## Dot Product and Normals to Lines and Planes

The equation of a line in the form ax + by = c can be written as a dot product:

(a,b) . (x,y) = c, or A . X = c,

where A = (a, b) and X = (x,y).

The equation of a line in the form ax + by + cz = d can be written as a dot product:

(a,b, c) . (x,y, z) = d, or A . X = d,

where A = (a, b, c) and X = (x,y, z).

### Normal Vector A

If P and Q are in the plane with equation A . X = d, then A . P = d and A . Q = d, so

A . (Q - P) = d - d = 0.

This means that the vector A is orthogonal to any vector PQ between points P and Q of the plane.

But the vector PQ can be thought of as a tangent vector or direction vector of the plane. This means that vector A is orthogonal to the plane, meaning A is orthogonal to every direction vector of the plane.

A nonzero vector that is orthogonal to direction vectors of the plane is called a normal vector to the plane. Thus the coefficient vector A is a normal vector to the plane.

This also means that vector OA is orthogonal to the plane, so the line OA is perpendicular to the plane.

Careful: It is NOT true that for any point P in the plane, A is orthogonal to P (unless d = 0).

Exercise: Show that if A is a normal vector to a plane, and k is a nonzero constant, then kA is also a normal vector to the same plane.

Debate: For any plane, is the 0 vector orthogonal to all the direction vectors of the plane?

Exercise on Lines in the Plane: The same reasoning works for lines. On graph paper plot the line m with equation 2x + 3y = 6 and also plot the point A = (2,3). Check that lines OA and m are perpendicular. Also plot the line 2x + 3y = 0. How is this line related to m and OA? Finally, find an equation for line OA. What is a normal vector for this line?

Example: Finding a plane when the normal is known. Suppose that A = (1, 2, 3). Find the equation of the plane through P = (1, -1, 4) with normal vector A.

Solution: The equation must be (1, 2, 3) . X = d for some constant d. But since P is on the plane, if we set X = P, we must get the correct value of d. Thus d = (1, 2, 3) . (1, -1, 4) = 1 -2 + 12 = 11. The equation is A . X = 11.

### Unit Normal Vector

A unit vector is a vector of length 1. Any nonzero vector can be divided by its length to form a unit vector. Thus for a plane (or a line), a normal vector can be divided by its length to get a unit normal vector.

Example: For the equation, x + 2y + 2z = 9, the vector A = (1, 2, 2) is a normal vector. |A| = square root of (1+4+4) = 3. Thus the vector (1/3)A is a unit normal vector for this plane. Also, (-1/3)A is a unit vector.

Unit normal vectors: (1/3, 2/3, 2/3) and (-1/3, -2/3, -2/3)

Exercise: Find a unit normal vector for the plane with equation -2x -4y -4z = 0. How is this related to the example? Could you use the example to find the unit normal in this case?

Exercise on Lines in the Plane: Continuing with line m with equation 2x + 3y = 6, find the unit normal vectors for this line. Also, find a unit normal vector for line OA. Look for relationships.

### Equations of Lines in Space

We have seen that one equation of the form A . X = h defines a line in the plane or a plane in 3-space. In each case, we can motivate this informally by saying that the space of solutions has dimension one less than the dimension of the containing space.

This intuitive idea is made rigorous in all dimensions in a linear algebra course.

A line in space cannot be given by one linear equation, since for any nonzero vector A, such an equation has a plane as a solution.

But a line is the intersection of two planes, so if we have two such planes, with two equations A . X = h and B. X = k, then the solution set of both equations togeteher is the line. Conversely, if we have two such equations, we have two planes. The two planes may intersect in a line, or they may be parallel or even the same plane.

The normal vectors A and B are both orthogonal to the direction vectors of the line, and in fact the whole plane through O that contains A and B is a plane orthogonal to the line.

### Normal Vectors and Cross Product

Given two vectors A and B, the cross product A x B is orthogonal to both A and to B. This is very useful for constructing normals.

Example (Plane Equation Example revisited) Given, P = (1, 1, 1), Q = (1, 2, 0), R = (-1, 2, 1). Find the equation of the plane through these points.

First, the normal vector is the cross product of two direction vectors on the plane (not both in the same direction!).

Let one vector be PQ = Q - P = (0, 1, -1) and the other be PR = R - P = (-2, 1, 0). The cross product

(Q - P) x (R - P) = (1, 2, 2) = normal vector A and the equation is A . X = d for some d

Using the method in the example above, we can find d = A . P = 5. Thus the equation is A . X = 5, which is the same as one of the equations in the earlier example.

Exercise. Check that for this A, A . Q and A . R also = 5. Why is this so?

Exercise. Find a unit normal for this plane. What is the equation if A is chosen to be the unit normal?

### Dihedral Angles and Normal Vectors

Given two planes, the measure of the dihedral angle between the two planes is defined as the measure of an angle formed by intersecting the two planes with another plane orthogonal to the line of interesection. (There are two angles - a pair of supplementary angles.)

The angle measure between the normal directions of the two planes is the same as the measure of the dihedral angles, so the dihedral angle can be measured by taking dot product of the normal directions and using the Cosine Theorem for Dot Products.

Reference:

Mathworld: Dihedral Angles

Mathforum: Octahedron

## Reference:

Normal vectors and equations at Texas A&M