# Thread: Vector and parametric equations of a plane

1. ## Vector and parametric equations of a plane

Hey!
I am having trouble with a question. I have NO IDEA how to go about finding the solution, so any help is greatly appreciated!

Question
Find a vector equation and parametric equations of the plane that contains the points M(2, 1, 3) and N(-1, 5, 7) and is perpendicular to the plane x + 2y + 3z + 4 = 0.

What I know
So I know how to find the vector equation of a plane if I am given 3 points (or 2 direction vectors and a point) on the plane. I also know how to find the parametric equation given the same thing.

My theory on how to solve this is using the two points given to get a direction vector (3, -4, -4).

I know that the Cartesian equation of a plane is in the format Ax + By + Cz + D = 0. The capital letters (A, B, C) are the normal to the plane the Cartesian vector is referring to (so, in this case, it would be the second direction vector I need). The x y z are a point on that plane. I do not know how to combine this information in order to solve the problem though!

Thank you for reading, and SUPER thank you if you help

2. Originally Posted by Kakariki
Hey!
I am having trouble with a question. I have NO IDEA how to go about finding the solution, so any help is greatly appreciated!

Question
Find a vector equation and parametric equations of the plane that contains the points M(2, 1, 3) and N(-1, 5, 7) and is perpendicular to the plane x + 2y + 3z + 4 = 0.

What I know
So I know how to find the vector equation of a plane if I am given 3 points (or 2 direction vectors and a point) on the plane. I also know how to find the parametric equation given the same thing.

My theory on how to solve this is using the two points given to get a direction vector (3, -4, -4).

I know that the Cartesian equation of a plane is in the format Ax + By + Cz + D = 0. The capital letters (A, B, C) are the normal to the plane the Cartesian vector is referring to (so, in this case, it would be the second direction vector I need). The x y z are a point on that plane. I do not know how to combine this information in order to solve the problem though!

Thank you for reading, and SUPER thank you if you help
$\begin{vmatrix}
i & j & k\\
3 & -4 & -4\\
x & y & z
\end{vmatrix}=(-4z+4y)\mathbf{i}-(3z+3x)\mathbf{j}+(3y+4x)\mathbf{k}=<1,2,3>$

3. Originally Posted by dwsmith
$\begin{vmatrix}
i & j & k\\
3 & -4 & -4\\
x & y & z
\end{vmatrix}=(-4z+4y)\mathbf{i}-(3z+3x)\mathbf{j}+(3y+4x)\mathbf{k}=<1,2,3>$
Thank you for responding! I do not understand what you put though. Can someone explain, please?

Thanks!

4. Originally Posted by Kakariki
Thank you for responding! I do not understand what you put though. Can someone explain, please?

Thanks!
The cross product of two vectors will produce a normal vector to the plane. You know two vectors in your problem and we know one of the vectors is normal to the plane; therefore, if you take the cross product of an arbitrary vector $$ with your vector you know, you will obtain a normal vector. However, you want you normal vector to be $<1,2,3>$

5. Originally Posted by dwsmith
The cross product of two vectors will produce a normal vector to the plane. You know two vectors in your problem and we know one of the vectors is normal to the plane; therefore, if you take the cross product of an arbitrary vector $$ with your vector you know, you will obtain a normal vector. However, you want you normal vector to be $<1,2,3>$
Took me a few minutes to understand, but I get it now. So I write out the cross product and solve for x y and z. I think you wrote out the cross product wrong though, it isn't following the formula I have.

Using the formula I have it writes out as:
$(-4z + 4y), (-4x - 3z), (3y + 4x) = (1,2,3)$

Which makes the system of equations to solve:
$-4z + 4y = 1$
$-4x - 3z = 2$
$3y + 4x = 3$

I can't seem to get a straight answer solving this system of equations!

Btw, I am using: $a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1$