Solving cross product

burt

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I keep getting the same answer for these cross products. Does that make sense?
\(\displaystyle B(1)=\left|\begin{matrix}i&j&k\\\frac{1}{\sqrt{14}}&\frac{2}{\sqrt{14}}&\frac{3}{\sqrt{14}}\\\frac{-3}{\sqrt{70}}&\frac{-6}{\sqrt{70}}&\frac{\sqrt5}{\sqrt{14}}\\\end{matrix}\right|=<\frac{2}{\sqrt5},\frac{-1}{\sqrt5},0>\) and \(\displaystyle B(0)=\left|\begin{matrix}i&j&k\\\frac{1}{\sqrt5}&\frac{2}{\sqrt5}&0\\0&0&1\\\end{matrix}\right|=<\frac{2}{\sqrt5},\frac{-1}{\sqrt5},0>\)
 

Harry_the_cat

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I haven't checked your answers but why do you think they can't give the same result?
 

burt

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@Harry_the_cat It just seems off to me, because they are the binormal product for a vector valued equation at two different points. The unit tangent vector and normal vector are not the same at those points.
 

Subhotosh Khan

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@Harry_the_cat It just seems off to me, because they are the binormal product for a vector valued equation at two different points. The unit tangent vector and normal vector are not the same at those points.
"...product for a vector valued equation at two different points..."

Those CAN be same mathematically. But there could be "physics" that make "equality" impossible - but we don't know that "physics" part of the problem.
 

burt

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"...product for a vector valued equation at two different points..."

Those CAN be same mathematically. But there could be "physics" that make "equality" impossible - but we don't know that "physics" part of the problem.
Here is the problem: 1585014804379.png1585014820894.png

Here is my work:
1585014996269.png
Actually, now that I'm asking I see I have another problem. Part of the question is to sketch the graph. How do I sketch the vectors? Is there initial point (0,0) or (1,1) (depending on if x=1 or if x=0) and the terminal point the point that I wrote down they are equal to?
 
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Jomo

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I checked it by hand and and they seem equal to me.

In the future can you please tell us the whole story. Possibly there is an obvious reason that someone would've seen but not if you don't tell us where the matrices came from. We are eager to help you but you sometime need to help us to do that.

I was naive to think that these are two random eterminants that have the same answer.
 

burt

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I checked it by hand and and they seem equal to me.

In the future can you please tell us the whole story. Possibly there is an obvious reason that someone would've seen but not if you don't tell us where the matrices came from. We are eager to help you but you sometime need to help us to do that.

I was naive to think that these are two random eterminants that have the same answer.
Yes, I'm sorry :)
 

Otis

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… Part of the question is to sketch the [curve r(t)] …
Hi Burt. The function r(t) plots as a curve in three dimensions. Are you allowed to use software, to visualize it?

😎
 

burt

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Hi Burt. The function r(t) plots as a curve in three dimensions. Are you allowed to use software, to visualize it?

😎
Yes, and I did that. I just am having a hard time putting in the vectors. This is my attempt. Does it look right?

1585027293846.png
 
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burt

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Hi Burt. The function r(t) plots as a curve in three dimensions. Are you allowed to use software, to visualize it?

😎
1585061323190.png
Something looks off to me, I think because the N(1) is so long - isn't it supposed to be a unit vector?[/QUOTE]
 

Otis

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… N(1) is so long - isn't it supposed to be a unit vector?
Hello. I didn't check your components, but I used them to calculate the magnitude of each vector. All of them are unit vectors. T(0), N(0) and B(0) appear to have the same length. I don't know why T(1), N(1) and B(1) appear to have different lengths. Also, B(0) and B(1) ought to point in the same direction (they're both the same vector). Something is not right.

\(\;\)
 

burt

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Hello. I didn't check your components, but I used them to calculate the magnitude of each vector. All of them are unit vectors. T(0), N(0) and B(0) appear to have the same length. I don't know why T(1), N(1) and B(1) appear to have different lengths. Also, B(0) and B(1) ought to point in the same direction (they're both the same vector). Something is not right.

