# Thread: equation of plane that contains P and perpendicular to line

1. ## equation of plane that contains P and perpendicular to line

Hi, can anyone help me with the following question?

Given P=(3, 4,3) and L1=(4-t,3t, -1+t), find the scalar equation of the plane that contains point P and perpendicular to the line L1.

I know how to figure out the equation that contains point P and L1, but I am not sure this one. So, please help. Thank you

2. Originally Posted by uyk
Hi, can anyone help me with the following question?

Given P=(3, 4,3) and L1=(4-t,3t, -1+t), find the scalar equation of the plane that contains point P and perpendicular to the line L1.

I know how to figure out the equation that contains point P and L1, but I am not sure this one. So, please help. Thank you
the direction vector for the line is <-1, 3, 1 >. use this as the normal vector for the plane. you have a point the plane passes through and the normal vector, you can find the plane

3. Originally Posted by uyk
Hi, can anyone help me with the following question?

Given P=(3, 4,3) and L1=(4-t,3t, -1+t), find the scalar equation of the plane that contains point P and perpendicular to the line L1.

I know how to figure out the equation that contains point P and L1, but I am not sure this one. So, please help. Thank you
The cartesian equation of a plane can be written ax + by + cz = d where the vector <a, b, c> is perpendicular to the plane.

A vector perpendicular to your plane is a vector parallel to the line: <-1, 3, 1>.

Get the value of d by substituting (3, 4, 3) into ax + by + cz = d.

4. I see.. Thanks a lot