# Thread: Planes in space

1. ## Planes in space

I need to find the plane $\displaystyle A$ that contains the line $\displaystyle r_A(t)=<3+2t , t , 8-t>$ and is parallel to the plane $\displaystyle B, 2x+4y+8z=17$

I know that the normal vectors $\displaystyle n_A=<a,b,c>,n_B=<2,4,8>$ should have a dot product equal to one. also know that the points $\displaystyle (3,1,7),(7,2,6)$ lie on the line $\displaystyle r_A(t)$, correpsonding to $\displaystyle t=1,t=2$. The points are also the terminal points of the vectors $\displaystyle <3,1,7>,<7,2,6>$. Substracting these vectors, I write the vector equation of the plane $\displaystyle A$ as

(1) $\displaystyle n_A\cdot<4,1,-1>=0$.

The problem is that I don't know how to find the vector $\displaystyle n_A$. I know that $\displaystyle n_A,n_B$ are parallel so

(2) $\displaystyle n_A\cdot n_B=\mid n_A n_B\mid\cos\theta_{A,B}$

I can only solve this for the magnitude of $\displaystyle n_A$. How do I get the vector? Secondly, I need to know if I'm setting this up right. Is equation (1) correct?

2. All you need for the equation of a plane is a point in the plane and vector normal to it.

They give you the normal vector by saying that there is a parallel plane with the corresponding normal vector: $\displaystyle <2, 4, 8>$

So all you need now is a point within the plane and you can use the equation:
$\displaystyle v_1(x - x_1) + v_2(y - y_1) + v_3(z - z_1) = 0$

Where $\displaystyle (x_1, y_1, z_1)$ are the coordinates of a point you need to find, and $\displaystyle <v_1, v_2, v_3> = <2, 4, 8>$

3. Originally Posted by adkinsjr
I need to find the plane $\displaystyle A$ that contains the line $\displaystyle r_A(t)=<3+2t , t , 8-t>$ and is parallel to the plane $\displaystyle B, 2x+4y+8z=17$
I know that the normal vectors $\displaystyle n_A=<a,b,c>,n_B=<2,4,8>$ should have a dot product equal to one. also know that the points $\displaystyle (3,1,7),(7,2,6)$ lie on the line $\displaystyle r_A(t)$, correpsonding to $\displaystyle t=1,t=2$. The points are also the terminal points of the vectors $\displaystyle <3,1,7>,<7,2,6>$. Substracting these vectors, I write the vector equation of the plane $\displaystyle A$ as
First of all $\displaystyle (3,1,7)$ is not on the line.
When $\displaystyle t=0$ the point $\displaystyle (3,0,8)$ is on the line.

Any plane parallel to $\displaystyle 2x+4y+8z=7$ looks $\displaystyle 2x+4y+8z=K$.
Use the point $\displaystyle (3,0,8)$ to find the value of $\displaystyle K$.

4. Originally Posted by Plato
First of all $\displaystyle (3,1,7)$ is not on the line.
When $\displaystyle t=0$ the point $\displaystyle (3,0,8)$ is on the line.

Any plane parallel to $\displaystyle 2x+4y+8z=7$ looks $\displaystyle 2x+4y+8z=K$.
Use the point $\displaystyle (3,0,8)$ to find the value of $\displaystyle K$.
Ok, so basically when ever I have two planes that are parallel, I can immediately assume that they have the same normal vector?

$\displaystyle n_A=<2,4,8>$

I should be able to write the equation of the plane as:

$\displaystyle n_A\cdot <r-r_o(0)>=0$

$\displaystyle <2,4,8>\cdot (<x,y,z> - <3,0,8>)=0$

5. Originally Posted by adkinsjr
Ok, so basically when ever I have two planes that are parallel, I can immediately assume that they have the same normal vector?

$\displaystyle n_A=<2,4,8>$

I should be able to write the equation of the plane as:

$\displaystyle n_A\cdot <r-r_o(0)>=0$

$\displaystyle <2,4,8>\cdot (<x,y,z> - <3,0,8>)\color{red}=0$
See the correction above. That gives the correct answer.

6. Originally Posted by adkinsjr
Ok, so basically when ever I have two planes that are parallel, I can immediately assume that they have the same normal vector?

$\displaystyle n_A=<2,4,8>$

I should be able to write the equation of the plane as:

$\displaystyle n_A\cdot <r-r_o(0)>=0$

$\displaystyle <2,4,8>\cdot (<x,y,z> - <3,0,8>)=0$

Bingo.

7. Cool, thanks for your help folks.