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Cross Products Verifying Third Vector Orthoginality?

  1. Mar 6, 2013 #1
    I have a question. Lets say you have three distinct, non zero vectors that lie in the same plane. Can I verify the othoginality of the third vector using strictly cross products?

    I know I can just do the dot product and if it equals to 0 then its orthoginal to the plane; however, I came across a method online (cannot find the source) where you can strictly use cross products to verify orthoginality. This had me and my professor thinking about whether or not this may be true such as below:

    If i were to take the cross product of lets say A X B. Then take the cross product of A X C. If the resultant vectors of both A X B and A X C are equal, doesn't it technically imply that the third vector C is orthogonal?

    For real number values: Lets say the 3 vectors were <1,-1,1>,<-2,3,4>,<0,1,6>.

    A X B = | 1 -1 1 | = <-7,-6,1>
    | -2 3 4 |

    A X C = |1 -1 1| = <-7,-6,1>
    |0 1 6|

    Since both have the same normal vectors that are perpendicular to the plane, we can conclude that both lie within the same plane.

    (Sorry if I do not make sense, its rather late. Perhaps ill revise this tomorrow morning)

    Best Regards,
  2. jcsd
  3. Mar 6, 2013 #2


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    Do you mean that given A,B,C you want to know if A is orthogonal to B and C?

    if A X (B X C)=0
    then A.B=A.C=0
    as is obvious
  4. Mar 6, 2013 #3


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    What is it supposed to be orthogonal to? The other two vectors? This is impossible if they're all in a plane that contains 0, as in your example.

    If the plane isn't supposed to contain 0, then I see no reason to mention a plane, since there's always a plane that contains the three vectors.

    Orthogonal to what? To A and B? In this example, C is not orthogonal to both A and B:
    $$(1,0,0)\times (1,1,0)=(1,0,0)\times (0,1,0).$$
  5. Mar 6, 2013 #4
    You could reason this way. Let A,B,C be three non zero vectors and we want to check if they are coplanar. Since we are talking about vectors (as opposed to points), then I assume that in this context, coplanar means that the origin must also lie in this plane. Otherwise, this does not make sense as Fredrik stated. In other words, the triple product (A X B)*C must be zero.

    Following the OPs reasoning, we instead calculate A X B and A X C. If the three vectors were coplanar, then both these vectors would be orthogonal to that common plane, so they would be parallel. Two non zero vectors are parallel if their cross product is zero. So the test would be

    (A X B) X (A X C)=0
    Last edited: Mar 6, 2013
  6. Mar 8, 2013 #5
    You have to normalize and directly compare the cross product vectors and allow for a case similar to when A = (1,0,0), B = (1,1,0), C = (-1,1,0) and your algorithm compares AxB and AxC. The cross product vectors are the same except one is scaled by -1.
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