Table of Contents
Miscellaneous Topics
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Courses

  1. Differential Equations
  2. Miscellaneous Topics

Miscellaneous Topics

  1. Vectors
    1. Fundamentals
    2. $\vec{a} \cdot \vec{b}$  Dot Product
    3. $\vec{a} \times \vec{b}$  Cross Product
    4. Normal $\vec{n}$ to Plane
  2. Coordinate Systems
    1. Polar $(r,\theta)$

Additional Resources

  1. WolframAlpha Examples
  2. Direction Field Plotter by Ariel Barton
  3. Phase Plane Plotter by Ariel Barton


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Normal Vector to Plane | youtube icon Topic Playlist

A quick way to create a vector perpendicular to a plane is to write the equation of the plane as a dot product of the coefficients on x, y, and z with vector [x,y,z] equal to a constant. The vector of coefficients is perpendicular to the plane, called a normal vector to the plane.

Creating a Normal Vector to $x - 2y + 3z = 0$

One way to create a normal vector to this plane is to use the properties of the dot product.

\begin{equation} x - 2y + 3z = 0 \qquad \langle 1,-2,3 \rangle \cdot \langle x,y,z \rangle = 0 \end{equation}

Since the RHS of this plane equation is 0 for this example, points that solve the equation of the plane are also vectors that lie in the plane because $\langle 0,0,0 \rangle$ is a solution to this plane equation. The tail of a vector starts at the origin and the head of a vector ends at a point in the plane. So if $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ solve the plane equation, meaning they are points in the plane, then the $\langle x_{1},y_{1},z_{1} \rangle$ and $\langle x_{2},y_{2},z_{2} \rangle$ vectors lie in the plane.

normal to plane

This means that the coefficients vector is orthogonal to the plane since the dot product is zero,

\begin{equation} \langle 1,-2,3 \rangle \cdot \langle x_{1},y_{1},z_{1} \rangle = 0 \qquad \langle 1,-2,3 \rangle \cdot \langle x_{2},y_{2},z_{2} \rangle = 0 \end{equation}

So a normal vector the this plane is simply,

\begin{equation} \vec{n} = \langle 1,-2,3 \rangle \end{equation}

Normal Vector to $ax + by + cz = d$

For a plane with equation,

\begin{equation} ax + by + cz = d \end{equation}

We can use the coefficients as in the above example for when $d=0$ to get the normal vector for this plane equation. This works because no matter what the constant $d$ is on the RHS, all the planes are parallel to each other for every value of $d$. Changing $d$ translates the plane but does not rotate it. To rotate the plane, you would have to change the coefficients $a,b,c$.

Then we can apply the same logic as the above example for when $d=0$ to get the normal vector for this plane equation.

\begin{equation} \vec{n} = \langle a,b,c \rangle \end{equation}