1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find all vectors in R^3 that are perpendicular to [1; 3; -1]

  1. Sep 12, 2006 #1

    The dot product is:

    [tex]\overrightarrow{x}\,=\,\left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right][/tex]

    [tex]\overrightarrow{y}\,=\,\left[ \begin{array}{c} y_1 \\ y_2 \\ \vdots \\ y_n \end{array} \right][/tex]

    in [tex]\mathbb{R}^n[/tex]:


    If the scalar [itex]\overrightarrow{x}\,\cdot\,\overrightarrow{y}[/itex] is equal to zero, the vectors are perpendicular.

    Find all vectors in [itex]\mathbb{R}^3[/itex] that are perpendicular to
    [tex]\left[ \begin{array}{c} 1 \\ 3 \\ -1 \end{array} \right][/tex].

    Draw a sketch as well.


    [tex]\left[ \begin{array}{c} 1 \\ 3 \\ -1 \end{array} \right]\,\cdot\,\left[ \begin{array}{c} x \\ y \\ z \end{array} \right]\,=\,0[/tex]



    Let s = x and t = y


    [tex]\left[ \begin{array}{c} 1 \\ 3 \\ -1 \end{array} \right]\,\cdot\,\left[ \begin{array}{c} s \\ t \\ s\,+\,3\,t \end{array} \right]\,=\,0[/tex]

    Does the above look right?
    Last edited: Sep 12, 2006
  2. jcsd
  3. Sep 12, 2006 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, that looks good- but where is your answer? Since the problem asks for the set of all vectors perpendicular to [1, 3, -1] your answer should be something like "All vectors satisfying" or "all vectors spanned by". You have already calculated that a vector [x, y, z] in that space must satisfy x+ 3y- z= 0; just say that. You have also calculated form that that z= x+ 3y and got [s, t, s+3t] as a "representative" vector. You could answer "all vectors of the form [s, t, s+ 3t] where s and t can be any real numbers. Finally, since [s, t, s+ 3t]= s[1, 0, 1]+ t[0, 1, 3], you could answer "the subspace spanned by [1, 0, 1] and [0, 1, 3]."

    Your picture, of course, would be the plane x+ 3y- z= 0.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Find all vectors in R^3 that are perpendicular to [1; 3; -1]
  1. Pi r^2 /3 (Replies: 3)

  2. Subspace in R^3 (Replies: 2)

  3. Subspace of R^3 (Replies: 1)