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Homework Help: Finding perpendicular unit vector using inner product

  1. Feb 10, 2014 #1
    1. The problem statement, all variables and given/known data
    You are given that with x = (x1,x2), y = (y1,y2), the formula

    (x,y) = [x1 x2] [2 1;1 2] [y1;y2] (where ; represents a new row).

    is a inner product for the vectors in R2

    Using this inner product, find a unit vector perpendicular to the vector (1,1)
    2. Relevant equations

    3. The attempt at a solution

    I dont particularly understand what a perpendicular unit vector has to do with the inner product, I can find a perpendicular unit vector in the normal dot product way however I dont know how to find one using the inner product, any help would be much appreciated,

    Thanks in advanced
  2. jcsd
  3. Feb 10, 2014 #2
    Two vectors are perpendicular if their inner product is equal to zero. The dot product is a common example of an inner product, but there are many others. Two vectors might be perpendicular when using one inner product, but perhaps not using another inner product. So just try taking the inner product of (1,1) with a vector (x1, x2) and setting the result equal to zero to come up with some dependency of x1 in terms of x2. Then you can just choose a value for x2 and TADA! you have a perpendicular vector. After that you can scale it to unit lenght (don't forget that you have a new inner product here! You should completely forget about the dot product during this problem).
    Last edited: Feb 10, 2014
  4. Feb 10, 2014 #3


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    Just work things out. Supposed p=(1,1) and the perpendicular unit vector is u=(u1,u2). Then, in terms of your inner product, to have u perpendicular to p means (u,p)=0, and u being a unit vector means (u,u)=1. Work both of those out in components and try to solve the resulting equations.
  5. Feb 10, 2014 #4


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    So the inner product is
    [tex]\begin{bmatrix}x_1 & x_2 \end{bmatrix}\begin{bmatrix}2 & 1 \\ 1 & 2 \end{bmatrix}\begin{bmatrix}y_1\\ y_2\end{bmatrix}= 2(x_1y_1+x_2y_2)+ (x_1y_2+ x_2y_1)[/tex]

    Really? The definition of "perpendicular" in terms of an inner product is that the inner product of two perpendicular vectors is 0.

    An "inner product" is just a generalization of "dot product". Here, with [itex](x_1, y_1)= (1, 1)[/itex] the inner product with (x, y) is 2((1)x+ (1)y)+ ((1)x+ (1)y)= 3x+ 3y= 0 which is the same as saying y= -x. Any vector perpendicular to (1, 1) is of the form (x, -x). Now you want to make that a "unit vector" by dividing by its length. Of course, you determine the length using that inner product: the length of a vector v is [itex]\sqrt{(v, v)}[/itex]. The inner product of (x, -x) with itself is [itex]2((x)(x)+ (-x)(-x))+ ((x)(-x)+ (-x)(x))= 2(x^2+ x^2)+ (-x^2- x^2)= 2x^2[/itex]. Set that equal to 1 and solve for x.

    Last edited by a moderator: Feb 10, 2014
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