Equation of a Plane

Two vectors are perpendicular if their directions are 90 degrees apart. The dot product of two perpendicular vectors is zero. 

A Normal vector to a plane

A normal vector to a plane is basically a vector that makes a ninety degree angle with the plane(perpendicular to the plane).

 

In the above picture, \(\vec{d}\) is a normal vector to the plane, P. A normal vector to the plane will be perpendicular to every other vector which is contained in the plane or parallel to the plane. 

Equation of a plane

To find the equation of a plane we need :

• Any non-zero vector normal to the plane

• Coordinates of any point in the plane

Consider a plane, P. Let there be a point A(-1, -1, 1) in the plane whose coordinates are known to us. Let point B(x, y, z) be an arbitrary point in the plane. It is an arbitrary point, so it could be anywhere on the plane. 

Let, \(\overrightarrow{d} = -2\hat{i} + 2\hat{j} + 0\hat{k}\) be a normal vector to the plane. The picture below roughly depicts how everything is supposed to look like. 

Define a vector from A to B; we will call it \(\overrightarrow{AB}\). This is a vector contained in the plane or parallel to the plane. Depending on where point B is in the plane, the size and/or direction of \(\overrightarrow{AB}\) will vary. 

\(\overrightarrow{AB} = (x - (-1))\hat{i} + (y - (-1))\hat{j} + (z - 1)\hat{k}\)

                       

                                 \(= (x + 1)\hat{i} + (y + 1)\hat{j} + (z - 1)\hat{k}\)

Wherever the location of point B is in the plane, \(\overrightarrow{AB}\) will always be parallel to the plane and hence will always be perpendicular to the normal vector \(\vec{d}\). 

So, 

\(\overrightarrow{AB} \cdot \vec{d}\) = 0

(x + 1)(-2) + (y + 1)(-2) + (z - 1)(0) = 0 

- 2x - 2 - 2y - 2 + z = 0

2x + 2y - z + 4 = 0

The above is the Cartesian equation of the plane P. Every point B(x, y, z) in the plane must satisfy this equation. 

Equation of a Plane from three given points

Equation of a plane can also be found from the given coordinates of any three non-collinear points in the plane. 

The first step is to find a normal vector to the plane from three given points. Define any two vectors parallel to the plane between the given points.

As seen in the above image, we have defined vectors \(\overrightarrow{AB}\) & \(\overrightarrow{AC}\) in the plane P. 

\(\overrightarrow{AB} = (3 - (-1))\hat{i} + (3 - (-1))\hat{j} + (2 - 1)\hat{k}\)

 

                               \( = 4\hat{i} + 4\hat{j} + \hat{k}\)\)

\(\overrightarrow{AC} = (5 - (-1))\hat{i} + (5 - (-1))\hat{j} + (2 - 1)\hat{k}\)

 

                               \( = 6\hat{i} + 6\hat{j} + \hat{k}\)

A normal vector to the plane P can be found by taking their cross product. Remember that cross product of two vectors is a vector perpendicular to every plane to which the two vectors are parallel. Hence, \(\overrightarrow{AB} \times \overrightarrow{AC}\) or \(\overrightarrow{AC} \times \overrightarrow{AB}\) is a normal vector to P(any one of them will do). It is important that the three points are non-collinear(cannot all lie on a single straight line). If they are collinear then any two vectors defined between them will be parallel to each other, and the cross product of parallel vectors will yield a zero normal vector(which will only get us to 0 = 0 in the end).

We could have also used the cross product \(\overrightarrow{AC} \times \overrightarrow{AB}\). This is also a normal vector to the plane but pointing in the opposite direction of \(\overrightarrow{AB} \times \overrightarrow{AC}\). We need only one normal vector to find the equation, any of them can do the job. 

Now that we have a normal vector, and from the coordinates of any point on the plane, we can find the equation of the plane P in the exact same way we did above. In fact this is the same example of the plane. It’s equation is :

2x + 2y - z + 4 = 0