First Year Mathematics A (Differentiation & Linear Algebra) (1 Viewer)

InteGrand · Jun 5, 2016

Re: MATH1131 help thread

turntaker said:
Stuck on this for 30mins. Help would be appreciated

$\noindent Let $G$ have position vector $\vec{g}$. You should be able to see by drawing a diagram that for $OEFG$ to be a parallelogram (with the points in that order), we will have: $\overrightarrow{OG}=\overrightarrow{EF}$ (because opposite sides of a parallelogram are equal in length and parallel, so those vectors are equal). This means that $\vec{g} = \vec{f}-\vec{e}$. Since we are given $\vec{e}$ and $\vec{f}$, we can calculate $\vec{g}$.$

$\noindent Once we know $G$, you should be able to find the lengths of the sides and diagonals (draw a diagram to see which vectors' lengths we need to find).$

Flop21 · Jun 6, 2016

Re: MATH1131 help thread

Does the point (-3,3,6) lie on the plane:

x = <2,1,-1> + lamda1<-1,2,4> + lamda2<3,2,1>.

If anyone wants to work that question out for practice and post the working out to it, that'd be awesome

I'm getting to the point where I reduce the matrix (before that I made the plane equation = the point, then rearranged), but it's not working out nicely at all.

leehuan · Jun 6, 2016

Re: MATH1131 help thread

>> A=[-1,3;2,2;4,1]

A =

-1 3
2 2
4 1

>> b=[-3;4;6]-[2;1;-1]

b =

-5
3
7

>> rref([A b])

ans =

1 0 0
0 1 0
0 0 1

>>

My Matlab output suggests that the point does not lie in the plane as there is clearly no solution to this linear system
I'll let someone else do the working

Flop21 · Jun 6, 2016

Re: MATH1131 help thread

Sorry I had a typo on the first point D:

Answers say 'yes' .

Flop21 · Jun 6, 2016

Re: MATH1131 help thread

Find a vector with a magnitude of 10 which is parallel to u. u = <2,1,7>.

*I know the answer, don't know how to get there.

Thanks

InteGrand · Jun 6, 2016

Re: MATH1131 help thread

Flop21 said:
Find a vector with a magnitude of 10 which is parallel to u. u = <2,1,7>.

*I know the answer, don't know how to get there.

Thanks

$\noindent The answer will be a vector $\vec{v}$ that is of the form $\vec{v} = \alpha \vec{u}$, for some real constant $\alpha$ (so that they're parallel). We just need to find an appropriate $\alpha$. Since we want $\left \| \vec{v}\right \|=10$, we want $\left \|\alpha \vec{u}\right \|=10\Rightarrow |\alpha | \left \| \vec{u}\right \|=10$. You should be able to find an $\alpha$ from here, since we know $\vec{u}$ (and can thus find its length).$

Flop21 · Jun 6, 2016

Re: MATH1131 help thread

InteGrand said:
$\noindent The answer will be a vector $\vec{v}$ that is of the form $\vec{v} = \alpha \vec{u}$, for some real constant $\alpha$ (so that they're parallel). We just need to find an appropriate $\alpha$. Since we want $\left \| \vec{v}\right \|=10$, we want $\left \|\alpha \vec{u}\right \|=10\Rightarrow |\alpha | \left \| \vec{u}\right \|=10$. You should be able to find an $\alpha$ from here, since we know $\vec{u}$ (and can thus find its length).$

Sweet got it, thanks!

turntaker · Jun 6, 2016

Re: MATH1131 help thread

InteGrand said:
$\noindent Let $G$ have position vector $\vec{g}$. You should be able to see by drawing a diagram that for $OEFG$ to be a parallelogram (with the points in that order), we will have: $\overrightarrow{OG}=\overrightarrow{EF}$ (because opposite sides of a parallelogram are equal in length and parallel, so those vectors are equal). This means that $\vec{g} = \vec{f}-\vec{e}$. Since we are given $\vec{e}$ and $\vec{f}$, we can calculate $\vec{g}$.$

$\noindent Once we know $G$, you should be able to find the lengths of the sides and diagonals (draw a diagram to see which vectors' lengths we need to find).$

thanks man!

i would rep but i need to spread it more

leehuan · Jun 6, 2016

Re: MATH1131 help thread

Flop21 said:
Sorry I had a typo on the first point D:

Answers say 'yes' .

