Hey bartleI have a question regarding Orthogonal projection in the linked picture there is the question and my teachers solution. i dont understand the process of question 2. i dont know how my teacher reach the answer (p2. p0) = 10, (p0. p0) = 4, (p2, p1) = 0 and (p1,p1) = 10i also dont understand question 3 PROBLEM 5.Let the vector space P2 have the inner product defined by evaluation at -2, -1, 1 and 2.Let po(t) = 1, pi(t) = t and p2(t) = t°.1. Compute the distance between po and p2.2. Compute the orthogonal projection of p2 onto the subspace spanned by po andAnother polynomium is given by pa(t) = 1+c, where c is a real number.3. Determine c so that p2 and pa are orthogonal.PROBLEM 5. SolutionP1.The distance between vectors po and p2 is given bydist(Po, P2) = ||po – P2|| = V(Po – P2. Po – P2).Computing po - P2 = 1-ť and using the standard trick of evaluating at the given pointsto turn the inner product computation into a dot product givesPo - P2 = 1-tH→dist(Po, P2) =18 = 4.24The orthogonal projection is given by(P2, P1).Pi(P1, P)"(P2, Po)(Po, Po Po +To ease the computation the following mapping is used1P2+Po +4.And the inner products becomes(P2, Po) = 10, (Po, Po) = 4,(P2, P1) = 0, (P1, P1) = 10.Hence(P2, Po)Po +(Po: Po)(P2, P1)10-1+= Id-(P1, P1)4Vectors p2 and p. to be orthogonal if (P2; Pa)= 0. Evaluating pa at the given pointsgives[1+c]1+cPa +1+c1+cThe inner product equation to be solved is then+c11+c(P2, Pa) = 0= 0 + 10(1+ c) = 0 + c= -1.1+c41+c- - - 1

Name: Hey bartleI have a question regarding Orthogonal projection in the lin
Uploaded: 2024-03-20T06:38:44.000Z
Channel: Bartleby
Description: Hey bartleI have a question regarding Orthogonal projection in the linked picture there is the question and my teachers solution. i dont understand the process of question 2. i dont know how my teacher reach the answer (p2. p0) = 10, (p0. p0) = 4, (p