# Find the vector algebraically

1. Mar 2, 2008

### chocolatelover

1. The problem statement, all variables and given/known data
Find a unit vector in the direction of the given vector w=-i-2j

2. Relevant equations

3. 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. Mar 2, 2008

### Shooting Star

Any vector A divided by magnitude of A is a unit vector.

3. Mar 2, 2008

### Schrodinger's Dog

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]

Last edited by a moderator: May 3, 2017
4. Mar 2, 2008

### chocolatelover

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?

Thank you

5. Mar 2, 2008

### Schrodinger's Dog

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 $|z|=\sqrt{x^2+y^2}.$

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

And then.

Like I say it's been a while but hope that helps.

Last edited: Mar 2, 2008
6. Mar 2, 2008

### tiny-tim

Some straight answers …

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

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?

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

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.

7. Mar 2, 2008

### chocolatelover

Thank you very much everyone

Regards

8. Mar 2, 2008

### Shooting Star

Right.

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.

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

9. Mar 3, 2008

### Schrodinger's Dog

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