Cross product (2 Viewers)

leehuan · Apr 28, 2016

Question is easy but I cannot find my mistake.

$$Find the shortest distance between the line through the points $(1,3,1), \, (1,5,-1)$ and the line through the points $(0,2,1), \, (1,2,-3)$

Not sure if this is the standard method but this is according to the method I was taught.

$\lambda, \mu \in \mathbb R\\ $The lines have equations:$ \\ \textbf{x}=\left[\begin{matrix}1\\3\\1\end{matrix}\right]+\lambda \left[\begin{matrix}0\\2\\-2\end{matrix}\right]\\\textbf{x}=\left[\begin{matrix}0\\2\\1\end{matrix}\right]+\mu\left[\begin{matrix}1\\0\\-4\end{matrix}\right]$

$\\$Vector parallel to the direction vectors is $\textbf{n}=\left[\begin{matrix}-8\\-2\\-2\end{matrix}\right]$

$\\$Shortest distance between the lines is $\\ d=\left|\text{proj}_\textbf{n} \left[\begin{matrix}-1\\-1\\0\end{matrix}\right]\right|=\frac{10}{6\sqrt{2}}$

Correct answer is 1/√2

Flop21 · Apr 28, 2016

leehuan said:
Question is easy but I cannot find my mistake.

$$Find the shortest distance between the line through the points $(1,3,1), \, (1,5,1)$ and the line through the points $(0,2,1), \, (1,2,-3)$

Not sure if this is the standard method but this is according to the method I was taught.

$\lambda, \mu \in \mathbb R\\ $The lines have equations:$ \\ \textbf{x}=\left[\begin{matrix}1\\3\\1\end{matrix}\right]+\lambda \left[\begin{matrix}0\\2\\-2\end{matrix}\right]\\\textbf{x}=\left[\begin{matrix}0\\2\\1\end{matrix}\right]+\mu\left[\begin{matrix}1\\0\\-4\end{matrix}\right]$

$\\$Vector parallel to the direction vectors is $\textbf{n}=\left[\begin{matrix}-8\\-2\\-2\end{matrix}\right]$

$\\$Shortest distance between the lines is $\\ d=\left|\text{proj}_\textbf{n} \left[\begin{matrix}-1\\-1\\0\end{matrix}\right]\right|=\frac{10}{6\sqrt{2}}$

Correct answer is 1/√2

First equation, you finding AB (the lambda part), looks like you have 1-1 = -2?

leehuan · Apr 28, 2016

Flop21 said:
First equation, you finding AB (the lambda part), looks like you have 1-1 = -2?

Oh my bad that was a typo. Fixing it now

leehuan · May 3, 2016

Need help with c)

$Points $A,B,C$ and $D$ have coordinate vectors $\left( \begin{matrix} 1 \\ 0 \\ 2 \end{matrix} \right) ,\left( \begin{matrix} -1 \\ 2 \\ 3 \end{matrix} \right) \left( \begin{matrix} 0 \\ 1 \\ 1 \end{matrix} \right)$ and $ \left( \begin{matrix} 2 \\ 0 \\ -1 \end{matrix} \right) $respectively.$
$\\$a) Find a parametric vector equation of the line through $A $ and $B$ and the line through $ C $and $ D\\\Rightarrow $Answers: \textbf{x}=\left( \begin{matrix} 1 \\ 0 \\ 2 \end{matrix} \right) +\left( \begin{matrix} -2 \\ 2 \\ 1 \end{matrix} \right); \quad \textbf{x}=\left( \begin{matrix} 0 \\ 1 \\ 1 \end{matrix} \right) ,\left( \begin{matrix} 2 \\ -1 \\ -2 \end{matrix} \right)$

$\\$b) Find the shortest distance between the lines $AB$ and $CD\\ \Rightarrow $Answer: $d_{min}=\left|{\text{proj}}_{\left( \begin{matrix} -2 \\ 2 \\ 1 \end{matrix} \right) \times \left( \begin{matrix} 2 \\ -1 \\ -2 \end{matrix} \right)} \left( \begin{matrix} 0 \\ 1 \\ 1 \end{matrix} \right) - \left( \begin{matrix} 1 \\ 0 \\ 2 \end{matrix} \right) \right| = \frac{3}{\sqrt{17}}$

$\\$c) Find the point $P$ on $AB$ and point $Q$ on $CD$ such that $PQ$ is the shortest distance between the lines $AB$ and $CD$

