Finding the Rotation Matrix for 60 Degree Rotation around (1,1,1) Axis

[cbarker1]

Dear Everybody,

I am having some problem with one exercise. And the question states:

Find the transformation Matrix R that describes a rotation by 60 degrees about an axis from the origin thru the pt (1,1,1). The rotation is clockwise as you look down toward the origin.

I know the standard basis is the identity matrix for 3x3 and the rotation matrix for a 3x3, the linear transformation $T: \mathbb{R}^{3}\mapsto \mathbb{R}^{3}$. I am stuck after these three factors.

 

[I like Serena]

Dear Everybody,

I am having some problem with one exercise. And the question states:

Find the transformation Matrix R that describes a rotation by 60 degrees about an axis from the origin thru the pt (1,1,1). The rotation is clockwise as you look down toward the origin.

I know the standard basis is the identity matrix for 3x3 and the rotation matrix for a 3x3, the linear transformation $T: \mathbb{R}^{3}\mapsto \mathbb{R}^{3}$. I am stuck after these three factors.

Hi Cbarker1,

The 'trick' is to find 3 independent vectors of which you know the image.
Then we can find the transformation matrix with linear combinations of these.
We can apply Gaussian elimination to facilitate the process.

For a rotation an obvious vector is a vector along the axis which has itself as image.
And additionally we need a vector that is perpendicular to the axis, which has an image that is also perpendicular to the axis and has an angle of 60 degrees in the appropriate direction.
We can construct the third vector and its image if we have difficulty coming up with it.

Can you find these vectors?
And preferably also a third perpendicular vector and its 60 degree image?

 

[cbarker1]

First, I need to Gaussian Elimination on the rotational matrix, right?
Second, I need to span the vectors from Gaussian Elimination so that the vectors are Linearly independent.

 

[I like Serena]

First, I need to Gaussian Elimination on the rotational matrix, right?
Second, I need to span the vectors from Gaussian Elimination so that the vectors are Linearly independent.

First we need a couple of vectors of which we know the image.

Let me illustrate with a 90 degree rotation matrix in 2D.
Let's pick the suboptimal choices $R\binom 1 1 = \binom {-1} 1$ and $R\binom {-1}1 = \binom {-1}{-1}$.
It follows that:
$$R\begin{pmatrix}1 & -1\\ 1 & 1\end{pmatrix} = \begin{pmatrix}-1 & -1\\ 1 & -1\end{pmatrix} \tag 1$$
We apply a variant of Gaussian elimination, bringing the left matrix in column echelon form.
Subtract the right column from the left column (on both sided of the equal sign) to get:
$$R\begin{pmatrix}2 & -1\\ 0 & 1\end{pmatrix} = \begin{pmatrix}0 & -1\\ 2 & -1\end{pmatrix}$$
Divide the left column by 2:
$$R\begin{pmatrix}1 & -1\\ 0 & 1\end{pmatrix} = \begin{pmatrix}0 & -1\\ 1 & -1\end{pmatrix}$$
Add the left column to the right column:
$$R\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix} = \begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}$$
And thus:
$$R = \begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}$$

To ease the process we can write the initial matrices of (1) in an augmented matrix form just like we do to apply Gaussian elimination to a set of equations:
$$\left[\begin{array}{cc|cc}1 & -1 & -1 & -1\\ 1 & 1 & 1 & -1\end{array}\right] \to \dots \to \left[\begin{array}{cc|cc}1 & 0 & 0 & -1 \\ 0 & 1 & 1 & 0 \end{array}\right]$$

To be fair, we could have deduced this immediately by picking the standard unit vectors: $R\binom 10=\binom 01$ and $R\binom 01=\binom {-1}0$, after which the result follows immediately.

 

[cbarker1]

First we need a couple of vectors of which we know the image.

Let me illustrate with a 90 degree rotation matrix in 2D.
Let's pick the suboptimal choices $R\binom 1 1 = \binom {-1} 1$ and $R\binom {-1}1 = \binom {-1}{-1}$.
It follows that:
$$R\begin{pmatrix}1 & -1\\ 1 & 1\end{pmatrix} = \begin{pmatrix}-1 & -1\\ 1 & -1\end{pmatrix} \tag 1$$
We apply a variant of Gaussian elimination, bringing the left matrix in column echelon form.
Subtract the right column from the left column (on both sided of the equal sign) to get:
$$R\begin{pmatrix}2 & -1\\ 0 & 1\end{pmatrix} = \begin{pmatrix}0 & -1\\ 2 & -1\end{pmatrix}$$
Divide the left column by 2:
$$R\begin{pmatrix}1 & -1\\ 0 & 1\end{pmatrix} = \begin{pmatrix}0 & -1\\ 1 & -1\end{pmatrix}$$
Add the left column to the right column:
$$R\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix} = \begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}$$
And thus:
$$R = \begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}$$

