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We discuss how to find implicit and explicit formulas for planes.

With our modern outlook, we think of slope as being synonymous with lines and derivatives. In single variable calculus, the derivative is the slope of the tangent line. However, this identification of “slope” as being the “key-player” is something of a new idea. Before calculus, people were often interested in finding normal lines to curves. While this might seem odd at first, consider this: A line is the set of all points normal to some vector. Check it out: From this we see that if $\vec {p}=\vector {x_0,y_0}$ is a point (denoted by a vector) on the line, and our normal vector is $\vec {n}=\vector {a,b}$, then

is an implicit formula for the line. This means that if points $(x,y)$ make the equation above true, then those points are on the line. As we will see, a nearly identical implicit formula defines a plane.

Give an implicit equation for a line that goes through $(-2,3)$ and is perpendicular to the vector $\vector {4,-5}$.
Quick, tell me a vector normal to the line $2x-3y = 4$.

### Implicit planes

We would like to know the implicit formula for a plane. Remember an implicit function in $\R ^3$ is one of the form: Here the dot product saves the day. Recall that if $\vec {n} = \vector {a,b,c}$ is any vector, and $\vec {x}= \vector {x,y,z}$, then the equation is solved by all vectors $\vec {x}$ that are orthogonal to $\vec {n}$. We plotted several such vectors below:

From this we see that

gives the formula for a plane. Since $\vec {0} = \vec {x}$ is a solution, this plane must pass through the origin. If we want our plane to be located anywhere in space, we must know a point on the plane, call it $\vec {p}=\vector {x_0,y_0,z_0}$. Putting this together, we can now see:

If you know

• a vector $\vec {n} = \vector {a,b,c}$ and
• a point (given by a vector) $\vec {p} = \vector {x_0,y_0,z_0}$

then,

is an implicit equation for a plane passing through the point $(x_0,y_0,z_0)$ with normal vector $\vec {n}$.

Find the implicit equation of a plane that passes through the point $(-1,-3,2)$ and with normal vector $\vector {3,-1,1}$. Check your answer by moving the sliders below:

Find an explicit formula for the plane above.

Normal vectors not only allow us to define equations for planes but also they help us describe properties of planes.

What is the (most obvious) normal vector for the plane
Which of the following planes are parallel to the plane $-6x+10y-2z = 4$?
$-2.46x+3.9y-0.82z=3$ $-2.07x+3.45y-0.69z = 8$ $10.38x-17.3y+3.46z=0$ $15.03x-25.1y+5.02z=0$
Which of the following planes are orthogonal to the plane $-6x+10y-2z = 4$?
$34x+16y-22z=0$ $-54x-32y+8z=-3$ $-36x-42y-102z=-17$ $54x+31y-2z=4$

### Parametric planes

Given any two nonzero vectors in $\R ^3$, $\vec {v}$ and $\vec {w}$, such that we can produce a parametric formula for a plane by writing where $\vec {p}$ is a vector whose “tip” is on the plane, and $\vec {v}$ and $\vec {w}$ are in the plane.

Given two nonzero vectors, $\vec {v}$ and $\vec {w}$, what does it mean for $\vec {v}\cross \vec {w} \ne \vec {0}$?
It means these vectors are parallel. It means these vectors are not parallel. It means these vectors are orthogonal. It means these vectors are not orthogonal.

The vector-valued formula for a plane is very similar to our formula for a line, where $\vec {v}$ is a vector that points in the direction of the line, both represent linear relationships, and hence we use similar notation for both.

Now that we have two methods of graphing planes, let’s use both of the representations at once!

Let $\vec {v} = \vector {1,-2,1}$ and $\vec {w} = \vector {-1,-2,1}$. Compute $\vec {v}\cross \vec {w}$.
Use your answer above to give an implicit equation for the plane that passes through the point $(3,2,1)$ that is normal to $\vec {v}\cross \vec {w}$. Check your answer by moving the sliders below:

Now give a vector-valued formula for the same plane using the vectors given above.