two linear algebra questions?

Question

1. if you have two vectors, how would you theoretically find the equation of the plane they span (you're not given a point in the plane)

2. if you have two planes that intersect at a line, how do you find the equation of the orthogonal vector from that line to a point?


thanks!!

Answer 1

A plane is determined by three points or a point and two vectors. Two vectors alone do not determine a single plane but a family of planes.

Let P be any point, let A and B be any two such that A x B (the cross product of A and B) is not 0.

Then P + xA + yB determines the unique plane through P (that is spanned by A and B.

There are lots of ways to convert this vector equation into an equation involving whatever coordinate system you'd like to use.

For question 2, let a point P and direction vector V determine the line L.

Let Q be any point off that line. Let T be any point on the line L. Then the line QT is perpendicular (orthogonal) to the line L if and only if:

(T-Q).V = 0

That is, the dot product of the vector from Q to T and the vector V must be 0.

Since T is on the line L, there must be a scalar t such that

T = P + tV

These two equations suffice. One way to look at this is with vector algebra:

(T - Q).V = (P + tV - Q).V = P.V - Q.V + tV.V

P.V and Q.V are both constants.

V.V = |V|^2 which is constant.

So you have a single equation that you can solve for t.

With t, you can compute T

So the problem becomes one of finding P and V.

One of representing a plane is with a point and two vectors that span the plane. Another is with a point and a vector perpendicular the plane.

The two are equivalent. For example, above we had the plane defined by the point P and the vectors A and B. Let C be the cross-product of A and B.

The the same plane is specified as the plane through P perpendicular to C, or the set of points Q such that

(Q - P).C = 0

Now let P and C determine one such plane, and Q and D determine another. Then the line of their intersection has to perpendicular to both vectors. This means the direction vector V for the line can be given by C x D, the cross-product of the two vectors.

Now we need any point on the intersection. Consider the vector E = C x V. It is perpendicular to vector C so the line

P + tE is in the first plane for all t. So lets find a t that puts it in the second plane as well:

(P + tE - Q).D = 0 = P.D - Q.D + (tE).D

Again:

P.D is a constant, as is Q.D.

(tE).D = t(E.D) and we can compute E.D so we have one equation in one unknown.

Computing for t and then P+tE we have a point on the intersection and a direction on the intersection. Given the equation for the line we can find the orthogonal vector for any other point.

All this seems like a lot of work, but it is really very simple once you understand it. Obviously, if you had understood before, you wouldn't have needed to ask the question in the first place, so this answer is very detailed.