Dot product (1 Viewer)

leehuan · Apr 26, 2016

$$Suppose I have these vectors:$\\ \overrightarrow { AB } =\left[ \begin{matrix} 2 \\ 2 \\ -1 \end{matrix} \right] \\ \overrightarrow { BC } =\left[ \begin{matrix} -1 \\ -1 \\ 5 \end{matrix} \right] \\$

Ignore CA for now

$I'm required to find the cosine of the internal angles of $\triangle ABC$. Suppose I want $\angle ABC$ first. Now, if I take $\overrightarrow { AB } \cdot \overrightarrow { BC }$ my expression is negative. Quite reasonably enough if I take $\overrightarrow { AB } \cdot\overrightarrow { CB }$ it comes out positive$

$How do I know which one is right?$

dan964 · Apr 26, 2016

leehuan said:
$$Suppose I have these vectors:$\\ \overrightarrow { AB } =\left[ \begin{matrix} 2 \\ 2 \\ -1 \end{matrix} \right] \\ \overrightarrow { BC } =\left[ \begin{matrix} -1 \\ -1 \\ 5 \end{matrix} \right] \\$

Ignore CA for now

$I'm required to find the cosine of the internal angles of $\triangle ABC$. Suppose I want $\angle ABC$ first. Now, if I take $\overrightarrow { AB } \cdot \overrightarrow { BC }$ my expression is negative. Quite reasonably enough if I take $\overrightarrow { AB } \cdot\overrightarrow { CB }$ it comes out positive$

$How do I know which one is right?$

The dot-product should be negative definitely for the first one. (-2-2-5 = -9) What do you get for CB in terms of the vector co-ordinates.

leehuan · Apr 26, 2016

dan964 said:
The dot-product should be negative definitely for the first one. (-2-2-5 = -9) What do you get for CB in terms of the vector co-ordinates.

Yes that's what I got for AB dot BC as well

Given that vec(CB) = -vec(BC), AB dot CB should just be -AB dot BC

So how do I know which one is the desired one? The answers say it's the positive and I can clearly see why but I don't know why I'm supposed to choose vec(CB) and not vec(BC)

InteGrand · Apr 26, 2016

leehuan said:
$$Suppose I have these vectors:$\\ \overrightarrow { AB } =\left[ \begin{matrix} 2 \\ 2 \\ -1 \end{matrix} \right] \\ \overrightarrow { BC } =\left[ \begin{matrix} -1 \\ -1 \\ 5 \end{matrix} \right] \\$

Ignore CA for now

$I'm required to find the cosine of the internal angles of $\triangle ABC$. Suppose I want $\angle ABC$ first. Now, if I take $\overrightarrow { AB } \cdot \overrightarrow { BC }$ my expression is negative. Quite reasonably enough if I take $\overrightarrow { AB } \cdot\overrightarrow { CB }$ it comes out positive$

$How do I know which one is right?$

$\noindent When finding the angle between two vectors, we line them up so that their tails are starting at the same place (or it is equivalent to have their heads start at the same place). So you should find the angle between $\overrightarrow{BA}$ and $\overrightarrow{BC}$ (draw a diagram of the triangle and you'll easily see these vectors have their tails starting at $B$).$

leehuan · Apr 26, 2016

InteGrand said:
$\noindent When finding the angle between two vectors, we line them up so that their tails are starting at the same place (or it is equivalent to have their heads start at the same place). So you should find the angle between $\overrightarrow{BA}$ and $\overrightarrow{BC}$ (draw a diagram of the triangle and you'll easily see these vectors have their tails starting at $B$).$

Right, is this a rule of thumb that I should remember?

dan964 · Apr 26, 2016

leehuan said:
Yes that's what I got for AB dot BC as well

Given that vec(CB) = -vec(BC), AB dot CB should just be -AB dot BC

So how do I know which one is the desired one? The answers say it's the positive and I can clearly see why but I don't know why I'm supposed to choose vec(CB) and not vec(BC)

Not entirely sure, maybe it has to do with the fact that both vectors use B as a reference/fixed point.

InteGrand · Apr 26, 2016

leehuan said:
Right, is this a rule of thumb that I should remember?

It's not really a rule of thumb, more like a rule. Like if you see the image here, the angle between two vectors is the angle when the vectors have their tails lined up:

https://upload.wikimedia.org/wikipedia/commons/0/05/Inner-product-angle.png .

If you lined up one head with one tail, the angle obtained would be the supplement of the one if you lined up the two tails (to see this, draw a diagram and it comes down to "co-interior angles are supplementary"). This is why the cosine becomes the negative of the other way (because cos(180 deg – theta) = -cos(theta)).

porcupinetree · Apr 26, 2016

leehuan said:
Right, is this a rule of thumb that I should remember?

