Relative positions of a straight line and a plane

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

$$r: \left\{\begin{array} {rcl} A_1x+B_1y+C_1z+D_1 & = & 0 \\ A_2x+B_2y+C_2z+D_2 &=& 0 \end{array}\right. \\ \pi: Ax+By+Cz+D=0$$

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

$$M=\begin{pmatrix} A & B & C \\ A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \end{pmatrix}$$

$$M'=\begin{pmatrix} A & B & C & -D \\ A_1 & B_1 & C_1 & -D_1\\ A_2 & B_2 & C_2 & -D_2 \end{pmatrix}$$

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

Compatible system

Determined

$$rank(M) = rank(M') = 3$$

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

Indeterminate

$$rank (M) = rank (M') = 2$$

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

Incompatible system

$$rank (M) = 2 \neq rank (M') = 3$$

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

Determine the relative position of the straight line $r: (x, y, z) = (2,-1, 0) + k \cdot (1, 2, 1)$ and the plane $ \pi: (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:

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

Therefore $rank (M) = 2$, 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 $\pi$.

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

We substitute in $\pi$:

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

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.