How to find the implicit ecuation of a plane that contains two lines

Question

So..I have 2 lines and I have to find the implicit ecuation of a plane that contains both. This are the lines

L1:(x,y,z)=(1,0,0)+t(2,0,-1)

L2:
x+3z=1
x+y+z=1

I know that both of the lines pass through P=(1,0,0) that will help me to calculate D. Ax+By+Cz=D

But I don't know what other vector should I use.

Sorry for my bad english and if the question is too basic. Thanks in advance.

Answer 1

Hint: take two vectors, one parallel to each line, and take their cross-product. This will give the normal to the plane. Use the normal as $(A,B,C)$. Then use $(A,B,C)\cdot (x-x_0,y-y_0,z-z_0)=0$.

$(2,0,-1)$ will do for the first. For the second use the difference of any two distinct points on the second line... so, say $(1,0,0)-(-5,4,2)=(6,-4,-2)$.

Now, $(2,0,-1)×(6,-4,-2)=\begin{vmatrix}i&j&k\\2&0&-1\\6&-4&-2\end{vmatrix} =(-4,-2,-8)$. Etc...

Answer 2

You have two lines with $$[x,y,z]=[1,0,0]+t[2,0,-1]$$ and $$[x,y,z]=[1,0,0]+s[-3,2,1]$$ and you want to compute the equation of a plane of the form $$Ax+By+Cz+D=0$$ You Can calculate Points which are located on both lines and plug them in the equation of the plane to Compute $$A,B,C,D$$ Can you do that? Both lines intersect each other in $P[1;0;0]$ then a Parameter form of youir plane is given by $$[x,y,z]=[1,0,0]+t[2;0;-1]+s[-3;2;1]$$

Answer 3

You need three points on your plane.

From the first line you have $(1,0,0)$ and $(2,0,-1)$ and from the second line you get $(1,1,0).$

From $(1,0,0)$ we get $$(z-1)=A(x)+B(y)$$

Plugging the other two points in the equation $$(z-1)=A(x)+B(y)$$ you find your $A,B$

I found the equation to be $$z=1-x$$