Weird? vectors question (1 Viewer)

leehuan · May 3, 2016

$$In two dimensions, let $\textbf{a}=\left[\begin{matrix}a_1\\a_2\end{matrix}\right]$ and $\textbf{b}=\left[\begin{matrix}b_1\\b_2\end{matrix}\right]$ and consider the parallelogram spanned by $\textbf{a}$ and $\textbf{b}$

$\textbf{Show}$ that a parametric vector form for the parallelogram is $\\\textbf{x}=\lambda_1\textbf{a}+\lambda_2\textbf{b}$ for $0\le\lambda_1\le 1, 0\le\lambda_2\le 1$

I'm sure it's easy but I'm a bit mind blanked as to how to prove it.

Drongoski · May 3, 2016

leehuan said:
$$In two dimensions, let $\textbf{a}=\left[\begin{matrix}a_1\\a_2\end{matrix}\right]$ and $\textbf{b}=\left[\begin{matrix}b_1\\b_2\end{matrix}\right]$ and consider the parallelogram spanned by $\textbf{a}$ and $\textbf{b}$

$\textbf{Show}$ that a parametric vector form for the parallelogram is $\\\textbf{x}=\lambda_1\textbf{a}+\lambda_2\textbf{b}$ for $0\le\lambda_1\le 1, 0\le\lambda_2\le 1$

I'm sure it's easy but I'm a bit mind blanked as to how to prove it.

Consider any point P on the parallelogram. Show that this point is on the parallelogram is expressible as that linear combination of a and b and any point outside the //gram is not. a and b spans the whole plane, defined by these 2 vectors, providing they are linearly independent (i.e. not parallel and nonzero) and the restriction on the linear combination gives the set of all points on the parallelogram.

You will now ask: but how?

InteGrand · May 3, 2016

leehuan said:
$$In two dimensions, let $\textbf{a}=\left[\begin{matrix}a_1\\a_2\end{matrix}\right]$ and $\textbf{b}=\left[\begin{matrix}b_1\\b_2\end{matrix}\right]$ and consider the parallelogram spanned by $\textbf{a}$ and $\textbf{b}$

$\textbf{Show}$ that a parametric vector form for the parallelogram is $\\\textbf{x}=\lambda_1\textbf{a}+\lambda_2\textbf{b}$ for $0\le\lambda_1\le 1, 0\le\lambda_2\le 1$

I'm sure it's easy but I'm a bit mind blanked as to how to prove it.

$\noindent Hint: Draw a picture. First consider it as $\lambda_2$ is held fixed in $[0,1]$ (say even $\lambda_2 = 0$) and $\lambda_1$ varies from 0 to 1: what set of points is traced out? Then consider the collection of all these points for each $\lambda_2$ in $[0,1]$. What is this collection?$

leehuan · May 3, 2016

I suppose yeah it's easier to explain than to "show". I need to stop being pedantic about my definitions.

Ok this question has me stuck. Because I actually don't know how to neatly to the derivation for the volume of the parallelepiped.

$\textbf{x}=\lambda_1\textbf{a}+\lambda_2\textbf{b}+\lambda_3\textbf{c}$ for similar conditions and $0\le \lambda_3 \le 1$

$\text{Show that the colume of the parallelipeped is equal to }\left|\det{A}\right|,$ where $A$ is the matrix with rows $\textbf{a}.\textbf{b}.\textbf{c}$

I kinda cheated for the 2D version by just saying well |a||b|sin(theta)=|axb|=det(A) but I don't know how to replicate the proof into R³

InteGrand · May 3, 2016

leehuan said:
I suppose yeah it's easier to explain than to "show". I need to stop being pedantic about my definitions.

Ok this question has me stuck. Because I actually don't know how to neatly to the derivation for the volume of the parallelepiped.

$\textbf{x}=\lambda_1\textbf{a}+\lambda_2\textbf{b}+\lambda_3\textbf{c}$ for similar conditions and $0\le \lambda_3 \le 1$

$\text{Show that the colume of the parallelipeped is equal to }\left|\det{A}\right|,$ where $A$ is the matrix with rows $\textbf{a}.\textbf{b}.\textbf{c}$

I kinda cheated for the 2D version by just saying well |a||b|sin(theta)=|axb|=det(A) but I don't know how to replicate the proof into R³

$\noindent If you can show that $\det \left(A\right) = \vec{a} \cdot \left(\vec{b}\times \vec{c}\right)$ (dot and cross products here), then you can use the fact that $\left \| \vec{b}\times \vec{c}\right \|$ is the area of the base parallelogram, and $\vec{v}\cdot \vec{w} = \left \|\vec{v} \right \|\left \|\vec{w} \right \|\cos \theta$ to see that this is the volume of the parallelipiped (up to sign, which is why we take the absolute value in the end). Note that the volume of the parallelipiped is the base area times the perpendicular height. Refer to this page and the picture there:$

http://mathinsight.org/scalar_triple_product.

leehuan · May 3, 2016

Dumb question that I don't feel any value in making a new thread for:

How do I prove two planes are (or are not) parallel?

Drongoski · May 3, 2016

2 planes are parallel if their normal vectors are parallel.

e.g. 2x-y+3z = 5 and -4x +2y - 6z = 1 are parallel, whereas

2x - y + 3z = 5 and x + 2y - z = 11 are not.

InteGrand · May 3, 2016

leehuan said:
Dumb question that I don't feel any value in making a new thread for:

How do I prove two planes are (or are not) parallel?

In R³, it can quickly be done by comparison of normals.

In higher dimensions (also applies to R³), the planes will be parallel iff the span of the direction vectors of plane 1 is equal to the span of the direction vectors of plane 2. (Note that in higher-than-3 dimensions, planes don't have a single normal, as they have more than one orthogonal direction. This is why we can't just use a normal vector in higher dimensions. In R³ though, planes have a single orthogonal direction. If you want an analogy, it's like how in R², a line will have a unique orthogonal direction, but in R³, lines will actually have more than one orthogonal direction.)

Weird? vectors question (1 Viewer)

leehuan

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