Find an equation of the line of intersection of the two planes \[ x-2y+z=-1\quad \hbox{and}\quad 3x+y-z=4 \]

Solution First notice that the normals \(\mathbf{N}_{1} = \mathbf{i}\, -\, 2\mathbf{j}\, +\, \mathbf{k}\) and \(\mathbf{N}_{2} = 3\mathbf{i}\, +\, \mathbf{j}\, -\, \mathbf{k}\) of the two planes are not parallel. To find an equation of the line of intersection, we need a point on the line and the direction of the line. Since the line of intersection is perpendicular to both \(\mathbf{N}_{1}\) and \( \mathbf{N}_{2}\), it is parallel to the vector \(\mathbf{D}= \mathbf{N} _{1}\times \mathbf{N}_{2}\). \[ \mathbf{D=N}_{1}\times \mathbf{N}_{2}= \left|\begin{array}{r@{\quad}r@{\quad}r} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] 1 & -2 & 1 \\[3pt] 3 & 1 & -1 \end{array}\right| =\mathbf{i}+4\mathbf{j}+7\mathbf{k} \]

The vector \(\mathbf{D}\) gives the direction of the line. We can find a point on the line by locating any point common to both planes. For example, if \( z=0\), then \(x-2y=-1\) and \(3x+y=4\). Solving these equations simultaneously, we find \(x=1\) and \(y=1\). So, the point \((1,1,0)\) is on the line, and symmetric equations of the line of intersection are \[ \dfrac{x-1}{1}=\dfrac{y-1}{4}=\dfrac{z}{7} \]