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Find the general form equation of a plane that contains the point $(2, -2, 1)$ and passes through a line $(x, y, z) = (1, 2, -3) + t(2, -3, 2)$.

I managed to use the fact that it contains a given point to write the equation of the plane as $A(x-2) + B(y+2) + C(z-1) = 0$ but got stuck there.

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Actually you want to use the point $(2, -2, 1)$ to set up the equation, so you can use the line to solve it: $$A(x-2) + B(y+2) + C(z-1) = 0$$ and then solve $$A((1+2t)-2) + B((2-3t)+2) + C((-3+2t)-1) = 0$$ by plugging in some useful values of $t$. So $$A(2t-1) + B(-3t+4) + C(2t-4) = 0$$ and try out $t=\frac{1}{2}$, then $t=\frac{4}{3}$, then $t=2$ and solve for $A,B,C$.

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  • $\begingroup$ It looks like I misunderstood the problem (I translated it). I interpreted this as a line cutting through a plane (intersecting it at one point) - does you assume in your solution that the plane contains the line i.e. the line lies in the plane? $\endgroup$ – Matt Jan 15 '18 at 21:07
  • $\begingroup$ @Matt Yes, my answer assumes that the line is contained in the plane. $\endgroup$ – The Phenotype Jan 15 '18 at 21:08
  • $\begingroup$ Well it would be impossible to find the solution if the author of the question meant line just intersecting the plane, right? That would give two points and three are needed to determine a plane, correct? $\endgroup$ – Matt Jan 15 '18 at 21:09
  • $\begingroup$ @Matt Correct, unless the line intersected the plane orthogonally, but then you only need $(2, -3, 2)$ as a normal vector, and not the whole line. $\endgroup$ – The Phenotype Jan 15 '18 at 21:11
  • $\begingroup$ Right, that would give the normal vector to the plane, and $(2, -2, 1)$ could be used to find complete the equation. Thanks a lot. :) $\endgroup$ – Matt Jan 15 '18 at 21:14
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Hint:

Take two points on the line (you can chose $t=0$ and $t=1$), so, since the first given point is not on the line, you have tree not collinear points..... and you can find the searched plane (see here or here).

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HINT...You need two non-parallel vectors which are parallel to the plane, such as $$\left(\begin{matrix}2\\-3\\2\end{matrix}\right)$$ and $$\left(\begin{matrix}2\\-2\\1\end{matrix}\right)-\left(\begin{matrix}1\\2\\-3\end{matrix}\right)=\left(\begin{matrix}1\\-4\\4\end{matrix}\right)$$

Normal to the plane is the cross-product of these vectors. Then use the dot product formula for the plane $$\underline{r}\cdot\underline{n}=\underline{a}\cdot\underline{n}$$

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Moving to a projective space, you can find three points on the plane by inspection. In homogeneous coordinates, they are $[2:-2:1:1]$, $[1:2:-3:1]$ and $[2:-3:2:0]$. The latter is the point at infinity that corresponds to the direction vector in the parametric equation of the line. The plane $\mathbf\pi$ that contains them satisfies the equation $\mathbf\pi^T\mathbf p=0$ for each of these points $\mathbf p$. This gives you a system of three homogeneous linear equations for the components of $\mathbf\pi$. Their solution is a null vector of the matrix $$\begin{bmatrix}2&-2&1&1\\1&2&-3&1\\2&-3&2&0\end{bmatrix}.$$ Row-reducing this matrix produces $$\begin{bmatrix}1&0&0&4\\0&1&0&6\\0&0&1&5\end{bmatrix},$$ from which $\mathbf\pi=[4:6:5:-1]$, which corresponds to the implicit Cartesian equation $4x+6y+5z=1$.

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