\(\;\)
Doesn't B(0) start at the point (0,0,0) and B(1) start at (1,2,1)?
 

Otis

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Doesn't B(0) start at the point (0,0,0) and B(1) start at (1,2,1)?
A vector's magnitude and direction are independent of its starting point. We can see that B(0) and B(1) are the same vector, by comparing their components (which are identical).

Consider wind that's blowing in a straight line at constant speed. We could make multiple copies of the vector having that magnitude and direction, and then group them in space, to model the wind. Positioning those vectors at different locations does not alter their direction or length; they are all the same vector (their components are identical).

The names B(0) and B(1) distinguish different positions of two identical vectors.

Did software construct the entire plot?

😎
 

burt

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A vector's magnitude and direction are independent of its starting point. We can see that B(0) and B(1) are the same vector, by comparing their components (which are identical).

Consider wind that's blowing in a straight line at constant speed. We could make multiple copies of the vector having that magnitude and direction, and then group them in space, to model the wind. Positioning those vectors at different locations does not alter their direction or length; they are all the same vector (their components are identical).

The names B(0) and B(1) distinguish different positions of two identical vectors.

Did software construct the entire plot?

😎
No. Do you know how to make software construct the entire plot? Then I could see where I went wrong. I have the Ti-nspire software - can you do it on there?
 

Otis

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No. Do you know how to make software construct the entire plot? Then I could see where I went wrong. I have the Ti-nspire software - can you do it on there?
For the programs I'm familiar with, I would need to look up the commands and syntax. I'll check tomorrow (I'm about to watch "21 Bridges"). If you haven't already, take a look at your software's manual, for plotting curves and vectors in 3d.

\(\;\)
 

burt

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For the programs I'm familiar with, I would need to look up the commands and syntax. I'll check tomorrow (I'm about to watch "21 Bridges"). If you haven't already, take a look at your software's manual, for plotting curves and vectors in 3d.

\(\;\)
Thanks for your help!
 

burt

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For the programs I'm familiar with, I would need to look up the commands and syntax. I'll check tomorrow (I'm about to watch "21 Bridges"). If you haven't already, take a look at your software's manual, for plotting curves and vectors in 3d.

\(\;\)
Geogebra can do it! This is my new graph:
1585174496389.png

How do I graph a vector that is undefined? Meaning, is there any way to graph N(0) and B(0)?
 

Otis

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Geogebra can do it! …

How do I graph a vector that is undefined? … any way to graph N(0) and B(0)?
Hi Burt. Glad to you see you figured out Geogebra (I'm short on time, today).

We cannot graph undefined vectors. (I can't think of much we could do with any undefined object.)

Can you explain why you think N(0) and B(0) are undefined? You wrote their components, in post #5.

\(\;\)
 

burt

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Hi Burt. Glad to you see you figured out Geogebra (I'm short on time, today).

We cannot graph undefined vectors. (I can't think of much we could do with any undefined object.)

Can you explain why you think N(0) and B(0) are undefined? You wrote their components, in post #5.

\(\;\)
First of all because geogebra said so:)
But also because T(0) is <0,0,0> so they should be undefined. (I think, someone else pointed that out).
 

Otis

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First of all because geogebra said so:)
Maybe geogebra didn't actually indicate that the 'zero vector' <0, 0, 0> itself is undefined (because it is). Instead, I'd expect the message meant we can't plot a zero vector as an arrow.

The zero vector is a special case; it's defined as a vector having no direction. Any vector whose components are all zero has no magnitude. Without any length, an arrow can't point toward anything.

The scalar number zero is known as "the additive identity" because the sum of adding 0 to a Real number is the same Real number. That's a useful identity, in algebra. A zero vector is like "the vector-additive identity" because adding a zero vector to another vector doesn't change anything. That's useful in vector arithmetic.

… T(0) is <0,0,0> …
That's not what you wrote in post #5. Each of the six vectors in post #5 have a defined magnitude and direction. None of them are zero vectors.

:confused:
 
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