Question 20758
 Recall that P3 is the space of all polynomials of degree less than three with real coefficients. Define an inner product < , > on P3 given by:
< p,q > = the sum from i=1 to 3 of p(i-1)*q(i-1), and let || || be the norm with respect to this inner product. 
(a) let p(x)=1+x-x^2 and q(x)=2-x
Find a polynomial f(x) in P3 which is orthogonal to both p(x) and q(x) with respect to the inner product.
(b) find the distance between p(x)+q(x) and f(x).

 [Note <p,q> as regular dot product in {{{R^3}}}]

 Sol: a) Let f = pxq (as cross product)
             = |i j k|  (i as 1, j as x, k as {{{x^2}}})
               |1 1 -1|
               |2-1 0|
            = -i - 2j + -3k = {{{-1 - 2x - 3x^2}}}
        Or
        Let f = a  + bx + cx^2 in P3, if <f,p> = <f,q> = 0, then
         <f,p> = (a,b,c). (1,1,-1) = a+b-c = 0 &
         <f,q> = (a,b,c). (2,-1,0) = 2a-b = 0
         then we have b=2a, c= a+b= 3a.
         f = a (1+2x + 3x^2) for some a.
     b) Let f(x) = {{{-1 - 2x - 3x^2}}}
||p(x)+q(x)- f(x)|| = || (3-x^2) + {{{1 + 2x + 3x^2}}}||
 = || {{{4 +2x +2x^2}}}||
 = {{{ sqrt(4^2+ 2^2 + 2^2)}}}
 = {{{2* sqrt(6)}}}

 Kenny
 PS: 1.Sorry for answering you late.
     2.It seems that you posted this question in wrong forum.
     3.You have to work hard.