Thread: Vector Troubles

Hello,

I am having trouble understanding vectors. I have attempted all of the questions and the results I ended up with seem (at least to me) like they could be correct. If someone could look over what I've got and tell me if I'm on the right track I would appreciate it. The upcoming lessons build on this stuff so I need to make sure I understand it. I apologize for the quality of the images.

For this one I have the following:

a) $\displaystyle \overrightarrow{EF} - \overrightarrow{FG} = \overrightarrow{EG}$

b) $\displaystyle \overrightarrow{EF} - \overrightarrow{FG} = \overrightarrow{EF} - \overrightarrow{EH} = \overrightarrow{FH}$

c) $\displaystyle \overrightarrow{FE} - \overrightarrow{FG} = \overrightarrow{EG}$

d) $\displaystyle \overrightarrow{FG} - \overrightarrow{EF} = \overrightarrow{FG} - \overrightarrow{GH} = \overrightarrow{FH}$







The second question asks:

The diagram below shows a cube, where $\displaystyle \overrightarrow{AB} = \overrightarrow{u}$, $\displaystyle \overrightarrow{AD} = \overrightarrow{v}$ and $\displaystyle \overrightarrow{AE} = \overrightarrow{w}$. Determine a single vector equivalent to each of the following.

a) u + v + w
b) u + v - w
c) u - v + w
d) u - v - w

For this I have:

a) $\displaystyle \overrightarrow{AB} + \overrightarrow{AD} + \overrightarrow{AE} = \overrightarrow{BD} - \overrightarrow{AE} = \overrightarrow{BD} + \overrightarrow{DH} = \overrightarrow{BH}$

b) $\displaystyle \overrightarrow{AB} + \overrightarrow{AD} - \overrightarrow{AE} = \overrightarrow{BD} - \overrightarrow{AE} = \overrightarrow{BD} - \overrightarrow{BF} = \overrightarrow{DF}$

c) $\displaystyle \overrightarrow{AB} - \overrightarrow{AD} + \overrightarrow{AE} = \overrightarrow{BD} + \overrightarrow{AE} = \overrightarrow{BD} + \overrightarrow{DH} = \overrightarrow{BH}$

d) $\displaystyle \overrightarrow{AB} - \overrightarrow{AD} - \overrightarrow{AE} = \overrightarrow{BD} - \overrightarrow{AE} = \overrightarrow{BD} - \overrightarrow{BF} = \overrightarrow{DF}$



Thank you.

Originally Posted by SDF

Hello,

For this one I have the following:

a) $\displaystyle \overrightarrow{EF} - \overrightarrow{FG} = \overrightarrow{EG}$

b) $\displaystyle \overrightarrow{EF} - \overrightarrow{FG} = \overrightarrow{EF} - \overrightarrow{EH} = \overrightarrow{FH}$

c) $\displaystyle \overrightarrow{FE} - \overrightarrow{FG} = \overrightarrow{EG}$

This is incorrect.

Note that $-\vec{u}-\vec{v}=-(\vec{u}+\vec{v})$ which you solved for in (a)

d) $\displaystyle \overrightarrow{FG} - \overrightarrow{EF} = \overrightarrow{FG} - \overrightarrow{GH} = \overrightarrow{FH}$

This is also incorrect.

Note that $\vec{v}-\vec{u}=-(\vec{u}-\vec{v})$ which you solved for in (b)

I don't think you got any of #2 correct.

For example the 4th one.

$\vec{u}$ takes us to (A+B, 0, 0)

$\vec{u}-\vec{v}$ takes us to (A+B, A-D, 0)

$\vec{u}-\vec{v}-\vec{z}$ takes us to (A+B, A-D, A-E)

This is off of the cube so it's useful to use a different starting point on the cube.

Since these vectors aren't tied to any particular point in space we can use H as the starting point as well.

Doing this we find that (H+B, H-D, H-E) = B

and so the vector that represents $\vec{u}-\vec{v}-\vec{z}$ is $\vec{HB}$

see if you can rework the first 3. Picking the correct starting point is the key to this.

