Finding the angle between two planes using normal vectors

[moca915]

Hi

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!

Peace, Love, & Happiness,

Monica

 

[moca915]

bump

 

[Leong]

$$\begin{multline*}<br /> \begin{split}<br /> &For\ plane\ 1:\ x+y+z=1:\\<br /> &Choose\ any\ 3\ points\ on\ the\ plane \ to\ find\ 2\ vectors\ on\ the\ plane.\\<br /> &A(0,0,1);\ B(0,1,0);\ C(1,0,0)\\<br /> &\vec{a}=\hat{k};\ \vec{b}=\hat{j}; \ \vec{c}=\hat{i}\\<br /> &\vec{d}=\vec{b}-\vec{a}=\vec{j}-\hat{k}: The\ first\ vector\\<br /> &\vec{e}=\vec{c}-\vec{a}=\vec{i}-\hat{k}: The\ second\ vector\\<br /> &\vec{f}=\vec{d}\times\vec{e}=-\hat{i}-\hat{j}-\hat{k}: Normal\ vector\ to\ plane\ 1.\\<br /> <br /> &For\ plane\ 2:\ x+2y+3z=6:\\<br /> &Choose\ any\ 3\ points\ on\ the\ plane \ to\ find\ 2\ vectors\ on\ the\ plane.\\<br /> &G(0,0,2);\ H(6,0,0);\ M(0,3,0)\\<br /> &\vec{g}=2\hat{k};\ \vec{h}=6\hat{i}\; \ \vec{m}=3\hat{j}\\<br /> &\vec{n}=\vec{h}-\vec{g}=6\vec{i}-2\hat{k}: The\ first\ vector\\<br /> &\vec{p}=\vec{m}-\vec{g}=3\vec{j}-2\hat{k}: The\ second\ vector\\<br /> &\vec{q}=\vec{n}\times\vec{p}=6\hat{i}+12\hat{j}+18\hat{k}: Normal\ vector\ to\ plane\\<br /> &\vec{r}=\vec{q}/6=\hat{i}+2\hat{j}+3\hat{k}: Another\ normal\ vector\ to\ plane\\<br /> &\vec{f}\bullet \vec{r}=frcos\ \theta; \ \theta=158^0<br /> \end{split}<br /> \end{multline*}$$

 

[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).