The plane

Problem

Given the plane \(E:3x_2+4x_3=5\),

1. explain the special location of the plane \(E\) in the coordinate system.

2. verify by calculation whether a sphere with center \(Z(1|6|3)\) and radius \(7\) intersects the plane \(E\).

Solution of part a

The equation for the plane is independent of the \(x_1\)-component. Therefore, the plane \(E\) is aligned parallel to the \(x_1\)-axis.

We can use Sage to depict the plane in three dimensions by solving the plane’s equation for \(x_3\).

Solution of part b

First, we will use Sage to get an overview of the problem.

It is easy to see that the sphere and the plane intersect.

In order to analytically verify this observation, we will determine the distance between the plane and the center of the sphere. The normal vector of the plane can be derived from its Hesse normal form:

\[\begin{split}\vec{n} = \left( \begin{matrix} 0\\ 3\\ 4 \end{matrix} \right),\end{split}\]

or in normalized form:

\[\begin{split}\vec{n}_0 = \left( \begin{matrix} 0\\ \frac{3}{5}\\ \frac{4}{5} \end{matrix} \right).\end{split}\]

In order to calculate the distance between the plane and the center of the sphere, we define points along the line aligned parallel to the normal \(\vec{n}_0\) and going through the center of the sphere as

\[\begin{split}\left( \begin{matrix} x_1\\ x_2\\ x_3 \end{matrix} \right) = \left( \begin{matrix} 1\\ 6\\ 3 \end{matrix} \right) + t\cdot \left( \begin{matrix} 0\\ \frac{3}{5}\\ \frac{4}{5} \end{matrix} \right).\end{split}\]

The value of \(|t|\) corresponds to the distance of the current point from the center of the sphere. The value \(t\) of the intersection point of the line and the plane can be determined by inserting the coordinates of the line into the plane’s equation. We find

\[\begin{split}&3 \cdot \left(6+\frac{3}{5}t\right) + 4 \cdot \left(3+\frac{4}{5}t\right)-5 =0 \quad\Leftrightarrow\\ &t=-5 \quad\Leftrightarrow\\ &|t|=5.\end{split}\]

Thus, the distance between the plane and the center of the sphere equals 5 and is smaller than the radius 7 of the sphere. Therefore, the sphere intersects the plane. The calculation can be verified by Sage: