find equation of plane through point that contains line in parametric form

Question

find equation of plane through point (3,5,1) and contains line x=4-t, y=2t-1, z=-3t.

I was able to see how to do this with a parallel and perpendicular plane but I'm not sure how to apply the fact that the plane contains this line to find the plane. I have a feeling I have to use the point vertices where the x0 y0 and z0 spots are, and i'm thinking I have to do something like plug x=4-t into the x spot, or convert the line into vector form but I have a feeling that would be wrong, if I could just have a little clarification on the process that would be helpful. there's no example in the book, and the online ones are confusing, I'm hoping these plane equations will come together in my head soon because they all seem like different methods i just have to memorize.

Answer 1

Let $A(3,5,1)$, $B(4,-1,0)$ and $\vec{n}(a,b,c)$ be a normal of the plane.

Thus, $\vec{AB}(1,-6,-1)$ and $$(1,-6,-1)(a,b,c)=0$$ and $$(-1,2,-3)(a,b,c)=0$$ or $$a-6b-c=0$$ and $$-a+2b-3c=0,$$ which gives $b=-c$, $a=-5c$ and we can assume that $\vec{n}(5,1,-1)$.

Thus, we got an equation of the plane: $$5(x-3)+(y-5)-(z-1)=0$$ or $$5x+y-z-19=0.$$

Answer 2

Well, think about this, if the plane does contain that straight line, it means

1. ANY point from that straight line belongs to the plane too.
2. The direction vector of the straight line is also a direction vector of the plane.

So you know that you can easily get one point and the direction vector in parametric form in a glanze: Q(4,-1,0), and $\vec{v}=(-1,2,-3)$.

So you already know a point and a direction vector.

Now, check that the other point $P(3,5,1)$ is not in the straight line. If it is inside the line, the information is redundant and incomplete. If it is not in the straight line, you can plot the vector $\vec{u}=\vec{PQ}$, and that's the 2nd direction vector you need.

So you have a point (two indeed) and 2 direction vectors. That's enough for building the plane equation.

You know you can use $(x,y,z)\cdot \left(\vec{u}\times\vec{v}\right)=0$ , and you get the equation from there.

Answer 3

Hint: The normal to your plane should be perpinducular to $(3,5,1)-(3,-1,0)=(1,-6,-1)$ and $(-1, 2,-3)$.