Need help, URGENT maths question: (5 Viewers)

leehuan · May 25, 2016

1008 said:
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

InteGrand · May 25, 2016

leehuan · May 25, 2016

^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 $R_i \leftarrow R_i + \alpha R_j$ always go for it

1008 · May 25, 2016

leehuan said:
It's not on me right now though

Thanks alright then.

1008 · May 25, 2016

leehuan said:
^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 $R_i \leftarrow R_i + \alpha R_j$ always go for it

Thanks will keep that in mind

1008 · May 25, 2016

InteGrand said:
$\noindent For the first one, since they're collinear, remember that $\overrightarrow{AB}$ is a multiple of $\overrightarrow{AC}$, so this imposes some conditions on the coordinates.$

$\noindent We don't need to expand the determinant either. We can just row-reduce a tiny bit. Doing $R_2 \to R_2 - R_2$ and $R_3 \to R_3 - R_1$, the determinant of that matrix is equal to $\det{\left( \begin{matrix} { x }_{ 1 } & { y }_{ 1 } & 1 \\ { x }_{ 2 } -x_1& { y }_{ 2 } -y_1& 0 \\ { x }_{ 3 }-x_1 & { y }_{ 3 }-y_1 & 0 \end{matrix} \right) }$. Now, the second row is a multiple of the third row, because $\overrightarrow{AB}$ is a multiple of $\overrightarrow{AC}$, which tells us that $x_2 - x_1 = \lambda \left(x_3 - x_1\right)$ for some real number $\lambda$, and similarly for the $y$-values with the same $\lambda$. The 0's are also going to satisfy $0 = \lambda \cdot 0$ of course. So row 3 is a multiple of row 2, so the determinant is 0.$

Thanks, any ideas on b)?

1008 · May 25, 2016

1008 said:
Thanks, any ideas on b)?

Is this right?

leehuan · May 25, 2016

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 · May 25, 2016

leehuan said:
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 · May 25, 2016

$Basically this system of equations should have a unique solution$

$\lambda_1 = 2\\ \lambda_1+\lambda_2 = 3\\ -\lambda_2=-1$

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

InteGrand · May 25, 2016

1008 said:
Is this right?

$\noindent There's an easier approach. The given line has direction vector $\bold{v}=\left(2,3,-1\right)$. By inspection, a normal to the plane is $\bold{n}=\left(1,-1,-1\right)$, because it gives 0 when dotted with each vector for the plane's span, so it is orthogonal to a basis for the plane's span, and is hence orthogonal to the plane. (If you couldn't spot this by inspection, another way to find a normal is via the cross product.)$

$\noindent Now, $\bold{n}\cdot \bold{v} = 2 -3+1 = 0\Rightarrow \bold{n} \perp \bold{v} $. Since the line's direction vector is orthogonal to the normal of the plane, the line is parallel to the plane.$

$\noindent (Actually, it's even easier to inspect that the direction vector to the line is equal to $2\bold{v}_1+\bold{v}_2$, where $\bold{v}_1$ and $\bold{v}_2$ are the vectors that span the plane. This shows that the line is parallel to the plane, as leehuan explained. The reason that it is easy to inspect the linear combination here is due to the presence of 0's in some components of the vectors. For instance, this tells us immediately that if the direction vector of the line can be written as $\alpha_1 \bold{v}_1 + \alpha_2 \bold{v}_2$ for some scalars $\alpha_1,\alpha_2$, then $\alpha_1$ has to be $2$, from the first components. Then knowing $\alpha_1=2$, we can easily find what $\alpha_2$ has to be from the second component. Then we just need to check that these values actually work (i.e. that the third components equal too), i.e. that the direction vector is in indeed in the span.)$