To ease the process we can write the initial matrices of (1) in an augmented matrix form just like we do to apply Gaussian elimination to a set of equations:
$$\left[\begin{array}{cc|cc}1 & -1 & -1 & -1\\ 1 & 1 & 1 & -1\end{array}\right] \to \dots \to \left[\begin{array}{cc|cc}1 & 0 & 0 & -1 \\ 0 & 1 & 1 & 0 \end{array}\right]$$

To be fair, we could have deduced this immediately by picking the standard unit vectors: $R\binom 10=\binom 01$ and $R\binom 01=\binom {-1}0$, after which the result follows immediately.

So, For a 3D rotation that has 60 degrees, the optimal matrices are the following:
$R\left[\begin{array}{ccc}1\\1\\1\end{array}\right]=?$
$R\left[\begin{array}{ccc}1\\0\\0\end{array}\right]=?$
$R\left[\begin{array}{ccc}1\\1\\0\end{array}\right]=?$

Is it the correct way to think about finding a transformation matrix?

 

[I like Serena]

So, For a 3D rotation that has 60 degrees, the optimal matrices are the following:
$R\left[\begin{array}{ccc}1\\1\\1\end{array}\right]=?$
$R\left[\begin{array}{ccc}1\\0\\0\end{array}\right]=?$
$R\left[\begin{array}{ccc}1\\1\\0\end{array}\right]=?$

Is it the correct way to think about finding a transformation matrix?

That's in the right direction.

However, instead of (1,0,0), I suggest to pick a vector that is perpendicular to (1,1,1).
For instance (0,1,-1), which has a dot product with (1,1,1) of zero, implying it is indeed perpendicular.
That is because we know that a perpendicular vector will have an image that is again perpendicular.
Another easy perpendicular vector is (1,0,-1), which happens to be the image of (0,1,-1).
We can verify by checking the dot product, and by checking the rotation direction.

 

[cbarker1]

That's in the right direction.

However, instead of (1,0,0), I suggest to pick a vector that is perpendicular to (1,1,1).
For instance (0,1,-1), which has a dot product with (1,1,1) of zero, implying it is indeed perpendicular.
That is because we know that a perpendicular vector will have an image that is again perpendicular.
Another easy perpendicular vector is (1,0,-1), which happens to be the image of (0,1,-1).
We can verify by checking the dot product, and by checking the rotation direction.

So the matrices are following
$R\left[\begin{array}{ccc}1\\1\\1\end{array}\right]=\left[\begin{array}{ccc}0\\1\\-1\end{array}\right]$
$R\left[\begin{array}{ccc}0\\1\\-1\end{array}\right]=\left[\begin{array}{ccc}1\\0\\-1\end{array}\right]$

$R\left[\begin{array}{ccc}1\\0\\-1\end{array}\right]=\left[\begin{array}{ccc}-1\\0\\1\end{array}\right]$

So a transformation matrix R is:
$\left[\begin{array}{ccc}1&0&1\\1&0&0\\1&-1&-1\end{array}\right]$

 

[I like Serena]

So the matrices are following
$R\left[\begin{array}{ccc}1\\1\\1\end{array}\right]=\left[\begin{array}{ccc}0\\1\\-1\end{array}\right]$

Isn't (1,1,1) supposed to be the axis of the rotation?
The axis consists of all points that have themselves as image.
So (1,1,1) has (1,1,1) as its image.

$R\left[\begin{array}{ccc}0\\1\\-1\end{array}\right]=\left[\begin{array}{ccc}1\\0\\-1\end{array}\right]$

Yes.

$R\left[\begin{array}{ccc}1\\0\\-1\end{array}\right]=\left[\begin{array}{ccc}-1\\0\\1\end{array}\right]$

Those are indeed 2 perpendicular vectors.
But their angle is not 60 degrees is it?
They are each other opposites, which implies that they have an angle of 180 degrees with each other instead of 60 degrees.