InteGrand · May 3, 2016

leehuan said:
Need help with c)

$Points $A,B,C$ and $D$ have coordinate vectors $\left( \begin{matrix} 1 \\ 0 \\ 2 \end{matrix} \right) ,\left( \begin{matrix} -1 \\ 2 \\ 3 \end{matrix} \right) \left( \begin{matrix} 0 \\ 1 \\ 1 \end{matrix} \right)$ and $ \left( \begin{matrix} 2 \\ 0 \\ -1 \end{matrix} \right) $respectively.$
$\\$a) Find a parametric vector equation of the line through $A $ and $B$ and the line through $ C $and $ D\\\Rightarrow $Answers: \textbf{x}=\left( \begin{matrix} 1 \\ 0 \\ 2 \end{matrix} \right) +\left( \begin{matrix} -2 \\ 2 \\ 1 \end{matrix} \right); \quad \textbf{x}=\left( \begin{matrix} 0 \\ 1 \\ 1 \end{matrix} \right) ,\left( \begin{matrix} 2 \\ -1 \\ -2 \end{matrix} \right)$

$\\$b) Find the shortest distance between the lines $AB$ and $CD\\ \Rightarrow $Answer: $d_{min}=\left|{\text{proj}}_{\left( \begin{matrix} -2 \\ 2 \\ 1 \end{matrix} \right) \times \left( \begin{matrix} 2 \\ -1 \\ -2 \end{matrix} \right)} \left( \begin{matrix} 0 \\ 1 \\ 1 \end{matrix} \right) - \left( \begin{matrix} 1 \\ 0 \\ 2 \end{matrix} \right) \right| = \frac{3}{\sqrt{17}}$

$\\$c) Find the point $P$ on $AB$ and point $Q$ on $CD$ such that $PQ$ is the shortest distance between the lines $AB$ and $CD$

A method for finding the points on two skew lines that are nearest each other is described here: https://en.wikipedia.org/wiki/Skew_lines#Nearest_Points .

leehuan · May 3, 2016

Say I want equations for the plane through (1,2,-2) perpendicular to (-1,1,2)^T

Cartesian and point-normal forms are easy
-x1+x2+x3 = dot product

(-1,1,2)^T.(x-(1,2,-2)^T)=0

All three forms are pretty easy. But what's the fastest way to go from any one of them to a parametric vector equation? (Just a simple answer will do, no need to finish the question)

Two methods I thought about:
1. Let x2 = lambda and x3 = mu
2. Cross (-1,1,2) with any arbitrary non-zero vector to get a right hand system and then take the two that aren't (-1,1,2)

InteGrand · May 3, 2016

leehuan said:
Say I want equations for the plane through (1,2,-2) perpendicular to (-1,1,2)^T

Cartesian and point-normal forms are easy
-x1+x2+x3 = dot product

(-1,1,2)^T.(x-(1,2,-2)^T)=0

All three forms are pretty easy. But what's the fastest way to go from any one of them to a parametric vector equation? (Just a simple answer will do, no need to finish the question)

Two methods I thought about:
1. Let x2 = lambda and x3 = mu
2. Cross (-1,1,2) with any arbitrary non-zero vector to get a right hand system and then take the two that aren't (-1,1,2)

If you're in Cartesian form to start with, your first method is good. If you're in point-normal form to start with, we can convert it to Cartesian form and then use your first method.

The second method won't work unless the vector you use to cross with the normal is one that lies in the plane (as in, is "submerged" in the plane), so you'd need to choose the vector carefully (which can be done by taking another point on the plane, i.e. other than the given one, and then connecting these and taking this connecting vector as the vector). This second method would be useful if we wanted to obtain an orthonormal or orthogonal basis for the plane (well the subspace that is the corresponding plane through the origin).

