# Need some help on calculus

1. Jun 30, 2004

### babipedes

i was having a trouble to find an equation on the planes through a known point (say : 1,2,1) perpendicular to other planes.. (say : x+y+z=1) ...
can anybody help me to get out of this ?

thanks

2. Jun 30, 2004

### Muzza

The plane you're looking for must have a normal vector that is orthogonal to the normal vector of x + y + z = 1.

Another way of saying that, is that the "new" plane must have a normal vector that is parallel to x + y + z = 1.

3. Jun 30, 2004

### babipedes

the normal vector is 1,2,1, but how to find its plane ?

4. Jun 30, 2004

### Muzza

I seriously doubt that's the normal vector. A normal vector of x + y + z = 1 is (1, 1, 1). The dot product of (1, 2, 1) and (1, 1, 1) is 4 (not 0), which means that the normals aren't perpendicular, so the planes can't be perpendicular either...

Anyway, say you have a normal vector, say (a, b, c). Then the plane will have the equation ax + by + cz + d = 0, where d can computed by using the known point on the plane.

5. Aug 25, 2008

### SallyGreen

does anyone guys advice me how to find the curvature at the corner of a rectangular,, cos I need to find it at any point of such geometry, and the curvature for flat side is just zero, but still sruggling with the corner.................

anyone could help.......

6. Aug 25, 2008

### HallsofIvy

Staff Emeritus
You do understand, don't you, that there are an infinite number of planes perpendicular to a given plane, through a given point? Your problem is not "well defined".

Draw a line from the given point perpendicular to the given plane. Any plane containing that line will satisfy those conditions.

As Muzza said, the normal vector to the new plane must be perpendicular to the normal vector to the given plane. In three dimensions, there exist an infinite number of (unit) vectors perpendicular to a given vector.