Labware - MA35 Multivariable Calculus - Two Variable Calculus

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Normal Vectors for Linear Function Graphs

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A normal vector to a linear function graph is any vector which is perpendicular to that function graph.

One way to find such a vector is to use the fact that two nonzero vectors with a dot product of zero must be perpendicular. If we know two vectors that are parallel to the plane but not parallel to each other, we can find a normal vector by finding a vector whose dot product with each of these two vectors is zero.

For example, for the graph of the function

f(x, y) = px + qy + k,

we can choose the vectors (1, 0, p) and (0, 1, q), both of which are parallel to the function graph. For any normal vector N, it must be true that

N⋅(1, 0, p) = 0 and

N⋅(0, 1, q) = 0.

This gives us a system of two linear equations for the three components of N, resulting in a range of possibilities. We can add the restriction Nz = 1 and we get

N= (-p, -q, 1).

We can use the same method for the graph of the implicit function

ax + by + cz = d.

Two vectors which are parallel to the graph but not to each other are (c, 0, -a) and (0, c, -b).

Again, we have a system of two linear equations for the three components of N. This time, we can add the arbitrary restriction x = a and we get the normal vector (a, b, c).

Demos

Exercises

  • 1. Find normal vectors for the graphs of the following explicit linear functions:
    • f(x, y) = 0
    • f(x, y) = 2x
    • f(x, y) = 3y
    • f(x, y) = 2x + 3y
    • f(x, y) = 6x + 11y
  • 2. Find normal vectors for the graphs of the following implicit linear functions:
    • x + y + z = 1
    • 3x + 4y + 5z = 12
    • 5x + 3y + 4z = 12
    • 4x + 3y + 5z = 1
    • y = 23
    • x = 7
    • z = 9
  • 3. How does any general expression for the normal vector to a plane depend on k for f(x, y) = px + qy + k and d for ax + by + cz = d. Why is this?