Find the equation of a plane containing a point and passing through a line

Question

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.

Answer 1

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 $A,B,C$.

Answer 2

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).

Answer 3

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 $$\underline{r}\cdot\underline{n}=\underline{a}\cdot\underline{n}$$

Answer 4

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 $4x+6y+5z=1$.