Having problem showing the vectors point in the same dir. its a 0 Vector?

[mr_coffee]

Hello everyone, I'm really stuck on this problem, the algebra is killing me. The professor said Show that the two vectors are in the same direction, but the fact is, its the zero vector he said. So it points to nothing or points everywhere. He told us to convert the diagram into Unit vectors. So I first found the distance d1 and d2. Then found the midpoint. I then attempted to find the Unit vectors by taking the (1/|A|) <x,y,z> but its getting really messy, and I'm stuck, any help would be great! Click the link for the picture:
http://www.imagehosting.us/index.php?action=show&ident=653183

 

[HallsofIvy]

"The professor said Show that the two vectors are in the same direction, but the fact is, its the zero vector he said."

Do you realize that that sentence makes no sense? You start of by talking about two and then say "it's the zero vector". What is "it"? What is the relation between the original two vectors and that "zero vector"?

". He told us to convert the diagram into Unit vectors. So I first found the distance d1 and d2. Then found the midpoint. I then attempted to find the Unit vectors by taking the (1/|A|) <x,y,z> but its getting really messy, and I'm stuck, any help would be great! Click the link for the picture!"

This makes no sense without knowing what d1 and d2 or, for that matter, what it is you are trying to do. I finally got the picture to enlarge but it doesn't say what you are trying to do! What is the question??

 

[Doc Al]

It sounds to me that you are being asked to prove that vectors D1 and D2 are parallel. Perhaps he asked you to demonstrate that their cross-product is zero?

 

[mr_coffee]

Its weird how this question became, sorry for the confusion. The orginal question was the following: (a) prove that the midpoint of the line segment from P1(x1,y1,z1) to P2(x2,y2,z2) is ([x1+x2]/2,[y1+y2]/2,[z1+z2]/2). So from this, he drew that picture I have posted, he labed the total length of the vector d1, he labled the midpoint d2 as shown in the drawing. So he found the distance of d1 and d2 as shown in the drawing. He then Stopped and said, now show that the vectors are pointing in the same direction. And someone said well its the 0 vector, and he didn't disagree with him, but he said why is it the 0 vector? So maybe it isn't hte 0 vector at all. So he then said, put that in Unit Vector form as a hint. The equation of a unit vector is the following: u = (1/|a|)*a = a/|a|; So he then said, bascially all you have to do is the following: 1/d1 * V1 and 1/d2 * U2. V1 is d1 and U2 = d2. and V1 and U2 are vectors. Does that help you any Doc?

 

[Doc Al]

From any vector you can create a unit vector in the same direction, using the formula given. ($\hat{a} = \vec{a}/{|a|}$). So, for each vector, D1 and D2, find the corresponding unit vector. Of course, these unit vectors will be identical, proving that the vectors are parallel. (Of course, this is overkill... but a good exercise.)

 

[mr_coffee]

I don't understand how I'm suppose to simplfy what i have though, I understand the concept i just have no idea how to simplfy the algebra

 

[Doc Al]

Here's a hint:
$$x_2 - \frac{(x_2 + x_1)}{2} = \frac{(x_2 - x_1)}{2}$$

 

[mr_coffee]

Thanks, But I'm not sure if i did this right...
http://show.imagehosting.us/show/655964/0/nouser_655/T0_-1_655964.jpg
This is just showing what I did with d2. I'm trying to do 1/d2 * U2;

 

[Doc Al]

You made a huge mistake:
$$\sqrt{a^2 + b^2} \ \ne \ a + b$$

(Test this with a few numbers to realize the problem. For example, let a = 3, b = 4.)

 

[mr_coffee]

Thanks for the help Doc. Can u see if I proved it or do i have to do more? Here is my work:
http://img238.imageshack.us/img238/7470/calciiii2mv.jpg

 

[Doc Al]

You are almost there. (You have the algebra part right) But the vectors are the differences in the coordinates. For example, the vector "d1" should be:
$$\vec{d_1} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$$

 

[mr_coffee]

Isn't my d1 the same as the one you just posted, mine is just under a square root sign and I didn't write the unit vectors besdies the differences, is that all your saying I need to do? or are you saying I need to try to simplfy it so d1 looks exactly like yours? I hope not because i have no idea how to get ride of that square root sign

 

[Doc Al]

As far as I can tell, you seem to take your unit vector representing "d1" to be:
$$\hat{d_1} = (1/d_1)[ (x_1)\hat{i} + (y_1)\hat{j} + (z_1)\hat{k}]$$

And the one representing "d2" to be:
$$\hat{d_2} = (1/d_2)[ (x_2)\hat{i} + (y_2)\hat{j} + (z_2)\hat{k}]$$

Not true. Good luck proving these vectors to be equal! (They're not.)

