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Two plane have equations:

$p_1:3x-y+z=17$

$p_2:x+2y-z=4$

where $p_1$ is perpendicular to $p_2$

and point $M(1,1,2)$

Find a vector equation of the line through $M$ which is parallel to both planes

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closed as off-topic by Nosrati, José Carlos Santos, Gibbs, Brahadeesh, Adrian Keister Feb 7 at 14:38

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The line has direction perpendicular to both plane normal vectors. $$\mathbf{e} = \mathbf{n}_1 \times \mathbf{n}_2$$

$$ \mathbf{e} = \pmatrix{3 \\ -1 \\1} \times \pmatrix{1\\2\\-1} = \pmatrix{-1 \\ 4 \\ 7} $$

A line with direction $\mathbf{e}$ that passes through the point M has equation

$$ \mathbf{e} \times \left( \pmatrix{x\\y\\z} - \mathbf{M} \right) = 0 $$

$$\pmatrix{-1 \\ 4 \\ 7} \times \left( \pmatrix{x\\y\\z} - \pmatrix{1 \\ 1\\ 2} \right) = \pmatrix{0\\0\\0} $$

or

$$ \pmatrix{ x \\ y \\ z} = \mathbf{M} + t\, \mathbf{e} $$

$$\pmatrix{ x \\ y \\ z} = \pmatrix{1\\1\\2} + t\,\pmatrix{-1\\4\\7} $$

which also solves the above system of equations, given an arbitrary value $t$ that moves the point along the line.

Here $\times$ is the cross product.

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Normal vectors of $p_1$ and $p_2$ are: $\vec n_1=(3,-1,1)$ and $\vec n_2=(1,2,-1)$

The line is parallel to both planes when it is perpendicular to both normal vectors of the planes. To find a vector which is perpendicular to 2 vectors we use outer product.

$\vec v=\vec n_1 \times \vec n_2=(3,-1,1)\times (1,2,-1)=(-1,4,7)$

The equation of a line with $v$ vector and passing the point $M=(1,1,2)$ will be like below: $\dfrac{x-1}{-1}=\dfrac{y-1}{4}=\dfrac{z-2}{7}$

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