It is also possible to define functions as being orthogonal if we can find the correct definition to replace the dot product. If we take two functions f and g, defined over the interval [0, 211'] , then we can define a fancy inner product 2 9y = [ f@le)is. This will play the same role as the dot product, taking two vectors as input and spitting out a number. Using this, we define two functions to be orthogonal if (fi.'}') =0, The same way we do for geometric vectors. Let's consider the functions f(z) =sin(7z), g(z) = sin(3z). Using the inner product definition above we can determine 2w -{fr9) =f sin(7z) sin(3z)dr = | Number 0 2w 9 . (f, f) = [ {SiJ:l{'T.'B:I) dx = | Number 0 2w 9 - {g:9) = [ (sin(3z))\"dz = Number 0 Showing that f(z) = sin(7z) and g(z) = sin(3z) are orthogonal with respect to this cool new product. Note: you may need to remember that the Maple notation for is Pi. Hint: - you might find it easier to evaluate the integrals above if you expand the product JATT _ oiTz\\ [oi3z _ idz in(7z)sin(3z) = . sin(7z) sin(3z) = = . this trick also works for {s.in('?"z)}2 and (3]11{3:::))2, and should yield familiar relations. A set of three vectors { V1, V2, v3} in a vector space is an orthogonal set if each is vector is a non-zero vector, and if every vector is orthogonal with the other two. This means explicitly that V1 . V2 = V2 . V3 = V3 . V1 = 0. So for example . ( 4 ) . (-> ) . (;.) is an orthogonal set in IR . Can you find a set of three orthogonal 3 vectors in IR which includes the vector -1 ? Such a set is 1 Note: . express your set of vectors using the Maple notation {,, } . don't forget to include in your set. 3 Hint: Once you have one vector orthogonal to -1 you can use the cross product to find a third