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: 6,542 Thanks: 592 
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  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,119 Thanks: 710 Math Focus: Physics, mathematical modelling, numerical and computational solutions  Quote:
$\displaystyle \vec{PQ} = 33i + 12j  6k$ 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 
Newbie Joined: Jul 2017 From: US Posts: 7 Thanks: 0  So this explicitly describes the head of the vector, Q, but does it describe its tail and direction as well? Thankyou both for the clarification! Last edited by pst; July 12th, 2017 at 08:09 AM. 
July 12th, 2017, 09:06 AM  #6  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,119 Thanks: 710 Math Focus: Physics, mathematical modelling, numerical and computational solutions  Quote:
$\displaystyle \vec{PQ} = 33i + 12j − 6k$ 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 xaxis, 12 units forwards along the yaxis and then 6 units backwards along the zaxis, 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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