The intersection between a line and a plane

Consider a straight line $g$ that passes through the point $U$ and has direction vector $\vec{v}$ , and a plane $E$ that contains the point $A$ and has normal vector $\vec{n}$ . What is the relationship between $g$ and $E$ ? There are basically two possibilities:

Parallel

$g$ is parallel to $E$ , in which case it is (see figure)

\vec{v} \perp \vec{n}

Assume now that $g || E$ . We can now check if $g$ is actually in the plane, or not. All we have to check is if $U\in E$ . If this is the case then the whole line has to be in $E$ (because $g$ is parallel to $E$ ). If $U\not\in E$ , then $g$ is not in $E$ .

Intersecting

$g$ is not parallel to $E$ , in which case there is an intersection point $S$ between $g$ and $E$ .

Let us find the intersection point $S$ . As $S$ is on $g$ as well on $E$ , we know that $S$ must fulfil the following condition:

\boxed{\overrightarrow{US} \parallel \vec v \text{ and } \overrightarrow{AS} \perp \vec n}

In principle, this is enough to determine $S$ , but let us make this more explicit by unpacking the equations that are hidden behind these two conditions. To do so, we first set $S(x\vert y \vert z)$ . As $S$ is on $E$ , it has to fulfil the normal equation of $E$ :

n_x x + n_y y + n_z z=d

and because $S$ is also on $g$ , it must fulfil the equation of the straight line as well:

\left(\begin{array}{ccc} x \\ y\\ z \end{array}\right) = \left(\begin{array}{ccc} U_x \\ U_y\\ U_z \end{array}\right)+c\cdot \left(\begin{array}{ccc} v_x \\ v_y\\ v_z \end{array}\right)

Thus, we end up with four equations:

\left\vert\begin{array}{rll} n_x\cdot x + n_y \cdot y + n_z \cdot z &=& d\\ x &=& U_x+c v_x\\ y &=& U_y+c v_y\\ z &=& U_z+c v_z\\ \end{array}\right\vert

Solve this system of equations to find $S$ . We demonstrate this using an example.

Example 1

Plane $E$ contains the point $A(0\vert 0\vert 9)$ and has normal vector $\vec{n}=\left(\begin{array}{r} 1\\ 4\\ -3 \end{array}\right)$ . Straight line $g$ passes through the point $U(7\vert 2\vert 9)$ and has direction vector $\vec{v}=\left(\begin{array}{r} -2\\ 3\\ 4 \end{array}\right)$ . Find the intersection point $S$ between $g$ and $E$ .

Solution

$g$ is not parallel to $E$ , as $\vec{v}\bullet \vec{n}=-2+12-12=2\neq 0$ , so there is a single intersection point $S(x\vert y\vert z)$ . The coordinates of $S$ must fulfil the equation of the plane and the equation of the straight line:

x+4y-3z=-27

and

\left(\begin{array}{ccc} x \\ y\\ z \end{array}\right) = \left(\begin{array}{ccc} 7 \\ 2\\ 9 \end{array}\right)+c\cdot \left(\begin{array}{ccc} -2 \\ 3\\ 4 \end{array}\right)

Thus we have the four equations

\left\vert\begin{array}{rll} x + 4y -3z &=& -27\\ x &=& 7-2c\\ y &=& 2+3c\\ z &=& 9+4c\\ \end{array}\right\vert

Let us first find out the value $c$ by inserting the lower three equations into the first one:

7-2c+4\cdot (2+3c)- 3\cdot (9+4c)=-27

And it follows $c=7.5$ . It follows

\begin{array}{lll} x &=&-8\\ y &=& 24.5\\ z &=& 39 \end{array}

The intersection point is $\underline{S(-8\vert 24.5\vert 39)}$ .

Exercise 1

Plane $E$ contains the point $A(4\vert 4\vert 2)$ and has normal vector $\vec{n}=\left(\begin{array}{r} -2\\ -2\\ 3 \end{array}\right)$ . Straight line $g$ passes through the point $U(1\vert 2\vert -3)$ and has direction vector $\vec{v}=\left(\begin{array}{r} 1\\ 1\\ 2 \end{array}\right)$ . Find the intersection point $S$ between $g$ and $E$ .

Solution

$g$ is not parallel to $E$ , as $\vec{v}\bullet \vec{n}=2\neq 0$ , so there is a single intersection point $S(x\vert y\vert z)$ . As $S$ is on both $g$ and $E$ , its coordinates must fulfil the equation of the plane and the equation of the straight line:

-2x-2y+3z=-10

and

\left(\begin{array}{ccc} x \\ y\\ z \end{array}\right) = \left(\begin{array}{ccc} 1 \\ 2\\ -3 \end{array}\right)+c\cdot \left(\begin{array}{ccc} 1 \\ 1\\ 2 \end{array}\right)

Thus we have the four equations

\left\vert\begin{array}{rll} -2x-2y+3z &=& -10\\ x &=& 1+c\\ y &=& 2+c\\ z &=& -3+2c\\ \end{array}\right\vert

Inserting $x, y$ and $z$ into the first equations, we get

-2(1+c)-2(2+c)+3(-3+2c)=-10

and thus $2c = 5$ , from which follows $c=2.5$ . We get

\begin{array}{lll} x &=&3.5\\ y &=& 4.5\\ z &=& 2 \end{array}

The intersection point is $\underline{S(3.5\vert 4.5 \vert 2)}$ .

Exercise 2

Plane $E$ contains the point $A(0\vert 0\vert 9)$ and has normal vector $\vec{n}=\left(\begin{array}{r} 0\\ 4\\ -3 \end{array}\right)$ . The straight line $g$ passes through the point $U(7\vert 2\vert 9)$ and has direction vector $\vec{v}=\left(\begin{array}{r} -2\\ 3\\ 4 \end{array}\right)$ . Find the intersection point between $g$ and $E$ .

Solution

There is no intersection point, as $g$ and $E$ are parallel ( $\vec{v}\bullet \vec{n}=0+12-12=0$ ) and $g$ is not in $E$ ( $\overrightarrow{AU} \bullet \vec{n} = 0+8+0 \neq 0$ ). Note that if you try to calculate the coordinates of the intersection point, you end up with the equation

0 + 12c+8 -12c = 0\rightarrow 8=0 \,???

In other words, there is no such $c$ , so no intersection point.

If $g$ is on $E$ , we would get an equation of the form

c=c

0=0

Thus, there is no condition for $c$ and we can choose any value we want (the equation is always correct). In other words, there are infinitely many intersection points. To find one, just choose an arbitrary value for $c$ .