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Consider the plane with general equation 3x+y+z=1 and the plane with vector equation (x, y, z)=s(1, -1, -2) + t(1, -2 -1) where s and t are real numbers. Describe the intersection of these two planes.

I started by substituting the parametric equations into the general equation and got 0=9. Does that imply the planes are parallel and hence do not intersect?

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Yes, the planes are parallel and therefore do not intersect at a unique line, we can see this as the normal vector of the first plane is $\bigr(\begin{smallmatrix}3 \\ 1 \\ 1\end{smallmatrix}\bigl)$, and the normal vector of the second can be found by computing:

$$\begin{pmatrix}1 \\ -1 \\ -2\end{pmatrix}\times\begin{pmatrix}1 \\ -2 \\ -1\end{pmatrix}=\begin{pmatrix}-3 \\ -1 \\ -1\end{pmatrix}=-1\begin{pmatrix}3 \\ 1 \\ 1\end{pmatrix}$$

Therefore the two planes are parallel as their normal vectors are anti-parallel.

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As a test to see if the planes are parallel you can calculate the normalvectors for the planes {n1, n2}.

If $abs\left (\frac{(n1\cdot n2)}{\left | n1 \right |*\left | n2 \right |} \right )==1$ the planes are parallel.

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