Vector Help :yuck:

[moca915]

Hi :smile:

I need a reminder on how to do this vector stuff... Here's the problem, please help:

(Planes)
Find the angle between x+y+z=1 and x+2y+3z=6.

So there are two planes, and I need to find the angle between the normals of these planes.

Any hints will help. Thanks! :biggrin:

Peace, Love, & Happiness,

Monica :wink:

[moca915]

bump :uhh:

[Leong]

\begin{multline*}
\begin{split}
&For\ plane\ 1:\ x+y+z=1:\\
&Choose\ any\ 3\ points\ on\ the\ plane \ to\ find\ 2\ vectors\ on\ the\ plane.\\
&A(0,0,1);\ B(0,1,0);\ C(1,0,0)\\
&\vec{a}=\hat{k};\ \vec{b}=\hat{j}; \ \vec{c}=\hat{i}\\
&\vec{d}=\vec{b}-\vec{a}=\vec{j}-\hat{k}: The\ first\ vector\\
&\vec{e}=\vec{c}-\vec{a}=\vec{i}-\hat{k}: The\ second\ vector\\
&\vec{f}=\vec{d}\times\vec{e}=-\hat{i}-\hat{j}-\hat{k}: Normal\ vector\ to\ plane\ 1.\\

&For\ plane\ 2:\ x+2y+3z=6:\\
&Choose\ any\ 3\ points\ on\ the\ plane \ to\ find\ 2\ vectors\ on\ the\ plane.\\
&G(0,0,2);\ H(6,0,0);\ M(0,3,0)\\
&\vec{g}=2\hat{k};\ \vec{h}=6\hat{i}\; \ \vec{m}=3\hat{j}\\
&\vec{n}=\vec{h}-\vec{g}=6\vec{i}-2\hat{k}: The\ first\ vector\\
&\vec{p}=\vec{m}-\vec{g}=3\vec{j}-2\hat{k}: The\ second\ vector\\
&\vec{q}=\vec{n}\times\vec{p}=6\hat{i}+12\hat{j}+18 \hat{k}: Normal\ vector\ to\ plane\\
&\vec{r}=\vec{q}/6=\hat{i}+2\hat{j}+3\hat{k}: Another\ normal\ vector\ to\ plane\\
&\vec{f}\bullet \vec{r}=frcos\ \theta; \ \theta=158^0
\end{split}
\end{multline*}

[moca915]

Thanks! That helped a lot :)

[Muzza]

Or more generally: if a plane has the equation ax + by + cz + d = 0, then a normal vector to that plane is (a, b, c).