Positional relationship: line and plane
We examine three cases in which a line can relate to a plane.
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Study Positional relationship: line and plane
Let a line \(g\) be given in parameter form \[g=\left\{\mathbf{p} + \lambda \mathbf{a}:\ \lambda\in\mathbb{R} \right\}\] and let a plane \(E\) be given in coordinate form.
Then one may put the 3 components of the general vector \(\mathbf{p} + \lambda \mathbf{a}\) of the line \(g\) into the equation of the coordinate form of \(E\). This results in an equation with variable \(\lambda\) for which there are three cases:
the equation has no solution: then \(g\) and \(E\) have no common point.
the equation has a unique solution \(\lambda\): then there is an intersection point. Its position vector can be determined by substituting the solution \(\lambda\) into the general vector \(\mathbf{p} + \lambda \mathbf{a}\).
the equation has infinitely many solutions: \(g\) is contained in \(E\).
Example 1. We check whether the following line \(g\) and the following plane \(E\) have an intersection point: \[g=\left\{\begin{pmatrix} 2\\-3\\0 \end{pmatrix} + \lambda \begin{pmatrix} 1\\-2\\1 \end{pmatrix}:\ \lambda\in\mathbb{R} \right\}\ \ \text{ and } \ \ E=\left\{\begin{pmatrix} x_1\\x_2\\x_3 \end{pmatrix}\in\mathbb{R}^3:\ 2x_1-x_2-2x_3=1 \right\}.\] Every point of \(g\) has a position vector of the form \[\begin{pmatrix} x_1\\x_2\\x_3 \end{pmatrix}=\begin{pmatrix} 2\\-3\\0 \end{pmatrix} + \lambda \begin{pmatrix} 1\\-2\\1 \end{pmatrix} =\begin{pmatrix} 2+\lambda\\ -3 -2\lambda \\ 0 + \lambda \end{pmatrix}.\] Such a point lies in \(E\) if and only if it satisfies \(2x_1-x_2-2x_3=1\); hence: \[2\cdot(2+\lambda)-(-3-2\lambda)-2\cdot(0+\lambda)=1.\] This equation has a unique solution: \(\lambda=-3\). Hence, there is an intersection point. Its position vector is \[\begin{pmatrix} 2\\-3\\0 \end{pmatrix} + (-3) \begin{pmatrix} 1\\-2\\1 \end{pmatrix} = \begin{pmatrix} -1\\3\\-3\end{pmatrix}.\] Hence, the intersection point is \(S=(-1,3,-3)\).
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