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.