It's part of the geometric definition of the dot product.

leehuan · Apr 26, 2016

Only just started using point normal form and I don't know how to do this.

$$What is the line of intersection of the planes $\\\textbf{x}\cdot\left(\begin{matrix}1\\-1\\3\end{matrix}\right)=0, \quad \textbf{x}=\lambda_1 \left( \begin{matrix}2\\1\\2\end{matrix} \right)+\lambda_2 \left( \begin{matrix}3\\ 1\\ -3 \end{matrix} \right)$

InteGrand · Apr 26, 2016

leehuan said:
Only just started using point normal form and I don't know how to do this.

$$What is the line of intersection of the planes $\\\textbf{x}\cdot\left(\begin{matrix}1\\-1\\3\end{matrix}\right)=0, \quad \textbf{x}=\lambda_1 \left( \begin{matrix}2\\1\\2\end{matrix} \right)+\lambda_2 \left( \begin{matrix}3\\ 1\\ -3 \end{matrix} \right)$

$\noindent Let $\vec{p} =\left[ \begin{matrix}a\\b\\c\end{matrix} \right]\in \mathbb{R}^{3}$. Suppose $\vec{p}$ lies on both planes. Then from the first plane's equation, we must have $a-b+3c=0 \quad (1)$.$

$\noindent From the second plane's equation, we have $\vec{p} = \lambda_1 \left[ \begin{matrix}2\\1\\2\end{matrix} \right]+\lambda_2 \left[ \begin{matrix}3\\ 1\\ -3 \end{matrix} \right]$ for some $\lambda_1,\lambda_2 \in \mathbb{R}$. That is (comparing coordinates of LHS and RHS),$

$a = 2 \lambda_1 + 3\lambda_2$

$b = \lambda_1 + \lambda_2$

$$c = 2\lambda_1 - 3\lambda_2$.$

$\noindent Substituting these relationships into (1), we have$

$\begin{align*}2 \lambda_1 + 3\lambda_2 - \left(\lambda_1 + \lambda_2\right) +3 \left(2\lambda_1 - 3\lambda_2\right) &= 0 \\ \Rightarrow 7\lambda_1 - 7\lambda_2 &= 0 \\ \Rightarrow \lambda_1 = \lambda_2. \end{align*}$

$\noindent So from the second plane's equation (since $\vec{p}$ lies on it), we have $\vec{p} =\lambda_1 \left[ \begin{matrix}2\\1\\2\end{matrix} \right]+\lambda_1 \left[ \begin{matrix}3\\ 1\\ -3 \end{matrix} \right] =\lambda_1 \left[ \begin{matrix}5\\2\\-1\end{matrix} \right]$, for some $\lambda_1 \in \mathbb{R}$.$

$\noindent Conversely, it is easy to see that any vector of the form $\lambda \left[ \begin{matrix}5\\2\\-1\end{matrix} \right]$ ($\lambda\in \mathbb{R}$) lies on both planes. Hence the line of intersection is the line $\vec{x} = \lambda \left[ \begin{matrix}5\\2\\-1\end{matrix} \right]$.$

leehuan · Apr 26, 2016

InteGrand · Apr 26, 2016

leehuan said:
$$Let $Q$ be a square $n\times n$ orthogonal matrix. Show that $\\(Q\textbf{x})\cdot(Q\textbf{y})=\textbf{x}\cdot\textbf{y} \, \forall \, \textbf{x}.\textbf{y}\in \mathbb R^n$

$\noindent Let $\bold{x},\bold{y}\in \mathbb{R}^{n}$. Recall that $\bold{a}\cdot \bold{b} = \bold{a}^{T}\bold{b}$ for any vectors $\bold{a},\bold{b} \in \mathbb{R}^{n}$ (basically the RHS refers to the vectors as matrices with one column (when it's not transposed), so the RHS is a matrix product).$

$\noindent Using this fact (and associativity of matrix multiplication where required), we have$

$\begin{align*}\left(Q\bold{x}\right)\cdot \left(Q\bold{y} \right)&= \left(Q\bold{x}\right) ^{T} \left(Q\bold{y} \right) \\ &= \bold{x}^{T} Q^{T} Q\bold{y} \quad (\text{using the standard result about the transpose of a matrix product being equal to the product of the transposes in reversed order}) \\ &= \bold{x}^{T} \bold{y} \quad (\text{since }Q^{T} Q = I\text{ as }Q \text{ is orthogonal}) \\ &= \bold{x}\cdot \bold{y}.\end{align*}$

leehuan · Apr 27, 2016

Can't draw the link to find part c).

