iPad FF514 74 webassignnet 5 20 points My Notes The given ve

iPad FF5:14 74% webassign.net 5. -20 points My Notes The given vectors form a basis for R3. Apply the Gram-Schmidt Process to obtain an orthogonal basis. Then normalize this basis to obtain an orthonormal basis. (Enter your answers as a comma-separated list. Enter each vector in the form [x1, x2, ...J.) r y3 Submit Answer Save Progress View Previous Question Question 5 of 8 View Next Question Home My Assignments WebAssign® 4.0 1997-2017 Advanced Instructional Systems, Inc. All rights reserved.

Solution

Let u₁ = y₁ = (1,-2,-2)^T

u₂ = y₂-Proj_u1(y₂) = y₂ - [(y₂.u₁)/(u₁ .u₁)]u₁ = (0,3,3)^T – [(0-6-6)/(1+4+4)] (1,-2,-2)^T = (0,3,3)^T+ (12/9) (1,-2,-2)^T =(0,3,3)^T+(4/3, -8/3, -8/3)^T = (4/3, 1/3, 1/3)^T and

u₃ =y₃- Proj_u1(y₃)- Proj_u2(y₃)= y₃-[(y₃.u₁)/(u₁ .u₁)]u₁-[(y₃.u₂)/(u₂ .u₂)]u₂= (3,2,4)^T –[(3-4-8)/( 1+4+4)] (1,-2,-2)^T – [(4+2/3+4/3)/(16/9+1/9+1/9)] (4/3, 1/3, 1/3)^T =(3,2,4)^T+(9/9)(1,-2,-2)^T –(6/2) (4/3, 1/3, 1/3)^T = (3,2,4)^T +(1,-2,-2)^T –(4,1,1)^T = (0, -1,1)^T.

Then {u₁,u₂,u₃} is an orthogonal basis for R³. We only need to convert u₁,u₂,u₃ into unit vectors e₁,e₂ and e₃.

We have ||u₁|| = ( 1+4+4) = 3 so thart e₁ = ( 1/3,-2/3,-2/3)^T.

Similarly, ||u₂|| = ( 16/9+1/9+1/9) = 2 so thart e₂ = (4/32 , 1/3 2, 1/32)^T =( 22/3, 2/6, 2/6)^T

and ||u₃|| = ( 0+1+1) = 2 so thart e₂ = (0 , -1/2, 1/2)^T

Then B = { e₁,e₂ , e₃} is the required orthonormal basis for R³.