Problems from Relative positions of a straight line and a plane

Considering the straight line $r : {\begin{array}{rcl} 2 x - y + z - 2 & = & 0 \\ x + y + 2 z - 7 & = & 0 \end{array}$ determine its relative position to the plane $π : 3 x - y - z = 5$ .

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Development:

To find the relative position of the straight line $r$ with the plane $π$ , we study the compatibility of the system formed by three equations: $| M | = | \begin{matrix} 2 & - 1 & 1 \\ 1 & 1 & 2 \\ 3 & - 1 & - 1 \end{matrix} | = - 2 - 6 - 1 - 3 - 1 + 4 = - 9 \neq 0$ Therefore the straight line and the plane are secant since $r a n k (M) = 3$ .

Solution:

The straight line $r$ and the plane $π$ are secant.

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