Plane and Line Equations for Points and Perpendicular Planes

[custer]

I've encountered this question but I do not know how to solve it? Help anyone? I need some clues.

Find a plane through the points P1 (1,2,3) and P2(3,2,1) and perpendicular to the plane 4x - y + 2z = 7

Another question, When given point A (1,2,3) and B (3,2,1) , I'm asked to find the line equation.
So If I got the direction of the line using B - A, I get the direction as (2,0,-2)
I should write the line equation as (1, 2, 3) + t (2, 0, -2). My question is, instead, can I write the equation as (3, 2, 1) + t (2, 0, -2) although (3, 2, 1) is the endpoint of the line?

 

[rock.freak667]

Another question, When given point A (1,2,3) and B (3,2,1) , I'm asked to find the line equation.
So If I got the direction of the line using B - A, I get the direction as (2,0,-2)
I should write the line equation as (1, 2, 3) + t (2, 0, -2). My question is, instead, can I write the equation as (3, 2, 1) + t (2, 0, -2) although (3, 2, 1) is the endpoint of the line?

Yes, both are correct.

I've encountered this question but I do not know how to solve it? Help anyone? I need some clues.

Find a plane through the points P1 (1,2,3) and P2(3,2,1) and perpendicular to the plane 4x - y + 2z = 7

For the plane 4x-y+2z=7, what is the equation of the normal,N?

If you visualize this plane being perpendicular to the one you want, what does it imply about N and the plane?

 

[custer]

I know that for 4x-y+2z=7 has normal vector (4,-1,2). I also know that 4x-y+2x=7 is perpendicular to the plane containing the P1 and P2, and thus the normal vector is parallel to the line P1P2. Therefore I should find P1P2 and cross with the normal vector (4,-1,2) and get a normal to the plane containing P1P2. After that I should substitute either P1P2 into the equation obtained to find d. then I got my answer? correct?

 

[rock.freak667]

I know that for 4x-y+2z=7 has normal vector (4,-1,2). I also know that 4x-y+2x=7 is perpendicular to the plane containing the P1 and P2, and thus the normal vector is parallel to the line P1P2. Therefore I should find P1P2 and cross with the normal vector (4,-1,2) and get a normal to the plane containing P1P2. After that I should substitute either P1P2 into the equation obtained to find d. then I got my answer? correct?

yes so do that and you will get the normal n of the plane you want.

then use the definition of the plane r.n=0 where r=(x,y,z) i.e. (x,y,z).n=0