Find the vector algebraically

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  • #1

Homework Statement


Find a unit vector in the direction of the given vector w=-i-2j


Homework Equations





The Attempt at a Solution



-i
_______

vertical -2j_________________

Could someone please show me how I would do it grafically and algebraically?

Do I just need to find the magnitude? Wouldn't any vector that has a negative x component and a negative y component work?

square root (-1)^2+(-2)^2=square root 5

Thank you very much
 
  • #2
Any vector A divided by magnitude of A is a unit vector.
 
  • #3
I haven't studied this in quite a while, but have you tried representing it with an Argand diagram?

http://scholar.hw.ac.uk/site/maths/topic11.asp?outline= [Broken]
 
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  • #4
Thank you very much

Isn't the magnitude, in this case, the square root of 5?

So, would the unit vector be 1/square root of 5i? Aren't i and j the unit vectors? I'm already given the unit vectors, right? Don't I just need to find another one in the same direction?:confused:

Thank you
 
  • #5
Well I had a quick check of old material, but the modulus of an Argand diagram or distance from 0 is the magnitude.

if z=x+yi then [itex]|z|=\sqrt{x^2+y^2}.[/itex]

http://www.clarku.edu/~djoyce/complex/abs.html

And then.

Like I say it's been a while but hope that helps.
 
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  • #6
Some straight answers …

Isn't the magnitude, in this case, the square root of 5?

So, would the unit vector be 1/square root of 5i?

Aren't i and j the unit vectors?

I'm already given the unit vectors, right? Don't I just need to find another one in the same direction?:confused:

No-one gives you a straight answer, do they? :smile:

Yes, the magnitude is √5. (btw, type alt-v and it gives you √)

No, you mean the unit vector is (-i - 2j)/√5.

Yes, i and j are always unit vectors in this sort of question.

Yes, you're right. Why so puzzled?

Wouldn't any vector that has a negative x component and a negative y component work?

No - I don't see why you'd think that. :confused:

graphically and algebraically?

Algebraically, you've understood it fine!

Graphically: -i -2j is on the circle of radius √5; join it to the origin by a line; then you want the point where that line cuts the circle of radius 1. :smile:
 
  • #7
Thank you very much everyone

Regards
 
  • #8
Isn't the magnitude, in this case, the square root of 5?

Right.

So, would the unit vector be 1/square root of 5i? Aren't i and j the unit vectors? I'm already given the unit vectors, right? Don't I just need to find another one in the same direction?:confused:

i and j are unit vectors along the positive x-axis and y-axis respectively, but you can find a unit vector in any direction. In this case, the given direction was along the vector w=-i-2j. So, you have to divide w by mod(w) to get the unit vector in the direction of w.

No-one gives you a straight answer, do they? :smile:

You got that right. :biggrin: Everybody wants to make sure the OP does learn a bit at least by doing something himself/herself.
 
  • #9
You got that right. :biggrin: Everybody wants to make sure the OP does learn a bit at least by doing something himself/herself.

Including me, I just learned how to calculate the magnitude of a vector using an Argand diagram. Again. :smile:
 

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