We can find the angle by using the identity of the dot product:
$$\mathbf a\cdot \mathbf b = a_xb_x+a_yb_y+a_zb_z = \|\mathbf a\| \cdot \|\mathbf b\| \cdot \cos\phi$$
where $\phi$ is the angle between the vectors $\mathbf a$ and $\mathbf b$.
In this case:
$$(1,0,-1)\cdot (-1, 0, 1) = 1\cdot -1 + -1 \cdot 1 = \|(1,0,-1)\|\cdot \|(-1,0,1)\|\cdot\cos\phi \quad\Rightarrow\quad
-2 = \sqrt 2 \cdot \sqrt 2\cdot \cos\phi \quad\\\Rightarrow\quad
\cos\phi = -1 \quad\Rightarrow\quad \phi = 180^\circ \ne 60^\circ$$
We need an image such that $\phi = 60^\circ$.

We can pick (1,0,-1), but we still need to find its image that has an angle of 60 degrees in clockwise direction.

 

[cbarker1]

Isn't (1,1,1) supposed to be the axis of the rotation?
The axis consists of all points that have themselves as image.
So (1,1,1) has (1,1,1) as its image.

Yes.
Those are indeed 2 perpendicular vectors.
But their angle is not 60 degrees is it?
They are each other opposites, which implies that they have an angle of 180 degrees with each other instead of 60 degrees.

We can find the angle by using the identity of the dot product:
$$\mathbf a\cdot \mathbf b = a_xb_x+a_yb_y+a_zc_z = \|\mathbf a\| \cdot \|\mathbf b\| \cdot \cos\phi$$
where $\phi$ is the angle between the vectors $\mathbf a$ and $\mathbf b$.
In this case:
$$(1,0,-1)\cdot (-1, 0, 1) = \|(1,0,-1)\|\cdot \|(-1,0,1)\|\cdot\cos\phi \quad\Rightarrow\quad
1\cdot -1 + -1 \cdot 1 = \sqrt 2 \cdot \sqrt 2\cdot \cos\phi \quad\\\Rightarrow\quad
\cos\phi = -1 \quad\Rightarrow\quad \phi = 180^\circ \ne 60^\circ$$
We need an image such that $\phi = 60^\circ$.

We can pick (1,0,-1), but we still need to find its image that has an angle of 60 degrees in clockwise direction.

I am trying the figure out the image of (1,0,-1). By using the dot product property which states $$\mathbf a\cdot \mathbf b = \|\mathbf a\| \cdot \|\mathbf b\| \cdot \cos\phi$$, let $$\mathbb{a}$$ be define as (1,0,-1) and let $$\mathbb{b}=(b_1,b_2,b_3)$$. Using the identity, $b_1-b_3=\sqrt2 \cdot \sqrt {b_1^2+b_2^2+b_3^2}\cdot \cos60$
$(b_1-b_3)^2=(\sqrt2 \cdot \sqrt {b_1^2+b_2^2+b_3^2}\cdot \cos60)^2$
$(b_1-b_3)^2=2(b_1^2+b_2^2+b_3^2)\cdot \cos^2(60)$
we know that $b_2=0$. We also know that $b_1-b_3=\sqrt2 \cdot \sqrt {b_1^2+b_2^2+b_3^2}\cdot \cos60$
$(b_1-b_3)^2=(\sqrt2 \cdot \sqrt {b_1^2+b_2^2+b_3^2}\cdot \cos60)^2$
$(b_1-b_3)^2=2(b_1^2+b_2^2+b_3^2)\cdot \cos^2(60)$
$b_1^2-2b_1b_3+b_3^2=2\cos^2(60)(b_1^2+b_2^2)$
Let $x=b_1$ and $y=b_3$.
$x^2-2xy+y^2=2\cos^2(60)(x^2+y^2)$
$(1-2\cos^2(60)x^2-2xy+(1-2\cos^2(60))y^2=0$
What to do next?
So the augemented matrix is: $\left[\begin{array}{ccc|ccc}1&0&1&1&1&x\\1&1&0&1&0&0\\1&-1&-1&1&-1&y\end{array}\right]$

 

[I like Serena]

I am trying the figure out the image of (1,0,-1). By using the dot product property which states $$\mathbf a\cdot \mathbf b = \|\mathbf a\| \cdot \|\mathbf b\| \cdot \cos\phi$$, let $$\mathbb{a}$$ be define as (1,0,-1) and let $$\mathbb{b}=(b_1,b_2,b_3)$$. Using the identity, $b_1-b_3=\sqrt2 \cdot \sqrt {b_1^2+b_2^2+b_3^2}\cdot \cos60$
$(b_1-b_3)^2=(\sqrt2 \cdot \sqrt {b_1^2+b_2^2+b_3^2}\cdot \cos60)^2$
$(b_1-b_3)^2=2(b_1^2+b_2^2+b_3^2)\cdot \cos^2(60)$
we know that $b_2=0$.

