Matrix equation for two perpendicular lines intersecting at an end point

Question

In my question there is a plane E and two lines (G, H) in the plane E. The lines are perpendicular and intersect at an end point.

E: x = (2 1 4) + u1(0 3 4) + u2(3 10 5)
G: x = (2 1 4) + v(3 7 1)
H: x = (2 1 4) + w(x, y, z)

now i know two things:

1. dot product of G and H is zero because they are orthogonal so i have equation:

   (x y z) . (3 7 1) = (0 0 0)

solving this i get equation:

   3x + 7y + z =0  ----- (1)

2. because line H lies in Plane E we can equate below determinant to zero:

\begin{pmatrix} {x} & 0 & 3 \\ {y} & 3 & 10 \\ z & 4 & 5\\ \end{pmatrix} = 0

solving above we get:

 -25x +  12y - 9z = 0 -----(2)

Now i need one more equation to solve for x, y and z. How can i get the third equation ?

Hint: Something to do with lines being perpendicular and intersecting at an endpoint.

Answer 1

You can't get a third equation. The vector $(x,y,z)$ is only determined up to a constant.

Consider $(x,y,z)$ as a direction for line $H$. This line needs to lie in plane $E$ and be perpendicular to line $G$. If you think about it, if $(x,y,z)$ is a vector parallel to plane $E$ and perpendicular to line $G$, then so is any multiple of $(x,y,z)$ (thus there are infinitely many solutions).

So to find an answer, just pick a random value for one of your variables (for example set $z=0$). Then you can solve. :)

If you know about cross products, there is a different approach. First cross $(0,3,4)$ and $(3,10,5)$ [parallel to the plane] to get a vector, $v$, which is perpendicular to plane $E$. Then take $v$ and cross with $(3,7,1)$ [parallel to line $G$]. The resulting vector will be perpendicular to both $(3,7,1)$ (thus perpendicular lines is satisfied) and perpendicular to $v$ (hence parallel to the plane).