Finding equation of a plane given a line and two points

Question

Find, if possible, a plane that contains the line $L: \frac{x-2}{-3}=y-1=z-4$ and the points $A(1;0;1)$ and $B(1;3;1)$.

Well, firstly I found the components of the vector $\vec{AB}$, which are $(0;3;0)$ and the direction vector of the line, which is $(-3;1;1)$. Then, the cross product between $\vec{AB}$ and the direction vector is equal to the normal of the plane. After computing that, the result is $(3;0;9)$, so the plane is $3x+9z+d=0$, and if I put the coordinates of the point $B$ I get that $d=-12$. Is that right?

Answer 1

I claim the plane containing $L$ and $A$ does not contain $B$;

i.e., there is no plane containing $L, A,$ and $B$.

Letting $y=t$, we find from the equations for $L$ that $x=-3t+5$ and $z=t+3$;

thus, $L$ can be parametrized as $(5,0,3)+t(-3,1,1)$.

So the plane containing $L$ and $A=(1,0,1)$

is normal to $(-3,1,1)\times[(5,0,3)-(1,0,1)]=(2,10,-4),$

so its equation could be $2x+10y-4z=-2$.

Note that $A=(1,0,1)$ is in this plane, and any point on $L$ including $(5,0,3)$ is,

but $B=(1,3,1)$ is not.

Answer 2

We can convert directly from the given symmetric equation of $L$ to the parametric form $(2,1,4)+t(-3,1,1)$. So, the point $(2,1,4)$ must certainly lie on the plane, if it exists. The equation of this plane can be given in determinant form as $$\begin{vmatrix}x&y&z&1\\2&1&4&1\\1&0&1&1\\1&3&1&1\end{vmatrix} = 0.$$ If this plane contains $L$, then if we plug its parametric form into this determinant it must vanish for all values of $t$: $$\begin{vmatrix}2-3t&1+t&4+t&1\\2&1&4&1\\1&0&1&1\\1&3&1&1\end{vmatrix} = \begin{vmatrix}-3t&t&t&0\\2&1&4&1\\1&0&1&1\\1&3&1&1\end{vmatrix} = -30t,$$ which only vanishes when $t=0$. We therefore conclude that there is no such plane.