How Can I Use GeoGebra to Calculate Vector Lengths and Projections?

[mathmari]

Hey!

We have the vectors $v=i+j+2k=(1,1,2)$ and $u=-i-k=(-1,0,-1)$.

I have calculated the following:
\begin{align*}&|v|=\sqrt{1^2+1^2+2^2}=\sqrt{1+1+4}=\sqrt{6} \\ &|u|=\sqrt{(-1)^2+0^2+(-1)^2}=\sqrt{1+0+1}=\sqrt{2} \\ &v\cdot u=(1,1,2)\cdot (-1,0,-1)=1\cdot (-1)+1\cdot 0+2\cdot (-1)=-1+0-2=-3 \\ &u\times v=\begin{vmatrix}i & j & k\\ -1 & 0 & -1 \\ 1 & 1 & 2\end{vmatrix}=i+j-k=(1,1-1) \\ &v\times u=-(u\times v)=-i-j+k=(-1-1,1) \\ &|v\times u|=\sqrt{(-1)^2+(-1)^2+1^2}=\sqrt{1+1+1}=\sqrt{3} \end{align*}

As for the angle between $u$ and $v$ in radians we have:
\begin{align*}|u\times v|=|-(v\times u)|=|v\times u|=|v|\cdot |u|\cdot \sin \theta &\Rightarrow \sin \theta=\frac{|v\times u|}{|v|\cdot |u|}=\frac{\sqrt{3}}{\sqrt{6}\cdot \sqrt{2}}=\frac{\sqrt{3}}{\sqrt{3}\cdot \sqrt{2}\cdot \sqrt{2}}=\frac{1}{2} \\ & \Rightarrow \theta=2\pi n+\frac{\pi}{6} \ \text{ or } \ \theta=2\pi n+\frac{5\pi}{6}, \ n\in \mathbb{Z}\end{align*}

The vector projection of $u$ on $v$ is:
\begin{equation*}\frac{u\cdot v}{|v|^2}v=\frac{v\cdot u}{|v|^2}v=\frac{-3}{\sqrt{6}^2}(1,1,2)=\frac{-3}{6}(1,1,2)=-\frac{1}{2}(1,1,2)=\left (-\frac{1}{2},-\frac{1}{2},-1\right )\end{equation*} Is everything correct so far? (Wondering)
Now I want calculate all the above using Geogebra. I need a little help at this.

Firstly, we have the length of the vectors $u$ and $v$. I found a manual and there is teh command Length[<vector>] but it didn't work, I must have done something wrong.
Could you help me? (Wondering)

 

[I like Serena]

We have the vectors $v=i+j+2k=(1,1,2)$ and $u=-i-k=(-1,0,-1)$.

As for the angle between $u$ and $v$ in radians we have:
\begin{align*}|u\times v|=|-(v\times u)|=|v\times u|=|v|\cdot |u|\cdot \sin \theta &\Rightarrow \sin \theta=\frac{|v\times u|}{|v|\cdot |u|}=\frac{\sqrt{3}}{\sqrt{6}\cdot \sqrt{2}}=\frac{\sqrt{3}}{\sqrt{3}\cdot \sqrt{2}\cdot \sqrt{2}}=\frac{1}{2} \\ & \Rightarrow \theta=2\pi n+\frac{\pi}{6} \ \text{ or } \ \theta=2\pi n+\frac{5\pi}{6}, \ n\in \mathbb{Z}\end{align*}

Hey mathmari!

Shouldn't we have 1 angle?
The angle between two vectors is usually defined as the unique shortest angle between them, which is between $0$ and $\pi$ radians.
Unfortunately the cross product does not allow us to distinguish whether the angle is bigger or smaller than $\frac\pi 2$. (Worried)

Perhaps we should use the dot product? (Wondering)

The vector projection of $u$ on $v$ is:
\begin{equation*}\frac{u\cdot v}{|v|^2}v=\frac{v\cdot u}{|v|^2}v=\frac{-3}{\sqrt{6}^2}(1,1,2)=\frac{-3}{6}(1,1,2)=-\frac{1}{2}(1,1,2)=\left (-\frac{1}{2},-\frac{1}{2},-1\right )\end{equation*}

Is everything correct so far?

Yep. (Nod)

Now I want calculate all the above using Geogebra. I need a little help at this.

Firstly, we have the length of the vectors $u$ and $v$. I found a manual and there is the command Length[<vector>] but it didn't work, I must have done something wrong.
Could you help me?

If I type [M]v=(1,1,2)[/M], I get the vector [M]v[/M].
And when I type [M]length(v)[/M] or [M]length((1,1,2))[/M], I get its length.
What do you get? (Thinking)

 

[mathmari]

Shouldn't we have 1 angle?
The angle between two vectors is usually defined as the unique shortest angle between them, which is between $0$ and $\pi$ radians.
Unfortunately the cross product does not allow us to distinguish whether the angle is bigger or smaller than $\frac\pi 2$. (Worried)

Perhaps we should use the dot product? (Wondering)

Ahh ok! Now I get only $\frac{5\pi}{6}$.

If I type [M]v=(1,1,2)[/M], I get the vector [M]v[/M].
And when I type [M]length(v)[/M] or [M]length((1,1,2))[/M], I get its length.
What do you get? (Thinking)

Now I got everything but I got stuck at the projection. Is there a command for the projection in geogebra? (Wondering)

 

[I like Serena]

Ahh ok! Now I get only $\frac{5\pi}{6}$.

Now I got everything but I got stuck at the projection. Is there a command for the projection in geogebra?

Good! (Happy)

There doesn't seem to be a command for a projection, but [M](u*v)/v^2 * v[/M] works for me.
It shows up as $w=\frac{u\,v}{v^2}v$. (Thinking)