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Vectors and Matrices

Hi guys, my answer to 1a is:

n^=u⃗×v⃗∣u⃗∣+∣v⃗∣\hat{n} = \frac{\vec{u} \times \vec{v}}{\sqrt{|\vec{u}| + |\vec{v}|}}

d=a⃗.(u⃗×v⃗)∣u⃗∣+∣v⃗∣d = \frac{\vec{a}.(\vec{u} \times \vec{v})}{\sqrt{|\vec{u}| + |\vec{v}|}}

Is that correct?

I'm not sure how to do 1b, would someone be able to explain it? Thanks.
(edited 8 years ago)

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Original post by r3l3ntl3ss
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I'm getting "Invalid Attachment".
Reply 2
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r3l3ntl3ss
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Original post by ghostwalker
I'm getting "Invalid Attachment".


Sorry, TSR is giving me problems and won't allow me to attach to the original post, here it is:

Screen Shot 2015-05-10 at 2.05.52 pm.png
Original post by r3l3ntl3ss
Sorry, TSR is giving me problems and won't allow me to attach to the original post, here it is:

Screen Shot 2015-05-10 at 2.05.52 pm.png


For part b, do you know how to do row reduction?
Reply 4
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r3l3ntl3ss
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Original post by rayquaza17
For part b, do you know how to do row reduction?


using Gaussian Elimination right? yeah I do
Original post by r3l3ntl3ss
Hi guys, my answer to 1a is:

n^=u⃗×v⃗∣u⃗∣+∣v⃗∣\hat{n} = \frac{\vec{u} \times \vec{v}}{\sqrt{|\vec{u}| + |\vec{v}|}}

d=a⃗.(u⃗×v⃗)∣u⃗∣+∣v⃗∣d = \frac{\vec{a}.(\vec{u} \times \vec{v})}{\sqrt{|\vec{u}| + |\vec{v}|}}

Is that correct?



Don't think your denominator is correct there. What's your thinking on that?
Reply 6
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r3l3ntl3ss
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Original post by ghostwalker
Don't think your denominator is correct there. What's your thinking on that?


Doesn't the unit vector n^\hat{n} represent the cross product between the two direction vectors divided by the square root of the magnitude of both of the direction vectors?
Original post by r3l3ntl3ss
Doesn't the unit vector n^\hat{n} represent the cross product between the two direction vectors divided by the square root of the magnitude of both of the direction vectors?


Wondering where you got the bit in bold from.

It's easy to construct a counterexample. If the two vectors are i and j, then the normal is k, but that formula gives k/sqrt(2), which isn't a unit vector.
Original post by r3l3ntl3ss
using Gaussian Elimination right? yeah I do


Actually are you familiar with the rank of a matrix?
You get at least one solution when the rank is 4, so you need to try different values of alpha and beta to get the rank as 4.
Reply 9
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Original post by ghostwalker
Wondering where you got the bit in bold from.

It's easy to construct a counterexample. If the two vectors are i and j, then the normal is k, but that formula gives k/sqrt(2), which isn't a unit vector.


ah I see, I guess I was getting confused.

The denominator of both should be ∣u⃗∣×∣v⃗∣|\vec{u}| \times |\vec{v}|, right?

Original post by rayquaza17
Actually are you familiar with the rank of a matrix?
You get at least one solution when the rank is 4, so you need to try different values of alpha and beta to get the rank as 4.


yup I'm familiar with rank, but shouldn't the rank be a maximum of 3 since the matrix has 3 rows?
Original post by r3l3ntl3ss

yup I'm familiar with rank, but shouldn't the rank be a maximum of 3 since the matrix has 3 rows?


Oops, yeah! I pressed the wrong button. :tongue:
Original post by rayquaza17
Oops, yeah! I pressed the wrong button. :tongue:


Haha fair enough - what I'm unsure of is the question saying at least one solution. If the rank is less than 3, doesn't that mean there's infinite solutions? I'd understand how to do it if it said just one solution.
Original post by r3l3ntl3ss
ah I see, I guess I was getting confused.

The denominator of both should be ∣u⃗∣×∣v⃗∣|\vec{u}| \times |\vec{v}|, right?


Nay.

You want ∣u⃗×v⃗∣|\vec{u} \times \vec{v}|,
Original post by r3l3ntl3ss
Haha fair enough - what I'm unsure of is the question saying at least one solution. If the rank is less than 3, doesn't that mean there's infinite solutions? I'd understand how to do it if it said just one solution.


Erm I think if the rank is 3 then you have one unique solution.
If the rank is less than 3, you have infinitely many solutions.

I think you might have to consider like rank 3, rank 2, etc then for the answer.
Original post by ghostwalker
Nay.

You want ∣u⃗×v⃗∣|\vec{u} \times \vec{v}|,


That's the same isn't it? :tongue:

Original post by rayquaza17
Erm I think if the rank is 3 then you have one unique solution.
If the rank is less than 3, you have infinitely many solutions.

I think you might have to consider like rank 3, rank 2, etc then for the answer.


ah yeah that's what I suspected, thanks!
Original post by r3l3ntl3ss
That's the same isn't it? :tongue:


I hope that's a joke.
Original post by ghostwalker
I hope that's a joke.


hah don't worry it was
Original post by rayquaza17
Erm I think if the rank is 3 then you have one unique solution.
If the rank is less than 3, you have infinitely many solutions.

I think you might have to consider like rank 3, rank 2, etc then for the answer.


I'm still a bit unsure on how to solve this question, I've reduced the matrix to this:

Unparseable latex formula:

\begin{pmatrix} 1 & 0 & {\frac{3}{4}} \\ 0 & 1 & -{\frac{5}{4}} \\ \alpha-1 & 0 & 0\end{vmatrix}

Unparseable latex formula:

\begin{matrix} \frac{\beta}{4} \\ -{\frac{3\beta}{4}} \\ 2+2\beta \end{pmatrix}

(edited 8 years ago)
Original post by r3l3ntl3ss
I'm still a bit unsure on how to solve this question, I've reduced the matrix to this:

Unparseable latex formula:

\begin{pmatrix} 1 & 0 & {\frac{3}{4}} \\ 0 & 1 & -{\frac{5}{4}} \\ \alpha-1 & 0 & 0\end{vmatrix}

Unparseable latex formula:

\begin{matrix} \frac{\beta}{4} \\ -{\frac{3\beta}{4}} \\ 2+2\beta \end{pmatrix}



Maybe I was right the first time and you didn't need to consider ranks, sorry!

You need to consider the bottom row of the matrix. To get solutions, we need to have a-1=2+2b.
If a=1, what is b? If a isn't 1, what's the relation between a and b?
Original post by rayquaza17
Maybe I was right the first time and you didn't need to consider ranks, sorry!

You need to consider the bottom row of the matrix. To get solutions, we need to have a-1=2+2b.
If a=1, what is b? If a isn't 1, what's the relation between a and b?


I'm still really confused, so if α=1\alpha = 1 then β=−1\beta = -1; would you just say that x=2+2βα−1x = \frac{2 + 2\beta}{\alpha - 1} as the relation?

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