# Vector and parametric equations of a plane

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• May 24th 2010, 04:51 PM
Kakariki
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 (Clapping)
• May 24th 2010, 06:17 PM
dwsmith
Quote:

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 (Clapping)

$\displaystyle \begin{vmatrix} i & j & k\\ 3 & -4 & -4\\ x & y & z \end{vmatrix}=(-4z+4y)\mathbf{i}-(3z+3x)\mathbf{j}+(3y+4x)\mathbf{k}=<1,2,3>$
• May 24th 2010, 07:09 PM
Kakariki
Quote:

Originally Posted by dwsmith
$\displaystyle \begin{vmatrix} i & j & k\\ 3 & -4 & -4\\ x & y & z \end{vmatrix}=(-4z+4y)\mathbf{i}-(3z+3x)\mathbf{j}+(3y+4x)\mathbf{k}=<1,2,3>$

Thank you for responding! I do not understand what you put though. Can someone explain, please?

Thanks!
• May 24th 2010, 07:24 PM
dwsmith
Quote:

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 $\displaystyle <x,y,z>$ with your vector you know, you will obtain a normal vector. However, you want you normal vector to be $\displaystyle <1,2,3>$
• May 25th 2010, 12:09 PM
Kakariki
Quote:

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 $\displaystyle <x,y,z>$ with your vector you know, you will obtain a normal vector. However, you want you normal vector to be $\displaystyle <1,2,3>$

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:
$\displaystyle (-4z + 4y), (-4x - 3z), (3y + 4x) = (1,2,3)$

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

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

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

HELP please!