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Some vector concepts

  1. Mar 1, 2009 #1
    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. jcsd
  3. Mar 1, 2009 #2
    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.
  4. Mar 1, 2009 #3
    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.
  5. Mar 1, 2009 #4

    Then how should I find a vector that orthogonal to both n1 and n2 ?
  6. Mar 1, 2009 #5
    Take their cross product. Don't confuse the statement and its contrapositive.
  7. Mar 2, 2009 #6


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    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.
  8. Mar 3, 2009 #7


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    I think you mean "converse" rather than "contrapositive".
  9. Mar 3, 2009 #8
    Yes, of course. You actually should confuse the statement and its contrapositive, since they're equivalent. Sorry for the confusion.
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