Need help, URGENT maths question: (1 Viewer)

leehuan

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That's alright :) and thanks. I guess the main trick is knowing where to get your determinant from in this question. Btw, do you still have your working for that particular question (proving all those properties of matrices)? Just wanna compare my working to yours...
It's not on me right now though
 

leehuan

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^Oh yeah that was the faster way that my tutor showed me

Yeah @1008 make a note of the fact that wherever 0s can be brought out by using always go for it
 

1008

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^Oh yeah that was the faster way that my tutor showed me

Yeah @1008 make a note of the fact that wherever 0s can be brought out by using always go for it
Thanks will keep that in mind
 

leehuan

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A line is parallel to the plane if the direction vector of the line (here it is [2,3,-1]T) can be written as a linear combination of the span of the plane

Which is lambda1(1,...
 

1008

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A line is parallel to the plane if the direction vector of the line (here it is [2,3,-1]T) can be written as a linear combination of the span of the plane

Which is lambda1(1,...
Yeah, I guess I could just say lambda1 = 2, lambda2 =1, but would both suffice by themselves?
 

leehuan

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Which by inspection lambda1 = 2 and lambda2 = 1 solve it
 

1008

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Thanks. Never thought of proving parallelism by proving they have a common normal vector! It would indeed be quicker to do this when figuring out values of the lambdas isn't so easy!

Do you have any textbook/resources in mind for vectors and matrices?

And how would you do this?


Any ideas on (b)?
 

1008

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What's the easiest way to convert the following equations of planes:





into parametric and point normal forms?

EDIT: for some reason, I can't see my LaTex code. Can you guys see it?
 
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InteGrand

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What's the easiest way to convert the following equations of planes:





into parametric and point normal forms?

EDIT: for some reason, I can't see my LaTex code. Can you guys see it?
For the first one (it's a line, not a plane), let (x-1)/3 = (z+1)/6 = t, where t will be a real parameter.

So x = 3t + 1 and z = 6t - 1. Also, y = 4. So on the curve,

x = (x,y,z) = (3t+1,4,6t-1) = (1,4,-1) + t(3,0,6) (parametric equation of the line).

For the second one (which is indeed a plane. It's a line that lies in the vertical plane y = 4.), set x = t = 4y, so y = t/4 and x = t. Also, z = s (another free parameter).

So on this surface, x = (x,y,z) = (t,t/4,s) = t(1,1/4,0) + s(0,0,1). This is the parametric vector form of the plane.

The Cartesian Equation of this plane is essentially given to us: it's just x - 4y = 0, z any real number.

And yes, I can see your LaTeX.
 
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1008

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For the first one (it's a line, not a plane), let (x-1)/3 = (z+1)/6 = t, where t will be a real parameter.

So x = 3t + 1 and z = 6t - 1. Also, y = 4. So on the curve,

x = (x,y,z) = (3t+1,4,6t-1) = (1,4,-1) + t(3,0,6) (parametric equation of the line).

For the second one (which is indeed a plane. It's a line that lies in the vertical plane y = 4.), set x = t = 4y, so y = t/4 and x = t. Also, z = s (another free parameter).

So on this surface, x = (x,y,z) = (t,t/4,s) = t(1,1/4,0) + s(0,0,1). This is the parametric vector form of the plane.

The Cartesian Equation of this plane is essentially given to us: it's just x - 4y = 0, z any real number.

And yes, I can see your LaTeX.
Thanks! Maybe internet's too slow at uni. It took a while to download and then the LaTex appeared slowly .

Also, is this a valid way to do this question:
Prove the associative law: if AB and BC exist, the A(BC) = (AB)C for matrices

 
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InteGrand

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Thanks! Maybe internet's too slow at uni. It took a while to download and then the LaTex appeared slowly .

Also, is this a valid way to do this question:
Prove the associative law: if AB and BC exist, the A(BC) = (AB)C for matrices

No not quite, because the kj entry of of BC isn't bc (after all, what even is bc?). Note the kj entry of BC is itself a sum. leehuan essentially proved it on one of his threads I think. It relies on being able to switch summation orders as I said a bit earlier on this thread.
 
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1008

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No not quite, because the kj entry of of BC isn't bc (after all, what even is bc?). Note the kj entry of BC is itself a sum. leehuan essentially proved it on one of his threads I think. It relies on being able to switch summation orders as I said a bit earlier on this thread.
Yeah, I'll have another look at that soon. Meanwhile, I've got a calculus q for a change:



Would the answer be zero, because q approaches 1 and n approaches infinity? (I think that would mean the width of every rectangle in the Riemann sum would be zero, so total area as n approaches infinity would be zero)
 

InteGrand

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Yeah, I'll have another look at that soon. Meanwhile, I've got a calculus q for a change:



Would the answer be zero, because q approaches 1 and n approaches infinity? (I think that would mean the width of every rectangle in the Riemann sum would be zero, so total area as n approaches infinity would be zero)
In Riemann sums, the width of rectangles do approach 0, but that doesn't necessarily mean the overall integral is 0, because the number of rectangles used is tending to infinity.

Since it says the function is x^j, where j is a positive integer, the answer should be [2^(j+1) – 1]/(j+1), i.e. what you'd get if you integrated 'normally' (using the Fundamental Theorem of Calculus). For this question, you need to show this by considering the given Riemann sum.
 

1008

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In Riemann sums, the width of rectangles do approach 0, but that doesn't necessarily mean the overall integral is 0, because the number of rectangles used is tending to infinity.

Since it says the function is x^j, where j is a positive integer, the answer should be [2^(j+1) – 1]/(j+1), i.e. what you'd get if you integrated 'normally' (using the Fundamental Theorem of Calculus). For this question, you need to show this by considering the given Riemann sum.
How would you do such questions? Because usually, the width of your partitions is the same, so you can factor that out, and all you're usually left with is a summation which can be easily figured out using standard arithmetic progression formulas, or formulas of summing the first n squares, cubes etc...That doesn't seem to be the case here...
 

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