Find all unit vectors in R^2 which are orthogonal to (−3, 1).

Answer 1

In R2, vectors orthogonal to (x,y) are trivially (y,-x). and (-y,x). Just normalize these if you want them unit length.

But since you seem more interested in using projections, which is good to know in a more general setting, here's a second approach:

Begin with the familiar projection formula

P(x,y)_(-3,1) = ((x,y)•(-3,1)/10) (-3,1)

This formula defines a mapping from R2 into R2. The mapping is further a linear transformation.

Apply this transformation to the standard ordered basis vectors [1,0] and [0,1] and you'll get the matrix of the transformation:

9/10 -3/10

-3/10 1/10

In this particular case, the transformation matrix would more commonly be referred to as a projection matrix, for obvious reasons.

Now note that the column space of a matrix is the orthogonal complement of the null space of its transpose. The column space of the matrix above is the space spanned by the vector [-3,1] because the matrix projects R2 onto the vector [-3,1]. So the null space of the transposed matrix with give a basis for everything orthogonal to [-3,1].

Reduce the transpose to RREF:

1 -1/3

0 0

which implies the null space has basis {[1,3]}. Now just normalize the basis vectors.

In a larger setting (R3 for instance), the orthogonal complement of a given set of vectors might have larger dimension, so that writing all the unit-length orthogonal vectors can only be done using a spanning set.

Answer 2

Let (x,y) be a vector which is orthogonal to (-3,1). So, (x,y)•(-3,1)=0. That is -3x+y=0. So the vectors that are orthogonal to (-3,1) must satisfy the equation y=3x. That is all vectors of the form (x,3x) are orthogonal to (-3,1). Now ||(x,3x||=√x²+9x²=√10x²=|x|√10. So, a unit vector orthogonal to (-3,1) will look like ±(1/√10,3/√10).