You can try out both methods to see which you find easier and faster.

leehuan · May 4, 2016

Can't draw the link again

$$a) Let $\textbf{a}$ and $\textbf{v}$ be two non-zero vectors in $\mathbb R^3$. Show how to write $\textbf{v}$ as $\textbf{c}+\textbf{d}$ where $\textbf{c}$ is parallel to $\textbf{a}$ and $\textbf{d}$ is perpendicular to $\textbf{a}.$

$\\$Answer: I have no idea how the question is properly addressed but I can see what they did$\\ \textbf{c}=$proj$_{\textbf{a}}\textbf{v}, \quad \textbf{d}=\textbf{v}-\textbf{c}$

$$b) Consider the plane$\\ \Pi=\left\{\left(\begin{matrix}x\\y\\z \end{matrix}\right):\, \left(\begin{matrix}x\\y\\z\end{matrix}\right)= \left(\begin{matrix}2\\1\\0\end{matrix}\right) + \lambda_1 \left(\begin{matrix}1\\2\\1\end{matrix}\right) + \lambda_2 \left(\begin{matrix}0\\1\\-1\end{matrix}\right), \, \lambda_1 , \lambda_2 \in \mathbb R \right\}\\$and the vector $\left(\begin{matrix}1\\1\\1\end{matrix}\right)$

$\\ $By using a) (or otherwise), express $\textbf{v}$ as $\textbf{c}+\textbf{d}$ where $\textbf{d}$ is parallel to $\Pi$ and $\textbf{c}$ is perpendicular to $\Pi$.$

$\\$Answers $\textbf{c}=\left(\begin{matrix}\frac{3}{11}\\-\frac{1}{11}\\-\frac{1}{11}\end{matrix}\right), \,, \textbf{d}=\left(\begin{matrix}\frac{8}{11}\\\frac{12}{11}\\\frac{12}{11}\end{matrix}\right)$

No idea where those vectors came from using part a

seanieg89 · May 4, 2016

Being orthogonal to the plane is the same as being parallel to the normal of the plane (our ambient space is three dimensional, so normals to a plane are uniquely defined up to scalar multiplication), so just apply a) using the normal to the plane.

leehuan · May 4, 2016

This could just be a dot product question but I'm too tired to understand the question right now so any advice please

$$Given a set of vectors $\left\{\textbf{v}_1,\dots,\textbf{v}_n\right\}$ such that none of the vectors forms a linear combination of ones earlier in the list, prove the following procedure constructs an orthornormal set $\left\{\textbf{u}_1,\dots,\textbf{u}_n\right\}$ with the property that for each $k, 1\le k \le n$

$$span$\left\{\textbf{u}_1,\dots,\textbf{u}_n\right\}=$span$\left\{\textbf{v}_1,\dots,\textbf{v}_n \right\}=\sigma_k.$

Procedure:

$$Set $\textbf{u}_1=\frac{\textbf{v}_1}{|\textbf{v}_1|}\\ $For $k=2, \dots, n $ in turn set $\\ \textbf{w}_k = \textbf{v}_k - {\text{proj}}_{{\sigma}_{k-1}}\textbf{v}_k\\ \textbf{u}_k=\frac{\textbf{w}_1}{|\textbf{w}_1|}$

InteGrand · May 4, 2016

leehuan said:
This could just be a dot product question but I'm too tired to understand the question right now so any advice please

$$Given a set of vectors $\left\{\textbf{v}_1,\dots,\textbf{v}_n\right\}$ such that none of the vectors forms a linear combination of ones earlier in the list, prove the following procedure constructs an orthornormal set $\left\{\textbf{u}_1,\dots,\textbf{u}_n\right\}$ with the property that for each $k, 1\le k \le n$

$$span$\left\{\textbf{u}_1,\dots,\textbf{u}_n\right\}=$span$\left\{\textbf{v}_1,\dots,\textbf{v}_n \right\}=\sigma_k.$

Procedure:

$$Set $\textbf{u}_1=\frac{\textbf{v}_1}{|\textbf{v}_1|}\\ $For $k=2, \dots, n $ in turn set $\\ \textbf{w}_k = \textbf{v}_k - {\text{proj}}_{{\sigma}_{k-1}}\textbf{v}_k\\ \textbf{u}_k=\frac{\textbf{w}_1}{|\textbf{w}_1|}$

This is known as Gram-Schmidt orthogonalisation (or orthonormalisation if we normalise the vectors too). Basically the idea is that given a set of vectors, from this we create a new set that has the same span and the vectors are orthogonal (so we can obtain an orthogonal basis for a vector space in this way given a basis). You can read about it here and see an animation: https://en.wikipedia.org/wiki/Gram–Schmidt_process#The_Gram.E2.80.93Schmidt_process

The proof will typically be by induction. So you may like to try induction and showing it for small cases to get a feel for it.

Cross product (2 Viewers)

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