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Homework Help: Vector question (using the cross product I think)

  1. Sep 6, 2011 #1
    1. The problem statement, all variables and given/known data
    Describe all unit vectors orthogonal to both of the given vectors:

    [itex]\vec{a} = 2\vec{i} - 4\vec{j} + 3\vec{k}[/itex]
    [itex]\vec{b} = -4\vec{i} + 8\vec{j} - 6\vec{k}[/itex]

    2. Relevant equations

    The cross product of two vectors using the determinant, then dividing by the magnitude of the vector solution to find the unit vector

    3. The attempt at a solution

    Alrighty, so my dumb self didn't recognize that these two vectors were parallel when I started the problem, so I went ahead and did the cross product and got 0, of course.

    I also know that the unit vector for the the two vectors is
    [itex]\pm 3 \sqrt{3}(\vec{i} - \vec{j} + \vec{k}) [/itex]

    So what I need is how to determine what are all the possible vectors orthogonal to those bad boys.

  2. jcsd
  3. Sep 6, 2011 #2
    The dot product of orthogonal vectors is zero. You can use this fact to find the required general vector. Also are you sure the unit vector you show is correct? The norm doesn't appear to be 1.
  4. Sep 6, 2011 #3
    Two vectors are orthagonal when their dot products are zero.

    Use the dot product to find [itex]c\cdot a=0[/itex] and [itex]c\cdot b=0[/itex] for some unknown vector c.

    The normal vector you find will be orthagonal to both a and b. But they threw you a curve. b=-2a. a and b lie on the same line. so you will get the family of unit othagonal vectors lying in the plane perpendicular to a and b.
  5. Sep 6, 2011 #4


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    Because these vectors are parallel (which I did not notice until I read Phrak's response), the set of all vector orthogonal to them- i.e. orthogonal to the single vector [itex]2\vec{i}- 4\vec{j}+ 3\vec{k}[/itex]- or to the single vector [itex]4\vec{i}- 8\vec{j}+ 6\vec{k}[/itex]- lie on the plane perpendicular to that vector. Use either of the given vectors to find the plane perpendicular to it and containing (0, 0, 0) and write a vector equation for any point in that plane. Note that a plane is two dimensional so the general such vector will depend on two parameters. After you have the general form, divide by the length to get unit vectors.

    (Which works out to be very similar to Phrak's "dot product= 0" suggestion.)
    Last edited by a moderator: Sep 6, 2011
  6. Sep 6, 2011 #5
    I would try and get unit vectors of a and b then cross-product them ( axb then bxa )
  7. Sep 7, 2011 #6
    Thanks. I wasn't thinking ahead. My idea leads to a messy problem. How did you know?
  8. Sep 7, 2011 #7


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    NO!!! That's what you suggested before and both Phrak and I told you that is wrong. Because [itex]\vec{a}[/itex] and [itex]\vec{b}[/itex] are parallel, their cross product (and the cross products of any multiples of them) will be the zero vector. Instead, do what both I and Phrak suggested.
  9. Sep 7, 2011 #8
    Then, can we conclude that the solution is a hollow cylinder in 3-D ?
  10. Sep 8, 2011 #9
    The solution consist of fixed vectors with their tales at the origin. There will be some plane, passing through the origin. The heads of the vectors lie on a circle in the plane, the circle concentric to the origin.
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