Actually, we do not know that $b_2=0$.
However, we do know that $\mathbf b\cdot (1,1,1)=0$, because $\mathbf b$ must be perpendicular to the axis.
Oh, and $\cos 60^\circ = \frac 12$.

So we have:
\begin{cases}(b_1-b_3)^2=2(b_1^2+b_2^2+b_3^2)\cdot \cos^2(60^\circ) \\
b_1+b_2+b_3=0 \\
\cos 60^\circ = \frac 12
\end{cases}

 

[cbarker1]

Actually, we do not know that $b_2=0$.
However, we do know that $\mathbf b\cdot (1,1,1)=0$, because $\mathbf b$ must be perpendicular to the axis.
Oh, and $\cos 60^\circ = \frac 12$.

So we have:
\begin{cases}(b_1-b_3)^2=2(b_1^2+b_2^2+b_3^2)\cdot \cos^2(60^\circ) \\
b_1+b_2+b_3=0 \\
\cos 60^\circ = \frac 12
\end{cases}

I can solve the equation (2) for $b_2$:

$b_2=-(b_1+b_3)$

Is that correct idea?

 

[I like Serena]

I can solve the equation (2) for $b_2$:

$b_2=-(b_1+b_3)$

Is that correct idea?

Yes. I suggest to keep going.

 

[cbarker1]

Yes. I suggest to keep going.

$b_2=-(b_1+b_3)$
I am plugging $b_2$ into the $(b_1-b_3)^2=2(b_1^2+b_2^2+b_3^2)\cos(60^{\circ})$
$\frac{1}{2}(b_1^2+(-b_1-b_3)^2+b_3^2)=(b_1-b_3)^2$
$\frac{1}{2}(2b_1^2+2b_1b_3+2b_3^2)=(b_1-b_3)^2$
$b_1^2+b_1b_3+b_3^2=b_1^2-2b_1b_3+b_3^2$
$3b_1b_3=0$
$b_1=0 or b_3=0$
Then the two cases for $b_2$ are:
Case I: if $b_1=0$, then $b_2=-b_3$
Case II: if $b_3=0$, then $b_2=-b_1$

 

[I like Serena]

$b_2=-(b_1+b_3)$
I am plugging $b_2$ into the $(b_1-b_3)^2=2(b_1^2+b_2^2+b_3^2)\cos(60^{\circ})$
$\frac{1}{2}(b_1^2+(-b_1-b_3)^2+b_3^2)=(b_1-b_3)^2$
$\frac{1}{2}(2b_1^2+2b_1b_3+2b_3^2)=(b_1-b_3)^2$
$b_1^2+b_1b_3+b_3^2=b_1^2-2b_1b_3+b_3^2$
$3b_1b_3=0$
$b_1=0 or b_3=0$
Then the two cases for $b_2$ are:
Case I: if $b_1=0$, then $b_2=-b_3$
Case II: if $b_3=0$, then $b_2=-b_1$

Correct.
And we actually have yet another equation.
For a rotation the image must have the same length as the original.
So $\sqrt{b_1^2+b_2^2+b_3^3}=\sqrt 2$.
(In retrospect we did not need to square.)

Furthermore, we introduced erroneous solutions due to squaring.
We need to check the solutions against the original equation and eliminate the ones that do not match.
Or alternatively redo the solution without squaring using the fact that the image must have the same length as the original.

Afterwards we should have 2 solutions, one that is clockwise, and one that is counter-clockwise.

 

[cbarker1]

what is the original equation?

Is the original equation, this one $(b_1-b_3)=\sqrt{2(b_1^2+b_2^2+b_3^2)\cos^2(60^{\circ})}$?

 

[I like Serena]

what is the original equation?

Is the original equation, this one $(b_1-b_3)=\sqrt{2(b_1^2+b_2^2+b_3^2)\cos^2(60^{\circ})}$?

That should be $(b_1-b_3)=\sqrt{2}\sqrt{b_1^2+b_2^2+b_3^2}\cos(60^{\circ})$.
The cosine should not be squared since then we could lose its sign.

