How to determine equation for plane given line and its reflected line?

Question

I need help to find the possible equations of the plane $\pi$ given that the line $l_1 : (x,y,z) = t(1,2-2)$ turns into the line $l_2: (x,y,z) = t(2,-1,2)$ when reflecting from the plane.

I have determined that the intersection point between the two lines is $(0,0,0)$ and thus, that's a point of the equation of the plane that then can be described as $\pi: Ax+By+Cz=0$

How can I from this go on to find the coefficients of the equation of the plane?

Thanks!

Answer 1

The unit vectors in the directions of the given lines are $$\left(\begin{matrix}\frac13\\\frac23\\-\frac23\end{matrix}\right)$$ and $$\left(\begin{matrix}\frac23\\-\frac13\\\frac23\end{matrix}\right)$$

Adding these gives the direction of the line through the origin which bisects the angle between the lines and is co-planar with them. This line lies on the bisecting plane.

This is $$\left(\begin{matrix}1\\\frac13\\0\end{matrix}\right)\parallel\left(\begin{matrix}3\\1\\0\end{matrix}\right)$$

Normal to both lines is $$\left(\begin{matrix}1\\2\\-2\end{matrix}\right)\times\left(\begin{matrix}2\\-1\\2\end{matrix}\right)=\left(\begin{matrix}2\\-6\\-5\end{matrix}\right)$$

Normal to the required plane is therefore

$$\left(\begin{matrix}2\\-6\\-5\end{matrix}\right)\times\left(\begin{matrix}3\\1\\0\end{matrix}\right)=\left(\begin{matrix}1\\-3\\4\end{matrix}\right)$$

Hence the Cartesian equation of the plane is $$x-3y+4z=0$$

Answer 2

Yes, the two lines intersect at (0,0,0). Also, when t= 1 on the first line we have the point (1, 2, -2) which is at distance $\sqrt{1+ 4+ 4}= 3$ from the point of intersection. The point on the second line at that same distance from the point of intersection satisifies $\sqrt{4t^2+ t^2+ 4t^2}= \sqrt{9t^2}= 3t= 3$ so again t= 1 and that point is (2, -1, 2). The point midway between those two points is ((1+ 2)/2, (2- 1)/2, (2- 2)/2)= (3/2, 1/2, 0) and that is another point on the plane. We must have (3/2)A+ (1/2)B= 0 or B= -3A. So far the plane is Ax- 3Ay+ Cz= 0.

To find a third point on the plane, and so determine the plane, use the fact that the plane is perpendicular to both lines so contains a point on the cross-product of the vectors i+ 2j- 2k and 2i- j+ 2k. That cross product is $\left|\begin{array}{ccc}i & j & k \\ 1 & 2 & -2 \\ 2 & -1 & 2 \end{array}\right|= 2i- 6j- 5k$. That line is (x, y, z)= t(2, -6, -5) and a point on that line is (2, -6, 5).

So the desired plane, which we so far have as Ax- 3Ay+ Cz= 0, contains the point (2, -6, -5) and we must have 2A+ 18A- 5C= 20A- 5C= 0. 5C= 20A. C= 4A.

The equation of the desired plane is 2Ax- 3Ay+ 4Az= 0, or, dividing by A, 2x- 3y+ 4z= 0.