Understanding unit vectors

[Clydey]

Hey all. I'm a beginner trying to wrap my head around unit vectors.

Let's say the unit vector of v is: (12/√ ̅ 169, 5/√ ̅ 169). That's how I've been shown to represent the answer. I'm having trouble with the fact that this doesn't return a magnitude of 1 unless you square it. You end up with 17/13, if I'm right. (144/169, 25/169) does return a magnitude of 1.

So I guess my question is why is the former used to represent a unit vector if it doesn return a magnitude of 1?

This is, of course, assuming I've understoof the subject correctly. I've no doubt missed something simple. Any help would be appreciated, though.

 

[pka]

If \(\displaystyle \vec{v}=<a,b>\) then its length is \(\displaystyle \sqrt{a^2+b^2}\) so to make a unit vector divide by length \(\displaystyle \left<\frac{a}{\sqrt{a^2+b^2}}, \frac{b}{\sqrt{a^2+b^2}}\right>\)

 

[Dr.Peterson]

How are you calculating magnitude? It sounds like you're using the wrong formula for that.

The magnitude of the vector [MATH]<a, b>[/MATH] is [MATH]\sqrt{a^2 + b^2}[/MATH]. In this case, the magnitude of [MATH]\left<\frac{12}{13},\frac{5}{13}\right>[/MATH] is [MATH]\sqrt{\frac{12^2}{13^2}+\frac{5^2}{13^2}} = 1[/MATH].

 

[Clydey]

If \(\displaystyle \vec{v}=<a,b>\) then its length is \(\displaystyle \sqrt{a^2+b^2}\) so to make a unit vector divide by length \(\displaystyle \left<\frac{a}{\sqrt{a^2+b^2}}, \frac{b}{\sqrt{a^2+b^2}}\right>\)

Thanks for responding. I might be missing something obvious. If I plug in the values a = 3, b = 4, I end up with (3/√ ̅ 25, 4/√ ̅ 25). Am I right in saying that equates to 7/5, which doesn't return a magnitude of 1? It only returns a magnitude of 1 if I square everything i.e. (3/√ ̅ 25, 4/√ ̅ 25)^2.

Do you see what I'm getting at? I might not be explaining myself properly, as I'm not particularly fluent in the subject.

 

[Clydey]

How are you calculating magnitude? It sounds like you're using the wrong formula for that.

The magnitude of the vector [MATH]<a, b>[/MATH] is [MATH]\sqrt{a^2 + b^2}[/MATH]. In this case, the magnitude of [MATH]\left<\frac{12}{13},\frac{5}{13}\right>[/MATH] is [MATH]\sqrt{\frac{12^2}{13^2}+\frac{5^2}{13^2}} = 1[/MATH].

My issue is with how I'm being asked to represent the answer (I'm following tutorials on Khan Academy). I wish I could explain myself better.

So as I understand it (12/13, 5/13) equates to 17/13, which isn't a magnitude of 1. It only becomes a magnitude of 1 if you square (12/13, 5/13). It becomes 169/169, whch is a magnitude of 1. Am I making any kind of sense? My apologies if I'm not.

 

[Dr.Peterson]

My issue is with how I'm being asked to represent the answer (I'm following tutorials on Khan Academy). I wish I could explain myself better.

So as I understand it (12/13, 5/13) equates to 17/13, which isn't a magnitude of 1. It only becomes a magnitude of 1 if you square (12/13, 5/13). It becomes 169/169, which is a magnitude of 1. Am I making any kind of sense? My apologies if I'm not.

No. You can't just add the two components of a vector. The vector (12/13, 5/13) goes 12/13 of a mile east and 5/13 of a mile north, so to speak, but its length is that of a straight line to that position, not the distance you travel if you go east and then north.

Both of us showed you the formula for magnitude of a vector, [MATH]\sqrt{a^2 + b^2}[/MATH]. You seem to be unfamiliar with it, or else to be misinterpreting what it means. Do you recognize it as the distance formula, or the Pythagorean Theorem? The squaring is not a separate thing you are doing to the vector, but part of the very process of finding the length of the vector.

 

[Clydey]

No. You can't just add the two components of a vector. The vector (12/13, 5/13) goes 12/13 of a mile east and 5/13 of a mile north, so to speak, but its length is that of a straight line to that position, not the distance you travel if you go east and then north.

Both of us showed you the formula for magnitude of a vector, [MATH]\sqrt{a^2 + b^2}[/MATH]. You seem to be unfamiliar with it, or else to be misinterpreting what it means. Do you recognize it as the distance formula, or the Pythagorean Theorem? The squaring is not a separate thing you are doing to the vector, but part of the very process of finding the length of the vector.

I recognise it as the Pythagorean Theorem.

What it boils down to is I don't understand how (12/13, 5/13) and (12/13, 5/13) ^2 are the same thing. In the tutorial I'm following, I'm told to use the former to represent my answer. I understand perfectly how the latter works. I understand that squaring it is necessary. What I don't understand is why I'm told to use (12/13, 5/13) to represent a unit vector, rather than (12/13, 5/13) ^2.

I don't have the mathematical vocabulary to be more clear. I'm sorry about that.

 

[Dr.Peterson]

I recognise it as the Pythagorean Theorem.

What it boils down to is I don't understand how (12/13, 5/13) and (12/13, 5/13) ^2 are the same thing. In the tutorial I'm following, I'm told to use the former to represent my answer. I understand perfectly how the latter works. I understand that squaring it is necessary. What I don't understand is why I'm told to use (12/13, 5/13) to represent a unit vector, rather than (12/13, 5/13) ^2.

I don't have the mathematical vocabulary to be more clear. I'm sorry about that.

They aren't the same thing! In fact, as I said, squaring a vector is meaningless. There is no such thing as (12/13, 5/13)^2.

Can you quote your tutorial exactly (maybe an image, or even a link) so we can see just what you are following and whether it is incorrect, or you are misinterpreting it?

 

[Clydey]

They aren't the same thing! In fact, as I said, squaring a vector is meaningless. There is no such thing as (12/13, 5/13)^2.

Can you quote your tutorial exactly (maybe an image, or even a link) so we can see just what you are following and whether it is incorrect, or you are misinterpreting it?

I can only assume I'm misinterpreting it. And I didn't mean squaring the vector. I meant squaring the calculation (if that's the correct term) to end up with √ ̅ 169/169. I'd link you to the tutorial, but I've sinced watched a different tutorial and they both have precisely the same explanation.

I won't waste your time, as I'm clearly just not getting something. I'll persist with the tutorials until it clicks. Thanks for the help.

 

[Clydey]

I can only assume I'm misinterpreting it. And I didn't mean squaring the vector. I meant squaring the calculation (if that's the correct term) to end up with √ ̅ 169/169. I'd link you to the tutorial, but I've sinced watched a different tutorial and they both have precisely the same explanation.

I won't waste your time, as I'm clearly just not getting something. I'll persist with the tutorials until it clicks. Thanks for the help.

And two minutes after posting this, it just clicked with me. You explained it perfectly.