 

[cbarker1]

That should be $(b_1-b_3)=\sqrt{2}\sqrt{b_1^2+b_2^2+b_3^2}\cos(60^{\circ})$.
The cosine should not be squared since then we could lose its sign.

This equation is where I plugin $(b_2)$ for the two different values, right?

 

[I like Serena]

This equation is where I plugin $(b_2)$ for the two different values, right?

There's no need since we already know that $\sqrt{b_1^2+b_2^2+b_3^2}=\sqrt 2$, which effectively eliminates $b_2$ from the equation.
When we have found $b_1$ and $b_3$ we can plug them into $b_2=-(b_1+b_3)$ to find $b_2$.

 

[cbarker1]

There's no need since we already know that $\sqrt{b_1^2+b_2^2+b_3^2}=\sqrt 2$, which effectively eliminates $b_2$ from the equation.
When we have found $b_1$ and $b_3$ we can plug them into $b_2=-(b_1+b_3)$ to find $b_2$.

I am struck with the algebraic manipulations:
$b_2=-(b_1+b_3)$
I am plugging $b_2$ into the $\sqrt{2}=\sqrt{(b_1^2+b_2^2+b_3^2)}$
$\sqrt{2}=\sqrt{2b_1^2+2b_1b_3+2b_3^2}$
Factor out $\sqrt{2}$
$1=\sqrt{b_1^2+b_1b_3+b_3^2}$
$1=b_1^2+b_1b_3+b_3^2$ -here is where I am struck.

 

[I like Serena]

How about eliminating $b_2$ first, and then eliminate $b_3$:
$$\begin{cases}b_1-b_3=\sqrt 2\sqrt{b_1^2+b_2^2+b_3^2}\cos 60^\circ \\
b_1+b_2+b_3=0 \\
\sqrt{b_1^2+b_2^2+b_3^2} = \sqrt 2
\end{cases} \quad\Rightarrow\quad

\begin{cases}b_1-b_3=\sqrt 2\cdot\sqrt 2\cdot \frac 12=1 \\
b_2 = -b_1-b_3 \\
b_1^2+b_2^2+b_3^2= 2
\end{cases} \quad\Rightarrow\quad \\

\begin{cases}b_3=b_1-1 \\
b_2 = -b_1-b_3 \\
b_1^2+(-b_1-b_3)^2+b_3^2= 2b_1^2+2b_1b_3+2b_3^2=2
\end{cases} \quad\Rightarrow\quad

\begin{cases}b_3=b_1-1 \\
b_2 = -b_1-(b_1-1) = -2b_1+1\\
b_1^2+b_1(b_1-1)+(b_1-1)^2=3b_1^2-3b_1+1=1
\end{cases} \quad\Rightarrow\quad \\

\begin{cases}b_3=b_1-1 \\
b_2 = -2b_1+1\\
b_1(b_1-1)=0
\end{cases} \quad\Rightarrow\quad

\begin{cases}b_1=0 \\
b_3=-1 \\
b_2 = 1\\
\end{cases} \lor \begin{cases}b_1=1 \\
b_3=1-1=0 \\
b_2 = -2\cdot 1+1 = -1\\
\end{cases}
$$

 

[cbarker1]

How about eliminating $b_2$ first, and then eliminate $b_3$:
$$\begin{cases}b_1-b_3=\sqrt 2\sqrt{b_1^2+b_2^2+b_3^2}\cos 60^\circ \\
b_1+b_2+b_3=0 \\
\sqrt{b_1^2+b_2^2+b_3^2} = \sqrt 2
\end{cases} \quad\Rightarrow\quad

\begin{cases}b_1-b_3=\sqrt 2\cdot\sqrt 2\cdot \frac 12=1 \\
b_2 = -b_1-b_3 \\
b_1^2+b_2^2+b_3^2= 2
\end{cases} \quad\Rightarrow\quad \\

\begin{cases}b_3=b_1-1 \\
b_2 = -b_1-b_3 \\
b_1^2+(-b_1-b_3)^2+b_3^2= 2b_1^2+2b_1b_3+2b_3^2=2
\end{cases} \quad\Rightarrow\quad

\begin{cases}b_3=b_1-1 \\
b_2 = -b_1-(b_1-1) = -2b_1+1\\
b_1^2+b_1(b_1-1)+(b_1-1)^2=3b_1^2-3b_1+1=1
\end{cases} \quad\Rightarrow\quad \\

\begin{cases}b_3=b_1-1 \\
b_2 = -2b_1+1\\
b_1(b_1-1)=0
\end{cases} \quad\Rightarrow\quad

\begin{cases}b_1=0 \\
b_3=-1 \\
b_2 = 1\\
\end{cases} \lor \begin{cases}b_1=1 \\
b_3=1-1=0 \\
b_2 = -2\cdot 1+1 = -1\\
\end{cases}
$$

Is the next step to choose the correct image for the linear transformation R such vector b satisfies the conditions in the exercise?
Do I need to use Gaussian Elimination to find the solution, right?