Originally Posted by romsek

I don't think you got any of #2 correct.

For example the 4th one.

$\vec{u}$ takes us to (A+B, 0, 0)

$\vec{u}-\vec{v}$ takes us to (A+B, A-D, 0)

$\vec{u}-\vec{v}-\vec{z}$ takes us to (A+B, A-D, A-E)

This is off of the cube so it's useful to use a different starting point on the cube.

Since these vectors aren't tied to any particular point in space we can use H as the starting point as well.

Doing this we find that (H+B, H-D, H-E) = B

and so the vector that represents $\vec{u}-\vec{v}-\vec{z}$ is $\vec{HB}$

see if you can rework the first 3. Picking the correct starting point is the key to this.

I'll be honest, I have no idea what any of that means. The only correlation I can see between the cube and the single vector for the fourth question being $\displaystyle \vec{HB}$ is that if you were to "slice" the cube from H to B you would get half of it, the same as if you were to mark AB, AD and AE, join them and remove them all at once. That is about the only way I can think about it without relying on trying to find opposites to remove, which is how I tried to solve them originally. (ie. for addition, if I had AB + BC I would shorten it to "AC" since the B's are together, and for subtraction I would look for opposites which would provide me with letter pairs that share either inner or outer letters. What I mean by this is for something like EF - EH I would know that the answer is FH because the E's are in the same position.) And yeah, I'm aware how messed up this approach is but it seemed to work until I got to the cube question.

Alright I have re-tried the first 3 from the cube question and here are my results:

a) $\displaystyle \overrightarrow{AB} + \overrightarrow{AD} + \overrightarrow{AE} = \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{AE} = \overrightarrow{AC} + \overrightarrow{AE} = \overrightarrow{AC} + \overrightarrow{CG} = \overrightarrow{AG}$

b) $\displaystyle \overrightarrow{AE} - \overrightarrow{AB} - \overrightarrow{AD} = \overrightarrow{EB} - \overrightarrow{AD} = \overrightarrow{EB} - \overrightarrow{EH} = \overrightarrow{BH}$

c) $\displaystyle \overrightarrow{AD} - \overrightarrow{AB} + \overrightarrow{AE} = \overrightarrow{DB} + \overrightarrow{AE} = \overrightarrow{DB} + \overrightarrow{BF} = \overrightarrow{DF}$

I tried to take into account what you said regarding the minus signs. With that, I approached the problems like this:

a) u + v + w (Remained the same)

b) u + v - w = -(w - u - v)

c) u - v + w = -(v - u + w)


With that in mind, I also tried d) on my own:

u - v - w = -(v + w - u)

$\displaystyle \overrightarrow{AD} + \overrightarrow{AE} - \overrightarrow{AB} = \overrightarrow{AD} + \overrightarrow{DH} - \overrightarrow{AB} = \overrightarrow{AH} - \overrightarrow{AB} = \overrightarrow{HB}$

It still makes very little sense to me but I ended up with HB like you did, though it may just be pure coincidence since I knew the answer I was looking for.

I've given it some time, and I think it makes sense now. If someone could verify whether or not questions A, B, and C are correct it would really help my understanding. If they aren't, it is probably a misinterpretation of the subtraction rules -- in which case, some reference material would be greatly appreciated as I have been unable to find any thus far.

Originally Posted by SDF

I've given it some time, and I think it makes sense now. If someone could verify whether or not questions A, B, and C are correct it would really help my understanding. If they aren't, it is probably a misinterpretation of the subtraction rules -- in which case, some reference material would be greatly appreciated as I have been unable to find any thus far.

a) correct
b) incorrect
c) correct

For (b) we start at A. We come over u. This puts us at B. We go over v. This puts us at C. We go down w (up -w).
This puts us at C - |AE|.

