Relative positions of a straight line and a plane

To determine the relative positions of a straight line $r (A^{'}; \vec{v})$ and a plane $π (P; \vec{u}, \vec{v})$ , we express the straight line by means of its implicit equations and the plane with its general equation:

$r : {\begin{array}{rcl} A_{1} x + B_{1} y + C_{1} z + D_{1} & = & 0 \\ A_{2} x + B_{2} y + C_{2} z + D_{2} & = & 0 \end{array} π : A x + B y + C z + D = 0$

Next we consider the system formed by three equations and write the matrix $M$ and the extended matrix $M^{'}$ associated with this system:

$M = (\begin{matrix} A & B & C \\ A_{1} & B_{1} & C_{1} \\ A_{2} & B_{2} & C_{2} \end{matrix})$

$M^{'} = (\begin{matrix} A & B & C & - D \\ A_{1} & B_{1} & C_{1} & - D_{1} \\ A_{2} & B_{2} & C_{2} & - D_{2} \end{matrix})$

According to the compatibility of the system we will determine their relative position:

Compatible system

Determined

$r a n k (M) = r a n k (M^{'}) = 3$

Determined Compatible system. The straight line and the plane are secant.

Indeterminate

$r a n k (M) = r a n k (M^{'}) = 2$

Indeterminate compatible system. The solutions depend on a parameter. The straight line is contained in the plane.

Incompatible system

$r a n k (M) = 2 \neq r a n k (M^{'}) = 3$

Incompatible system. The straight line and the plane are parallel.

Example

Determine the relative position of the straight line $r : (x, y, z) = (2, - 1, 0) + k \cdot (1, 2, 1)$ and the plane $π : (x, y, z) = (5, 0, 0) + l \cdot (3, 0, 1) + m \cdot (4, - 1, 1)$

We start by considering the matrix which columns are the components of the three director vectors (2 of the plane and 1 of the straight line) and we find its rank:

$| M | = | \begin{matrix} 1 & 3 & 4 \\ 2 & 0 & - 1 \\ 1 & 1 & 1 \end{matrix} | = 0$

Therefore $r a n k (M) = 2$ , and the straight line will be contained or it will be parallel to the plane.

To see what case we are faced with, we can take a point of the straight line $P$ and look to see if it belongs to the plane $π$ .

$P = (2, - 1, 0)$

We substitute in $π$ :

$\begin{array}{rcl} 2 & = & 5 + 3 \cdot l + 4 \cdot m \\ - 1 & = & - m \\ 0 & = & l + m \end{array}$

Therefore $m = 1, l = - 1$ , and we see that the point does not belong to the plane.

Thus, the straight line and the plane are parallel.