Direction vector of intersecting straight lines and equation of the plane passing through it

Question

Problem:

Find the equation of the plane passing through these intersecting straight lines

\begin{align*} \begin{cases} x = 2z + 1\\ y=3z+2 \end{cases}\quad\text{and}\quad \begin{cases} 2x = 2 - z\\ 3y=6+z \end{cases} \end{align*}

How do I find the direction vector?

Answer 1

The direction of the line defined by the intersection of the planes $x - 2z = 1$ and $y - 3z = 2$ is given by the cross product $d_{1} = (1,0,-2)\times(0,1,-3)$.

Similarly, the direction of the line defined by the intersection of the planes $2x + z = 2$ and $3y - z = 6$ is given by the cross product $d_{2} = (2,0,1)\times(0,3,-1)$.

Consequently, the normal direction of the plane passing through both lines is given by $d = d_{1}\times d_{2}$.

Since you have the normal direction of the plane, you just need to determine one of its points in order to describe it properly.

More precisely, one has the following system of equations \begin{align*} \begin{cases} x = 2z + 1\\\\ y = 3z + 2\\\\ 2x = 2 - z\\\\ 3y = 6 + z \end{cases}\Longleftrightarrow P = (x_{0},y_{0},z_{0}) = (1,2,0) \end{align*}

Finally, the equation of the plane is given by

$$\langle (x,y,z) - (x_{0},y_{0},z_{0}),d\rangle = 0$$

Answer 2

Well, since both lines lie on the plane, its normal will be perpendicular to them, so you can take the cross product of their direction vectors. Each line is defined as the intersection of two planes, so you take the cross product of their normals to get the direction vector of the line of intersection. Once you’ve done all of this, you’ll still need to find a point on the plane in order to construct an equation for it.

Alternatively, here’s a way to construct the equation of the plane without explicitly computing either a point on it or its normal. Every plane that includes the first line has an equation that’s a linear combination of the two plane equations that define the line, i.e., $$\lambda(x-2z-1)+\mu(y-3z-2)=0.\tag{*}$$ We want this plane to include the other line, which means that the system of equations formed from this combined equation and the two plane equations that define the second line must have an infinite number of solutions. For this to occur, the determinant of the coefficient matrix $$\begin{bmatrix}1-\lambda & \mu-2\lambda & -3\mu \\ 2&0&1\\0&3&-1\end{bmatrix}$$ must vanish. Expanding this determinant produces a simple homogeneous linear equation in $\lambda$ and $\mu$, any nontrivial solution of which you can plug back into (*) to obtain an equation for the plane defined by the two lines.