It's not immediately clear what vector this corresponds to however if we started at point E instead of point A we'd have

Start at E. Over u puts us at F. Over v puts us at G. Down w puts us at C. So the vector we are after is $\vec{EC}$

Hi romsek,

I am not sure what you are doing regarding "moving along" the vectors. The way I solved them was using the 'letter rules'. For example, question (c) said $\displaystyle \vec{u}-\vec{v}+\vec{w}$, which I turned into $\displaystyle -(\vec{v}-\vec{u}+\vec{w})$ and then substituted the letters into the equation. When I did this, it became $\displaystyle \vec{AD} - \vec{AB} + \vec{AE}$ and I changed AD - AB into DB because the outer A's matched. Afterwards, it was $\displaystyle \vec{DB} + \vec{AE}$, so I looked for an equivalent side that would give me an inner 'B' so that I could cancel them out, and ultimately got an answer of $\displaystyle \vec{DF}$

Come on....

I start at the point A. I add the vector u to it. This is identical to "moving along" the vector and in this case is the easiest way to visualize it.

You can't see that "moving along" vector u from A puts you at B? Maybe some examples will firm it up

Adding u to A puts you at B
Adding u to D puts you at C
Adding u to E puts you at F
Adding u to H puts you at G

Adding v to A puts you at D
Adding v to B puts you at C
Adding v to E puts you at H
Adding v to F puts you at G

Adding w to A puts you at E
Adding w to B puts you at F
Adding w to C puts you at G
Adding w to D puts you at H

Study that until this pretty simple concept becomes clear.

No need to be condescending. Not everyone's mind works the same as yours and learning circumstances are different. In my case I was handed possibly the worst text-book (if you can call it that) ever created and left to complete the assignments with the aid of piss poor examples and no teacher to ask for help. On top of that, math concepts are incredibly difficult for me to grasp, especially anything spatial. Please do not interpret my failing to understand something as me not being serious about carrying my weight. Gaps in knowledge are the reason I am here in the first place and it is hard to "know what it is I don't know" before formulating a question. With that said, I appreciate the help you have provided. You are not obligated to help in any way and I am grateful for the time you have taken to explain things.

As for your examples, yes, I can see that moving from A gets you to B, however I think the likely source of confusion lies in the wording. For example, "Adding u to A puts you at B". In my mind, $\displaystyle \vec{u}$ IS the length/area between A and B, thus adding it to A would imply something different. I will attempt these questions again with the new information you've given.

Part of the problem may be in your understanding of $\vec{u}$. It is a length and a direction. It has nothing to do with area.

Adding $\vec{u}$ is just like someone telling you to walk $|\vec{u}|$ units in the direction of $\vec{u}$. If you orient your cube along the usual x,y,z axes, and face the y axis, you can see that this is like walking $|\vec{u}|$ units to the right. Adding $\vec{v}$ is like moving $|\vec{v}|$ units forward, and adding $\vec{z}$ is like moving $|\vec{z}|$ units up.

You can do these one after the other just like you'd follow a set of directions.

Alright I'm pretty sure I've got this down.

(a) u + v + w
$\displaystyle \vec{AB}+\vec{BC}+\vec{CG} = \vec{AC} + \vec{CG} = \vec{AG}$

(b) u + v - w
$\displaystyle \vec{EF} + \vec{FG} - \vec{GC} = \vec{EG} - \vec{GC} = \vec{EC}$

(c) u - v + w
$\displaystyle \vec{DC} - \vec{CB} + \vec{BF} = \vec{DB} + \vec{BF} = \vec{DF}$

(d) u - v - w
$\displaystyle \vec{HG} - \vec{GF} - \vec{FB} = \vec{HF} - \vec{FB} = \vec{HB}$


You were right in that the problem was not understanding the vector properly. I knew that anything I did had to conform to the vector rules (same side-length and direction), hence why I was looking for equivalent sides that were nearby/attached. This was the root cause of all the confusion and attempts to match letters. After you clarified that the vector was merely a property and was not dependent on certain corners (ie. $\displaystyle \vec{AB}=\vec{u}$ didn't HAVE to only represent A to B), things got a lot clearer.

Thank you for all your help! (and patience )