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?