b2e703…: “Subsequent to the interesting circle”

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Given a normal vector $\vec {n} = \vector {a,b,c}$ and a point $(x_0,y_0,z_0)$ , the equation of the plane with normal vector $\vec {n}$ that passes through $(x_0,y_0,z_0)$ is given by

$a(x-x_0)+b(y-y_0)+c(z-z_0) = 0.$

Find the equation of a plane with normal vector $\vector {2,-1,9}$ that passes through $(2,-3,4)$ . Express your final answer in the form $ax+by+cz=d$ .

$2x+\left (\answer {-1}\right )y+\left (\answer {9}\right )z = \answer {43}.$

One important observation is that it’s not too difficult to find a normal vector to a plane once we have an equation that describes it.

Which of the following vectors is a normal vector for the plane $2x-y+4z=7$ ?

$\vector {0,1,2}$ $\vector {4,1,0}$ $\vector {2,-1,4}$ more than one of these none of these

We can easily extract a normal vector from the coefficients of $x$ , $y$ , and $z$ , but how do we verify that this vector is indeed normal to the plane?

We take the dot product of the $\vector {2,-1,4}$ and the plane. We take the dot product of the $\vector {2,-1,4}$ and a single vector parallel to the plane. We take the dot product of the $\vector {2,-1,4}$ and an arbitrary vector parallel to the plane.

To get started, let’s find a vector parallel to the plane. We can do this by finding two points on the plane. Since the plane gives a single constraint between $x$ , $y$ , and $z$ , we can freely choose any two of them, then use the equation of the plane to find the third.

For instance, if $x=2$ , $y=1$ , then $z= \answer {1}$ , and if $y=1$ and $z=2$ , then $x=\answer {0}$ , so the points $\left (2,1,\answer {1}\right )$ and $\left (\answer {0}, 1, 2\right )$ lie on the plane.

Thus, a vector parallel to the plane is $\vec {v} = \vector {2,\answer {0},\answer {-1}}$ .

Now, we check $\vec {n} \dotp \vec {v} = \answer {0}$ .

To establish that $\vec {n}$ is normal to the plane, we must establish that $\vec {n}$ is orthogonal to any vector that is parallel to the plane. To do this, pick two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ that lie on the plane.

A vector that is parallel to the plane that starts at $(x_1,y_1,z_1)$ and ends at $(x_2,y_2,z_2)$ is $\vec {v} = \vector {x_2-x_1,y_2-y_1,z_2-z_1}$ .

Now, $\vec {n} \dotp \vec {v} = \vector {x_2-x_1,y_2-y_1,z_2-z_1} \dotp \vector {2,-1,4} = 2(x_2-x_1) -(y_2-y_1) +4(z_2-z_1)$ , and we must establish that this is $0$ .

We can rearrange the equation.

$\begin{align*} 2(x_2-x_1) -(y_2-y_1) +4(z_2-z_1) &= 2x_2-2x_1 -y_2+y_1 +4z_2-4z_1 \\ &= \left[2x_2 -y_2+4z_2\right]-\left[2x_1 -y_1 +4z_1\right] \end{align*}$

Now, since $(x_1,y_1,z_1)$ and ends at $(x_2,y_2,z_2)$ lie on the plane, $2x_2 -y_2+4z_2 = \answer {7}$ and $2x_1 -y_1+4z_1 = \answer {7}$ , so

$\vec {n} \dotp \vec {v} = \left [2x_2 -y_2+4z_2\right ]-\left [2x_1 -y_1 +4z_1\right ] = \answer {7} -\answer {7} = 0.$

Thus, $\vec {n}$ is orthogonal to any arbitrary vector that is parallel to the plane, and thus $\vec {n}$ is a normal vector to the plane.