Question

In space R^3, we define a scalar product by regulation
〈(x1, y1, z1), (x2, y2, z2)〉 = 2x1x2 + y1y2 + 2z1z2 + x1z2 + x2z1.
(a) [10] Calculate the perpendicular projection of the point T (1, 1, 1) on the plane U in R3 with the equation x + 2y + 2z = 0 with respect to the given scalar product.
(b) [10] Let φ: R^3 → R be a linear functional with φ (x, y, z) = x + 2y + 2z. Find the vector belonging to the functional φ according to Riesz's theorem with respect to a given scalar product.

Answer 1

In R3 I we the at +22=0 Qy -22. Y, ZER iie. u = .. . define ine a scalar product by regulation <(2494, 921) "(*29/29 Zz)>= 2x, % + Y, G₂ + 221 Zz + x Z2 +X Z, ay we have to calculate the perpendicular projection of point (ted,d) on the plane u in LR3 with the equation -2y +22=0, with respect to the given scalar product. x+2y+22 Any point on u can be represented as : (-2y - 22 , y27), { (-2y-22, 2y922) • y»ZER? {cay984, 0)+(-2290922) : yazer? { -Ry.(49–190) + (-22)(905-1) = yo? ER} span { les {(1,-1,0);(420, (1,-1,0) and (190,-)) are linearly independent, {(4,-1,0); (1,0,-1)} forms a basis of u. Heuce perpendicular projection of ded, 1) on the plane u 344,1,1), (1,-1,0)> (1,-1,0) + ((1,1,1),(1,0,-1)(11054) (2-1 +0+0+1) (1,-1,0) + (2+0 - 2-1 + 1)(110,-1) 2.(1,-1,0) + O.(1,0,-1) (2,-2,0) (Ans ) 0,-1)} Since 2 hence by $: R37 R is a linear furetional, 41x141Z) = x+2y +22 According to Rieses Representation theorem, there 29 = (2247 22 923)) Sito $(2972) = < (X,Y, 2), (2609 283293 ) vbxny, 2)ER 3. exists unique 20 t LR3

Now {(1,0,0), (Ond,0), (0,0,1)} is a basis of R3. From this basis we will find a basis {w,, W2, W3} of (R3 sito the basis is orthogonal. We will use Grom - Schmidth method for this. define W, = (1,0,0) W2 = (0,2,0) - <0,1,0), (1,0,0) (410,0) 11 (190,00112 (on1,0) - (1,0,0) ll (190, 02112 = (0,2,0): W3 = (0,0,1) -- <(0,0,1),(2,0,0) (1,0,0) + <<0.00 ll (1,0,0) 11 <(0,0,1),(0,1,0) I cout,o) 112 (0,1,0)] O.CO,1,0) (11010) + (0,0,1)- [choose <0,0,0),(1,0,0) <(0,1,0),(0,130) - (0,0,1) - (1,0,0) =(- 2 90 Ond) 2 1 :{(1,010 0), 1091,0), (- 320 091) } is an orth an orthogonal basis of R² Now, 11(1,0,0)11 (2,0,0), (1,0,03>]" [0,1,0),(0,490)>]'2 111- 2011) || = [<63 2014), (-3,0,1)>]"2 = (2+2-2-3) Il cost,0) || 12 1312 (

{(1,0,0), Cond90), E (1.014)} of R3 is an orthonormal basis with respect to the defined scalar proceet. belonging to the function & according ple;) Hence the rector ze be to Riesz's theorem a ei ial 2 ar3 12 253 2 3 PC 20.0) (ano, c) +$10,1,0) (0,1,0) + 10:13).( 2017 al 990,0) + 2(0,2,0) + ( 10 *(5,090)+(012,0) + f ( 0 -,010)+(0,2,0) + (-$20,4) 10,291) (Ang) 2