Determine the equation for a plane P that goes through a line D1 and is parallel to another line D2

Question

Looking forward to seeing if any of you can help me with this.

Determine the equation for a plane P that goes through a line $ D_1 = \frac{x+1}{2} = \frac{y-1}{-1} = \frac{z-2}{3}$ and is parallel to another line $ D_2 = \frac{x}{-1} = \frac{y+2}{2} = \frac{z-3}{-3}$

I found out so far that the direction of D1 is $ ( 2,-1,3) $ and the direction of D2 is $ (-1,2,-3) $

Please explain to me theoretically what happens next and why, and PLEASE give me an example so I can finally figure this out.

Answer 1

Hint:

There results the cross-product $\vec n=\vec v_1\times \vec v_2$ of these directing vectors is normal to the plane. Considering any point on $D_1$, say $M_1(-1,1,2)$, if $(a,b,c)$ are the coordinates of $\vec n$, an equation is $$ax+by+cz=-a+b+2c.$$

Answer 2

If you work in homogeneous coordinates, you can view this problem as one of finding the plane defined by three points: the direction vectors of the two lines (represented by points at infinity) and a finite point on the first line, all of which you can read from the equations of the two lines.

You’ve found the two direction vectors, and a point on $D_1$ is, by inspection $(-1,1,2)$. The plane is represented by a homogeneous vector $\mathbf p$ with the property that its dot product with any point on the plane vanishes. (It’s components are just the coefficients of the implicit plane equation $ax+by+cz+d=0$.) So, applying this condition to the three points gives you a system of homogeneous linear equations in the components of this vector, therefore it lies in the null space of the matrix $$\begin{bmatrix} -1&1&2&1 \\ 2&-1&3&0 \\ -1&2&-3&0 \end{bmatrix}.$$ Such a vector can be solved by any of the usual methods, such as row reduction, producing $[1,-1,-1,4]^T$, which corresponds to the implicit Cartesian equation $x-y-z+4=0$.

Equivalently, the homogeneous coordinate vectors of these three known points are linearly independent, and any point on the plane can be written as a linear combination of them, so an equation of the plane is $$\begin{vmatrix} x&y&z&1 \\ -1&1&2&1 \\ 2&-1&3&0 \\ -1&2&-3&0 \end{vmatrix} = 0,$$ which simplifies to the same equation as derived above. This approach uses the fact that the determinant of a matrix vanishes iff its rows are linearly dependent. (This determinant is a four-dimensional analogue of the cross product.)

Answer 3

If you work in homogeneous coordinates, you can view this problem as finding the plane defined by three points: the direction vectors of the two lines (represented by points at infinity) and a finite point on the first line, all of which you can read from the equations of the two lines.

You’ve found the two direction vectors, and a point on $D_1$ is, by inspection $(-1,1,2)$. The plane is represented by a homogeneous vector $\mathbf p$ such that the dot product of $\mathbf p$ with any point on the plane is zero. The components of $\mathbf p$ are just the coefficients of the Cartesian equation $ax+by+cz+d=0$. This vector is thus the solution to a system of linear equations in those coefficients, so the plane vector is a null vector of the matrix $$\begin{bmatrix} -1&1&2&1 \\ 2&-1&3&0 \\ -1&2&-3&0 \end{bmatrix}.$$ This can be found by any of the usual methods, such as row reduction. The null space is the span of $[1,-1,-1,4]^T$, which corresponds to the Cartesian equation $x-y-z+4=0$.