1008 · May 25, 2016

InteGrand said:
$\noindent There's an easier approach. The given line has direction vector $\bold{v}=\left(2,3,-1\right)$. By inspection, a normal to the plane is $\bold{n}=\left(1,-1,-1\right)$, because it gives 0 when dotted with each vector for the plane's span, so it is orthogonal to a basis for the plane's span, and is hence orthogonal to the plane. (If you couldn't spot this by inspection, another way to find a normal is via the cross product.)$

$\noindent Now, $\bold{n}\cdot \bold{v} = 2 -3+1 = 0\Rightarrow \bold{n} \perp \bold{v} $. Since the line's direction vector is orthogonal to the normal of the plane, the line is parallel to the plane.$

$\noindent (Actually, it's even easier to inspect that the direction vector to the line is equal to $2\bold{v}_1+\bold{v}_2$, where $\bold{v}_1$ and $\bold{v}_2$ are the vectors that span the plane. This shows that the line is parallel to the plane, as leehuan explained. The reason that it is easy to inspect the linear combination here is due to the presence of 0's in some components of the vectors. For instance, this tells us immediately that if the direction vector of the line can be written as $\alpha_1 \bold{v}_1 + \alpha_2 \bold{v}_2$ for some scalars $\alpha_1,\alpha_2$, then $\alpha_1$ has to be $2$, from the first components. Then knowing $\alpha_1=2$, we can easily find what $\alpha_2$ has to be from the second component. Then we just need to check that these values actually work (i.e. that the third components equal too), i.e. that the direction vector is in indeed in the span.)$

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!

1008 said:
Do you have any textbook/resources in mind for vectors and matrices?

And how would you do this?

Any ideas on (b)?

InteGrand · May 25, 2016

1008 said:
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!

Any ideas on (b)?

For (b), you may want to check out the page here: http://mathforum.org/library/drmath/view/55063.html .

1008 · May 26, 2016

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

$\\\frac{x-1}{3}=\frac{z+1}{6},y = 4$

$x = 4y$

into parametric and point normal forms?

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

InteGrand · May 26, 2016

1008 said:
What's the easiest way to convert the following equations of planes:

$\\\frac{x-1}{3}=\frac{z+1}{6},y = 4$

$x = 4y$

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.

1008 · May 26, 2016

InteGrand said:
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

InteGrand · May 26, 2016

1008 said:
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.

1008 · May 28, 2016

InteGrand said:
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:

$\\\text{Consider the partition }\mathcal{P}_{n}\text{ of [1,2], given by }\mathcal{P}_{n} = \{q^{0},q^{1},q^{2},...,q^{n},\}\text{ where } q^{n} = 2 \text{. Notice that (i) the divisions are not of equal width and (ii) }1<q<2\text{ and }q \rightarrow 1\text{ as }n \rightarrow \infty .)\text{If }f(x) = x^{j}\text{ for some positive integer }j,\text{ then evaluate the integral }\\ \intop\nolimits_{1}^{2}f(x) dx \\\text{ by calculating the limit } \lim_{n\to \infty} \underline{S}_{\mathcal{P}_{n}}(f) \text{ of the corresponding lower Riemann sums}$

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 · May 28, 2016

1008 said:
Yeah, I'll have another look at that soon. Meanwhile, I've got a calculus q for a change:

$\\\text{Consider the partition }\mathcal{P}_{n}\text{ of [1,2], given by }\mathcal{P}_{n} = \{q^{0},q^{1},q^{2},...,q^{n},\}\text{ where } q^{n} = 2 \text{. Notice that (i) the divisions are not of equal width and (ii) }1<q<2\text{ and }q \rightarrow 1\text{ as }n \rightarrow \infty .)\text{If }f(x) = x^{j}\text{ for some positive integer }j,\text{ then evaluate the integral }\\ \intop\nolimits_{1}^{2}f(x) dx \\\text{ by calculating the limit } \lim_{n\to \infty} \underline{S}_{\mathcal{P}_{n}}(f) \text{ of the corresponding lower Riemann sums}$

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 · May 28, 2016

InteGrand said:
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...

Need help, URGENT maths question: (5 Viewers)

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