# Finding plane equation given two lines

Determine an equation of the plane containing the lines $\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-5}{6}$; $r=\lt1,-1,5\gt+t\lt1,1,-3\gt$.

I calculated the cross product between the directional vector of both lines to find the normal vector $n$, but when I looked for an intersection point $r_0$ to apply the formula: $\lt r-r_0\gt \bullet \ \ n$, I did not find any.

Can I use the point $\lt1,-1,5\gt$ given in the line $r$? Or is it not possible to find an equation of the plane containing two lines that do not intersect?

• Yes, you can use the $(1,-1,5)$. See my answer below. – Juniven Feb 12 '17 at 0:43
• Please correct the title of this question. The lines DO intersect at (1,-1,5). Plug in x=1,y=-1 and z=5 into the first line, and you get 0=0=0. – Ameet Sharma Feb 12 '17 at 6:09

The first line is given by $$L_1:\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-5}{6}$$ and the symmetric form of the other line is $$L_2:\frac{x-1}{1}=\frac{y+1}{1}=\frac{z-5}{-3}.$$ Clearly, $L_1\cap L_2$ is the point $(1,-1,5)$. Clearly, $L_1$ and $L_2$ are not parallel. Write $$V_1=2\hat{i}-\hat{j}+6\hat{k}$$ and $$V_2=\hat{i}+\hat{j}-3\hat{k}.$$ Then, $$V_1\times V_2= \begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\ 2&-1&6\\ 1&1&-3\\ \end{vmatrix}=\hat{i}(3-6)-\hat{j}(-6-6)+\hat{k}(2+1)=-3\hat{i}+12\hat{j}+3\hat{k}.$$ Thus, the equation of the plane containing $L_1$ and $L_2$ is given by $$-3(x-1)+12(y+1)+3(z-5)=0.$$ That is, $$-3x+12y+3z=0.$$
• @Hydrous Caperilla The cross product is by definition perpendicular to both $V_1$ and $V_2$ and hence this cross product serves as a normal vector to the plane that contains $L_1$ and $L_2$ – Juniven Apr 13 '18 at 8:24
You should go back and check your work. The point $(1,-1,5)$ obviously satisfies the first equation, and the lines aren’t parallel, so if you’re not getting that as the point of intersection you’re making a mistake somewhere along the way.