Science:Math Exam Resources/Courses/MATH152/April 2012/Question 01 (c)/Solution 2

From 1(a) we have that the normal directions to the two planes are

${\begin{aligned}\mathbf {n} &=[2,1,-1]\\\mathbf {m} &=[1,1,0]\end{aligned}}$

We know that the line must lie on both planes and therefore the direction of the line must be orthogonal to both normals. Therefore we can find it by taking the cross product,

$\mathbf {n} \times \mathbf {m} =[2,1,-1]\times [1,1,0]={\textrm {det}}{\begin{pmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\2&1&-1\\1&1&0\end{pmatrix}}=[1,-1,1].$

We then must identify a point on the line. To do this we attempt to find one of the intercepts with the coordinate planes (one of $x_{1}$, $x_{2}$, or $x_{3}$ being zero). Set $x_{1}$ to be zero, from the plane equation for $S_{2}$, this tells us that $x_{2}=1$. Then using the plean equation for $S_{1}$, we get that $x_{3}=0$. Therefore, the point $(0,1,0)$ is on the line. This tells us that we can write the equation of the line as

${\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}={\begin{bmatrix}0\\1\\0\end{bmatrix}}+t{\begin{bmatrix}1\\-1\\1\end{bmatrix}}$

which is the parametric equation of the line.