# Some vector concepts

1. Mar 1, 2009

### nns91

I am not sure about these things about vector:

- Is it true to say that a vector n orthogonal to vector n1 and n2 is the cross product of n1 and n2 ?

- Is it true to say that if a line is parallel to a plane, the normal vector of the plane is orthogonal to the parallel vector of the line ?

- Is the acute angle between two planes the angle created by two normal vectors of those planes ?

2. Mar 1, 2009

### phreak

This isn't true, because there are many vectors orthogonal to both n1 and n2 (since the orthogonal complement of {n1,n2} is an entire subspace). In the case of R^3, the subspace is a "line" of vectors, and you can find the vector that is the cross product of n1 and n2 by specifying its norm as |n1|*|n2| and then finding its direction.

3. Mar 1, 2009

### csprof2000

1.) No. There are always an infinite number of vectors which are orthogonal to any two vectors n1 and n2. Only one of these will be the cross product.

2.) Yes. Because a line parallel to a plane would lie in the plane after a suitable translation, and if it lies in the plain, well, it's orthogonal to the plane's normal.

3.)Yes. Assume intersecting but not coincident planes. Their intersection is a line lying in both planes. Now imagine rotating these pictures so that you're looking down this line; the cross section you see will be two intersecting lines, one for each of the two planes. Now, go out a little from their intersection point (in the cross section) and draw two intersecting normal vectors, one originating from one line, one from the other. Then it follows from a simple geometric construction that the acute angle formed by the intersecting normal vectors is, in fact, equal to the angle formed by the plane-lines. I assume this is what you were asking.

4. Mar 1, 2009

### nns91

Thanks.

Then how should I find a vector that orthogonal to both n1 and n2 ?

5. Mar 1, 2009

### phreak

Take their cross product. Don't confuse the statement and its contrapositive.

6. Mar 2, 2009

### mathman

In 3 space, the cross product gives a vector perpendicular to the given pair of vectors. Any scalar multiple of the cross product is also perpendicular to the given pair.

7. Mar 3, 2009

### HallsofIvy

I think you mean "converse" rather than "contrapositive".

8. Mar 3, 2009

### phreak

Yes, of course. You actually should confuse the statement and its contrapositive, since they're equivalent. Sorry for the confusion.