$$Let $B$ be a point in $\mathbb R^n$ with coordinate vector $\textbf{b}.$ Let $\textbf{x}=\textbf{a}+\lambda\textbf{d}, \lambda \in \mathbb R $ be the equation of a line.$\\ $a) Show that the square of the distance from $B$ to an arbitrary point $x$ on the line is given by$ \\ q(\lambda)={\left|\textbf{b}-\textbf{a}\right|}^{2}-2\lambda(\textbf{b}-\textbf{a})\cdot\textbf{d}+\lambda^2{\left|\textbf{d}\right|}^2$
Easy

$$b) Find the shortest distance between the point $B$ and the line by minimising $q(\lambda)$\\ Answer: $q_{min}(x)={\left|\textbf{b}-\textbf{a}\right|}^2-\frac{{\left( (\textbf{b}-\textbf{a}) \cdot \textbf{d}\right)}^2}{{|\textbf{d}|}^2}$

$$c) If $P$ is the point on the line closest to $B$, show that $\\\overrightarrow{PB}=\textbf{b}-\textbf{a}-$proj$_{\textbf{d}}(\textbf{b}-\textbf{a})\\$ and show that $\overrightarrow{PB}$ is orthogonal to the direction $\textbf{d}$ of the line$

InteGrand · Apr 27, 2016

leehuan said:
Can't draw the link to find part c).

$$Let $B$ be a point in $\mathbb R^n$ with coordinate vector $\textbf{b}.$ Let $\textbf{x}=\textbf{a}+\lambda\textbf{d}, \lambda \in \mathbb R $ be the equation of a line.$\\ $a) Show that the square of the distance from $B$ to an arbitrary point $x$ on the line is given by$ \\ q(\lambda)={\left|\textbf{b}-\textbf{a}\right|}^{2}-2\lambda(\textbf{b}-\textbf{a})\cdot\textbf{d}+\lambda^2{\left|\textbf{d}\right|}^2$
Easy

$$b) Find the shortest distance between the point $B$ and the line by minimising $q(\lambda)$\\ Answer: $q_{min}(x)={\left|\textbf{b}-\textbf{a}\right|}^2-\frac{{\left( (\textbf{b}-\textbf{a}) \cdot \textbf{d}\right)}^2}{{|\textbf{d}|}^2}$

$$c) If $P$ is the point on the line closest to $B$, show that $\\\overrightarrow{PB}=\textbf{b}-\textbf{a}-$proj$_{\textbf{d}}(\textbf{b}-\textbf{a})\\$ and show that $\overrightarrow{PB}$ is orthogonal to the direction $\textbf{d}$ of the line$

$\noindent If $P$ is the point on the line closest to $B$, then $P$ is the point corresponding to the vector $\vec{p}=\vec{a}+\lambda_{\text{vert.}}\vec{d}$, where $\lambda_{\text{vert.}}$ is the $\lambda$ value where the quadratic in part (a) is minimised. Using standard vertex of parabola formula, we find that $\lambda_{\text{vert.}}=-\frac{-2\left(\vec{b}-\vec{a}\right)\cdot \vec{d}}{2 \left \| \vec{d}\right \|^2}=\frac{\left(\vec{b}-\vec{a}\right)\cdot \vec{d}}{\left \| \vec{d}\right \|^2}$.$

$\noindent Now to complete the problem we just need to show that $\overrightarrow{PB}\equiv \vec{b}-\vec{p}$ simplifies to the expression the provided, and that $\overrightarrow{PB} \cdot \vec{d}=0$ (i.e. the vectors are orthogonal).$

leehuan · Apr 27, 2016

InteGrand said:
$\noindent Now to complete the problem we just need to show that $\overrightarrow{PB}\equiv \vec{b}-\vec{p}$ simplifies to the expression the provided, and that $\overrightarrow{PB} \cdot \vec{d}=0$ (i.e. the vectors are orthogonal).$

I think I'm not fully focused. I can't tell why vec(PB) . d = 0

InteGrand · Apr 27, 2016

leehuan said:
I think I'm not fully focused. I can't tell why vec(PB) . d = 0

Hints: Just write b – a as vec(AB) for simplicity, recall that the dot product is linear (can expand it), and d•d = ||d||².

More generally though, for any vectors u, v^† in Rⁿ, it is true that u – proj_v u is orthogonal to v. This is essentially how the vector projection is defined (i.e. note the right angles in the diagrams at the start here: https://en.wikipedia.org/wiki/Vector_projection. The vector a₂ there is simply a – proj_b a, and is orthogonal to b). Using this, it is immediate that vec(PB) is orthogonal to d.

^†Assuming v is not the zero vector of course (in that case, the projection is undefined).

leehuan · Apr 28, 2016

InteGrand said:
Hints: Just write b – a as vec(AB) for simplicity, recall that the dot product is linear (can expand it), and d•d = ||d||².

More generally though, for any vectors u, v^† in Rⁿ, it is true that u – proj_v u is orthogonal to v. This is essentially how the vector projection is defined (i.e. note the right angles in the diagrams at the start here: https://en.wikipedia.org/wiki/Vector_projection. The vector a₂ there is simply a – proj_b a, and is orthogonal to b). Using this, it is immediate that vec(PB) is orthogonal to d.

^†Assuming v is not the zero vector of course (in that case, the projection is undefined).

Lol ok I see now. For some reason I had it in my mind that (b-a).d wasn't going to cancel out. After using it so many times I suddenly forgot what d.d was

Dot product (1 Viewer)

leehuan

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dan964

what

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what

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porcupinetree

not actually a porcupine

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