 

[I like Serena]

Is the next step to choose the correct image for the linear transformation R such vector b satisfies the conditions in the exercise?

Yes. I suggest to make a drawing to figure out what the orientation is.

Do I need to use Gaussian Elimination to find the solution, right?

Yes.

 

[cbarker1]

Yes. I suggest to make a drawing to figure out what the orientation is.
Yes.

So the last image for the transformation is
$R\left[\begin{array}{ccc}1\\0\\-1\end{array}\right]=\left[\begin{array}{ccc}0\\-1\\1\end{array}\right]$.

Then the augmented matrix is the following:
$\left[\begin{array}{ccc|ccc}1&0&1&1&1&0\\1&1&0&1&0&-1\\1&-1&-1&1&-1&1\end{array}\right]$

 

[I like Serena]

So the last image for the transformation is
$R\left[\begin{array}{ccc}1\\0\\-1\end{array}\right]=\left[\begin{array}{ccc}0\\-1\\1\end{array}\right]$.

The coordinates are in the wrong order. Note that the solution of the equation has the coordinates in an unusual order.
It should be:
$$R\left[\begin{array}{ccc}1\\0\\-1\end{array}\right]=\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]$$

Then the augmented matrix is the following:
$\left[\begin{array}{ccc|ccc}1&0&1&1&1&0\\1&1&0&1&0&-1\\1&-1&-1&1&-1&1\end{array}\right]$

So the augmented matrix should be:
$$\left[\begin{array}{ccc|ccc}1&0&1&1&1&1\\1&1&0&1&0&-1\\1&-1&-1&1&-1&0\end{array}\right]$$

 

[I like Serena]

This is how it looks.
\begin{tikzpicture}[>=stealth', shorten >=2pt]
%preamble \usepackage{tikz-3dplot}
%preamble \usetikzlibrary{arrows}
\tdplotsetmaincoords{80}{110}
\begin{scope}[scale=3,tdplot_main_coords]
\coordinate (O) at (0,0,0);

\draw[ultra thick,->] (0,0,0) -- (2,0,0) node[anchor=north east]{$x$};
\draw[ultra thick,->] (0,0,0) -- (0,2,0) node[anchor=north west]{$y$};
\draw[ultra thick,->] (0,0,0) -- (0,0,2) node[anchor=south]{$z$};

\draw[thick,color=red] (-2,-2,-2) -- (2,2,2);
\draw[ultra thick,->,color=red] (O) -- (1,1,1) node[anchor=west] {(1,1,1)};

\draw[->] (0,0,0) -- (0,1,-1) node[anchor=west] {(0,1,-1)}; \draw[dashed] (0,1,0) -- (0,1,-1) -- (0,0,-1) -- (O);
\draw[->] (0,0,0) -- (1,0,-1) node[anchor=north] {(1,0,-1)}; \draw[dashed] (1,0,0) -- (1,0,-1) -- (0,0,-1);
\draw[->] (0,0,0) -- (1,-1,0) node[anchor=east] {(1,-1,0)}; \draw[dashed] (1,0,0) -- (1,-1,0) -- (0,-1,0) -- (O);
\draw[->] (0,0,0) -- (0,-1,1); \draw[dashed] (0,-1,0) -- (0,-1,1) -- (0,0,1);
\draw[->] (0,0,0) -- (-1,0,1); \draw[dashed] (0,0,1) -- (-1,0,1) -- (-1,0,0) -- (O);
\draw[->] (0,0,0) -- (-1,1,0); \draw[dashed] (0,1,0) -- (-1,1,0) -- (-1,0,0);
\end{scope}
\end{tikzpicture}

 

[karush]

how did you create this image?
with transparency?

 

[I like Serena]

how did you create this image?
with transparency?

It's a $\LaTeX$ TikZ image, which MHB supports natively, and which includes transparency if desired.
Click on the image and you'll see.
Or alternatively click Reply With Quote, and you'll see how it was constructed.

 

[karush]

very informative thread btw..