1. ## Planes in space

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

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

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

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

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

I can only solve this for the magnitude of $n_A$. 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: $<2, 4, 8>$

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

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

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

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

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

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

$n_A=<2,4,8>$

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

$n_A\cdot =0$

$<2,4,8>\cdot ( - <3,0,8>)=0$

Ok, so basically when ever I have two planes that are parallel, I can immediately assume that they have the same normal vector?

$n_A=<2,4,8>$

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

$n_A\cdot =0$

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

Ok, so basically when ever I have two planes that are parallel, I can immediately assume that they have the same normal vector?

$n_A=<2,4,8>$

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

$n_A\cdot =0$

$<2,4,8>\cdot ( - <3,0,8>)=0$

Bingo.

7. Cool, thanks for your help folks.