartesian plane of P = (1,-2), Q = (6, 10), we find that $\displaystyle \vec{PQ}=\><5,12>$, and $\displaystyle |\vec{PQ}| = \sqrt{25+144}=13$. By definition of unit vectors, we can find the unit vectors parallel to $\displaystyle \vec{PQ}$ by doing $\displaystyle \frac{<5,12>}{13} = <\frac{5}{13},\frac{12}{13}>$, then following the textbook, we can use unit vectors $\displaystyle i=<1,0>$, and $\displaystyle j=<0,1>$ to get $\displaystyle \frac{5}{13}i + \frac{12}{13}j$. If we're asked to generate two parallel vectors of length 7. It would suffice to multiply $\displaystyle \frac{<5,12>}{13}$ by $\displaystyle \pm7$. But the book format shows the two vectors of length 7 parallel to $\displaystyle \vec{PQ}$ to be: $\displaystyle \frac{35}{13}i+\frac{84}{13}j$ and $\displaystyle -\frac{35}{13}i-\frac{84}{13}j$. What I don't understand is, why do we have the vectors in the format of i and j? Does it signify anything? Would it be incorrect to simply state that the two vectors are $\displaystyle <\frac{35}{13},\frac{84}{13}>$ and $\displaystyle <-\frac{35}{13},-\frac{84}{13}>$? Why do we notate unit vectors of $\displaystyle \vec{PQ}$ in terms of i and j?
 July 11th, 2017, 01:44 PM #2 Global Moderator   Joined: May 2007 Posts: 6,607 Thanks: 616 The use of i and j is simply a notation convention. It gives labels to the unit vectors (1,0) and (0,1). A vector of the form (u,v) can then be expressed as (ui,vj). To some extent it depends on whether you are working in math (i and j omitted) or physics (i and j used).
 July 11th, 2017, 02:39 PM #3 Newbie   Joined: Jul 2017 From: US Posts: 7 Thanks: 0 So i and j can be treated as multiplicative identities for vectors? And convention wise, expressing the vectors in form of addition (or subtraction) is valid?
July 12th, 2017, 01:28 AM   #4
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Quote:
 Originally Posted by pst So i and j can be treated as multiplicative identities for vectors? And convention wise, expressing the vectors in form of addition (or subtraction) is valid?
I have no idea what you mean by "multiplicative identities for vectors". As Mathman said, the use of i and j is a particular convention and they are called unit vectors (i.e. they are vectors with length 1). They're used to denote directions on the Cartesian plane. Note that a scalar multiplied by a vector is a vector, so if we multiply our unit vectors with scalars and then add them together, like this:

$\displaystyle \vec{PQ} = 33i + 12j - 6k$

then the result of that series of operations is a single vector that points from the coordinate P to the coordinate Q. If P is the origin, Q will be found at the coordinate (33, 12, -6) in Cartesian coordinates.

Last edited by Benit13; July 12th, 2017 at 01:39 AM.

July 12th, 2017, 07:41 AM   #5
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Quote:
 Originally Posted by Benit13 $\displaystyle \vec{PQ} = 33i + 12j - 6k$
So this explicitly describes the head of the vector, Q, but does it describe its tail and direction as well?

Thank-you both for the clarification!

Last edited by pst; July 12th, 2017 at 08:09 AM.

July 12th, 2017, 09:06 AM   #6
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Quote:
 Originally Posted by pst So this explicitly describes the head of the vector, Q, but does it describe its tail and direction as well? Thank-you both for the clarification!
I'm really not sure what you're trying to say, but the vector

$\displaystyle \vec{PQ} = 33i + 12j − 6k$

describes how to get from point P to point Q. There's no explicit information about the coordinates of P or Q in that vector; it's a relative quantity that maps one point to the other point.

If you start at P and then travel 33 units forwards along the x-axis, 12 units forwards along the y-axis and then 6 units backwards along the z-axis, you'll arrive at Q.

To travel the reverse (Q to P), you can multiply the whole vector by -1:

$\displaystyle \vec{QP} = -\vec{PQ} = -( 33i + 12j − 6k) = -33i - 12j + 6k$

You can get the distance between P and Q by finding the magnitude of the vector (using Pythagoras' theorem)

$\displaystyle |\vec{PQ}| = \sqrt{33^2 + 12^2 + 6^2} = 35.623$

You can get the unit vector that describes the direction you need to go in to reach Q from P by normalising the vector. This is done by dividing each component's magnitude by the total magnitude of the vector:

$\displaystyle \hat{PQ} = \frac{33}{35.623}i + \frac{12}{35.623}j - \frac{6}{35.623}k$
$\displaystyle = 0.926 i + 0.337j - 0.168k$

So another way of describing the journey from P to Q is to say "Travel 35.623 along the direction 0.926 i + 0.337j - 0.168k". We're now getting into polar coordinates territory now

So... yeah! Vectors are useful for things involving directions or maps between points. In physics, vectors are used for things like force, velocity, acceleration, momentum, moment of inertia, torque, electric field, magnetic field, light rays... all sorts of stuff

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