The real "d1" unit vector should be:
$$\hat{d_1} = (1/d_1) [(x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}]$$

And the real "d2" unit vector should be:
$$\hat{d_2} = (1/d_2) [((x_2 + x_1)/2 - x_1)\hat{i} + ((y_2 + y_1)/2 - y_1)\hat{j} + ((z_2 + z_1)/2 - z_1)\hat{k}]$$

These are the unit vectors that you should prove are equal. (Using the same algebraic manipulations that you've been doing.)

 

[mr_coffee]

THanks again Doc , I totally forgot I had to make them into vector form. Does this prove it?
http://img209.imageshack.us/img209/7140/blee9el.jpg

 

[Doc Al]

I think you are using $\vec{v}$ and $\vec{u}$ to represent the unit vectors that I called $\hat{d_1}$ and $\hat{d_2}$. If so, your job is to show that $\vec{v} = \vec{u}$.

That means you need to show that your fourth equation (for $\vec{u}$) is identical to your second equation (for $\vec{v}$). Since your fourth equation still is in terms of d2, while your second equation is in terms of d1, you aren't done yet.

 

[mr_coffee]

hm...I showed that d1 = 4*d2, doesn't that make d1 == d2? so having d1 under V and d2 under U shouldn't make a difference or isn't that the way its going to work?

 

[Doc Al]

For one thing, d1 = 2*d2. (You forgot a square root in your work from post #10.) And when doing a proof, you just can't leave out the last step!

 

[mr_coffee]

ahh good call, sorry my bad. I'm not very good at math, I came to this conclusion, luckly I'm almost done :) Thanks for the help as usual Doc!

 

[mr_coffee]

why would d1 = 2*d2? If i have d2 = 2*sqrt(junk); wouldn't d2/2 = sqrt(junk), so d1 = d2/2 ?

 

[Doc Al]

If i have d2 = 2*sqrt(junk); wouldn't d2/2 = sqrt(junk),

Sure... if!

so d1 = d2/2 ?

Where did this come from?

Review your work in post #10.

 

[mr_coffee]

Sorry i'll be more specific...
http://show.imagehosting.us/show/664511/0/nouser_664/T0_-1_664511.jpg
Now if i wanted to say d2 is 2 times as big as d1, that makes sense. But if I wanted to use d2 in an expression, wouldn't I have to solve d2, and make it .5*d2? So when I take 1/d2 * U, would it be 2*d2 * U?

 

[Doc Al]

Sorry i'll be more specific...
http://show.imagehosting.us/show/664511/0/nouser_664/T0_-1_664511.jpg

In going from line 1 to line 2 you make an error. That 2 in front of the square root should be a 1/2.

 

[mr_coffee]

thanks, so now if i wanted to use d2 in an expression, would I bring that 1/2 on the left side so it would be 2*d2 = sqrt((x2-x1)^2 + (y2-y1)^2 +(z2-z1)^2);
So when I use it in this expression:
1/d2 * [(x2-x1)/2 i + (y2-y1)/2 j + (z2-z1)/2) k];
1/(2*d2) * [(x2-x1)/2 i + (y2-y1)/2 j + (z2-z1)/2) k];
see what I'm asking?

 

[Doc Al]

thanks, so now if i wanted to use d2 in an expression, would I bring that 1/2 on the left side so it would be 2*d2 = sqrt((x2-x1)^2 + (y2-y1)^2 +(z2-z1)^2);

There's nothing wrong with doing that since it's a perfectly legitimate mathematical move. (But I don't see why you would.)

So when I use it in this expression:
1/d2 * [(x2-x1)/2 i + (y2-y1)/2 j + (z2-z1)/2) k];
1/(2*d2) * [(x2-x1)/2 i + (y2-y1)/2 j + (z2-z1)/2) k];

Realize that in going from line 1 to line 2 you replaced d2 by 2*d2. What makes you think you can get away with that? (And what does that have to do with what you asked above?)

Assuming you are still working on the given problem, what you need to be doing is expressing both unit vectors in terms of d1. (That way you'll be able to prove them equal.)

see what I'm asking?

Not yet.