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

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.

• Hint: Pick two points on the line. You now have three points on the plane. Can you proceed from there? Sep 27, 2017 at 20:46
• yes. could I simply plug in 0 for t, then 1 for t, and get points Q(4,-1,0) and then R(3,1,-3), have (3,5,1) be P, then take PQ x PR to get the coeficients, and plug P in for the p values, and simplify to get my equation? Sep 27, 2017 at 20:59
• that leaves me with 8x +y -2z = 31 Sep 27, 2017 at 21:07

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 $$5x+y-z-19=0.$$

• im sorry im in a hurry and couldn't test all methods, can you tell me if my method i used in the comments above is right? Sep 27, 2017 at 21:14
• @2316354654 Your equation of the plane is wrong because the point $(3,5,1)$ don't belong to the plane. Sep 27, 2017 at 21:42
• then why does it say "find equation of plane through point (3,5,1) "? Sep 27, 2017 at 21:43
• @2316354654 $8\cdot3+5-2\cdot1-31\neq0$. Sep 27, 2017 at 21:44
• is the process at least right? maybe I made an error taking the cross product Sep 27, 2017 at 21:46

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.

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

• im sorry im in a hurry and couldn't test all methods, can you tell me if my method i used in the comments above is right? Sep 27, 2017 at 21:15
• or perhaps a better question would be how did you get (3,−1,0)=(1,−6,−1)and (−1,2,−3) Sep 27, 2017 at 21:29