Question

1.1. Sample Problems. 1. In R3, (a) Find an orthonormal basis which contains the vector ( 1, 0); (b) Find the component of the vector (1,1,1) in the direction of (1,0,1).

Answer 1

solution : (a). We Bolve this problem by cross product Let Vector VE = 1 7 het. 141 where ulil our first goal to find the vector U, and yg such that $41, 42, 43} is an orthogonal basis foro Let x = 7 be vector perpendicular to ut Then we have 7. U=o and hence the relation, LZ / [:][:)- ZJU 5 x+y = 0 vector 4, let L otiefie relation U2:4,-0 Now, let us detine the cross product of us 42 third vector u to be the 4 = uix ua "Qxy figky 2 O I 10 1-10 L6J T : U2 =

:. by property of cross pooduct the vector ų is perpendicular to both U, R42 Therefore the set {41,42,43} sri] 1] lol.107:1-2, is an orthogonal basis for IRS at it consist of I those nonzero orthogonal lectors" How To obtain ortho normal basis for IR3 we have to normalize the length of ree these rector 110,11 ] 0192+12 + (0) 1) + CO2 114211 = 1 60276-2276032 = 2 11411 - Cos?760)2 46-232 = 54 : 2 - Then we have vi = 41 - to (1,1,0)' ne uza (1-1,0) (orthorormal) vas (0,0,-2) Side basis of 1R3 containing vi 14211 Jlyll {*, *. *} = {[], ta[:] [:] which is required solution

We have to find component of rector (1,1,1) in the direction (1,0,1) for this let, v = (1,1,1) & V2 =(1,0,1) we know that component of vector v inth direction of of V₂ = Vue 1121 - pvz = (1,1,1)-(1,0,1) - (1 +0+1)=2, Tuzla 160927109278092 0 0 0 U U با U با component of v = - Veis2 = 2 component of a = 52 با ر ر ر ر ر ر