- I have found two planes trough the origin that meet the given plane
at right angles.
I found three points in the plane, getting the vectors between those points, and using that as a normal vector for the equation of the plane I am looking to find.
I don't understand how I am to describe all of these planes though. Would finding a basis of all vectors lying in the plane and using an arbitrary linear combination of these as our normal to the plane we are looking for be the right way?
- I took $(4, -1, 1)$ and $(1, -2, -3)$ as normal vectors to each plane.
Their cross product is $(5, 13, -7)$ is in the direction of the line
of intersection of the two planes, thus parallel to both.
$(x, y, z) = (x_0 + 5t, y_0 + 13t, z_0 -7t)$ is a parametric equation for all lines parallel to both. Is this correct? And what do they mean by refining this such that each line is listed once?
Any help is greatly appreciated!
Best Answer
Your approach to (a) will work, but there’s no reason to compute a new pair of basis vectors since you’ve already got one: the normals to the two distinct planes that you’ve already found. Indeed, all of the sought-after planes have equations that are linear combinations of the two equations that you’ve already found.
It sounds like you did more work than necessary to solve the first part of (a), though. You just need to find two linearly-independent vectors that are perpendicular to $(4,-8,11)$. A simple way to do this is to swap and negate vector entries. That is, given a nonzero vector $(a,b,c)$, its dot products with $(0,c,-b)$, $(-c,0,a)$ and $(b,-a,0)$ are all zero, and at least two of them are nonzero. In this case, you can pick any two of $(0,11,8)$, $(-11,0,4)$ and $(-8,-4,0)$ as the normals to the two planes you’re asked to find. Taking the first and third gives the plane equations $11y+8z=0$ and $2x+y=0$, and per the previous paragraph the one-parameter family of planes $(1-\lambda)(11y+8z)+\lambda(2x+y)=0$ contains all of the perpendicular planes through the origin.
For part (b) notice that for any two fixed points $(x_0,y_0,z_0)$ and $(x_1,y_1,z_1)$, if their difference is a multiple of the direction vector that you computed, then they generate the same line. To ensure that each line is only listed once, you need a way to generate a set of points that are on distinct lines that share this fixed direction vector. That describes, among other things, a plane perpendicular to these lines, so find a convenient parameterization of such a plane. (The plane doesn’t have to be perpendicular to the lines, but it shouldn’t be parallel to them.)