A=[-1,3;2,2;4,1]

A =

-1 3
2 2
4 1

>> b=[-3;3;6]-[2;1;-1]

b =

-5
2
7

>> rref([A b])

ans =

1 0 2
0 1 -1
0 0 0
With the adjusted point MATLAB suggests yes as well
_________________
The augmented matrix [A | b] is

-1 3 | -5
2 2 | 2
4 1 | 7

Upon R2 <- R2 + 2*R1
And R3 <- R3 + 4*R1

-1 3 | -5
0 8 | -8
0 13 | -3

The rest of the row reduction process is trivial

turntaker · Jun 6, 2016

Re: MATH1131 help thread

$\text{Find \mathit{s} and \mathit{t} if: } (3,s,t)\parallel (2,1,5)$

this is probably really simple: they are vectors

InteGrand · Jun 6, 2016

Re: MATH1131 help thread

turntaker said:
$\text{Find \mathit{s} and \mathit{t} if: } (3,s,t)\parallel (2,1,5)$

this is probably really simple: they are vectors

Since the two vectors are parallel, their components are in the same ratio, so

3:2 = s:1 = t:5.

Since 3:2 = s:1 (i.e. 3/2 = s/1), we have s = 3/2.

Since t:5 = 3:2 (t/5 = 3/2), we have t = 5*3/2 = 15/2.

leehuan · Jun 6, 2016

Re: MATH1131 help thread

turntaker said:
$\text{Find \mathit{s} and \mathit{t} if: } (3,s,t)\parallel (2,1,5)$

this is probably really simple: they are vectors

Haha liking the LaTeX (y)

(Yeah just remember that two parallel vectors are scalar multiples of each other)

turntaker · Jun 6, 2016

Re: MATH1131 help thread

leehuan said:
Haha liking the LaTeX (y)

(Yeah just remember that two parallel vectors are scalar multiples of each other)

OH. SCALAR PRODUCT!

that makes sense lmaoo

InteGrand · Jun 6, 2016

Re: MATH1131 help thread

It's not really scalar "product" (the term "scalar product" commonly refers to the dot product). Better to say scalar "multiple".

leehuan · Jun 6, 2016

Re: MATH1131 help thread

InteGrand said:
It's not really scalar "product" (the term "scalar product" commonly refers to the dot product). Better to say scalar "multiple".

I edited already lol. I realised.

turntaker · Jun 6, 2016

Re: MATH1131 help thread

InteGrand said:
It's not really scalar "product" (the term "scalar product" commonly refers to the dot product). Better to say scalar "multiple".

TRUU

this makes sense too, thanks.

Flop21 · Jun 6, 2016

Re: MATH1131 help thread

How do I know what interval to use for this question (I get how to use it after that)?

x^3 + root(3)x - 6

use intermediate value theorem to prove that f has at least one real root

leehuan · Jun 6, 2016

Re: MATH1131 help thread

Flop21 said:
How do I know what interval to use for this question (I get how to use it after that)?

x^3 + root(3)x - 6

use intermediate value theorem to prove that f has at least one real root

If you're just showing that it has a zero you can pick any arbitrary interval. We just care that you know to pick one that conveniently has one side negative and the other positive.

Oh and of course it's continuous on that interval as well.

Flop21 · Jun 6, 2016

Re: MATH1131 help thread

leehuan said:
If you're just showing that it has a zero you can pick any arbitrary interval. We just care that you know to pick one that conveniently has one side negative and the other positive.

Oh and of course it's continuous on that interval as well.

Yes that's what I'm asking how do I do that?

leehuan · Jun 6, 2016

Re: MATH1131 help thread

Flop21 said:
Yes that's what I'm asking how do I do that?

That's not even hard. Just put random numbers (for x) into your calculator and get something >0 and something <0 ?

E.g. prove your function has a root.
I could pick 0 and 10

Or I could pick -1000 and 1000

First Year Mathematics A (Differentiation & Linear Algebra) (1 Viewer)

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