Question

Consider a set ?=(?⃗ 1,?⃗ 2,?⃗ 3), where

?⃗ 1=(12√,0,−12√)
?⃗ 2=(12,12√,12)
?⃗ 3=(12,−12√,12)

Mark all true statements

The set ? is an orthogonal set

The set ? is an orthonormal set

?⃗ 1,?⃗ 2,?⃗ 3 are all unit vectors

?⃗ 1,?⃗ 2,?⃗ 3 are linearly dependent

?⃗ 1,?⃗ 2,?⃗ 3 are linearly independent

?⃗ ?⋅?⃗ ?=0 where (??,??∈?∧?≠?)

?⃗ ?⋅?⃗ ?=1 where (??,??∈?∧?≠?)

?⃗ ?⋅?⃗ ?=0 where (??,??∈?∧?=?)

?⃗ ?⋅?⃗ ?=1 where (??,??∈?∧?=?)

? is an orthonormal basis for ? where ?=????(?⃗ 1,?⃗ 2,?⃗ 3)

Answer 1

gures set B= (ui, Io, G) whose ūs = 62, 0, -VT) U2 = 118, VT2,12) 43 = 112, tiz, 12) Ads -1 Big an or thogonal set ie who have to show cujouito & iti solution - 11>-<ū ūg) = < (VT2, 0, -V12),. 119, Víz, 12) - 12 Viz +0 - 12 V2 = 0 <u, ūg>=0 <ū,.> - < (V12, 0,- Viz) : (12,-V12,12)) A. 1 - 12 V12 to +12 V12 = 0 - <ū,. us >=0 - and cü uz >= < (12,012, 12). (12,-V12, 12) 3 144 -12 +144 = 266 40 .. pe <uza) = 26680 so above Bet B= (0, I, P) is not orthogenal - B. is not aloo orthosomal . . To if its because : <uli, 44 ) =; if éj) is not follow from abowe .

Aim - Now check Dependent and Independent sed 12 Viz epiz 7 As Soluton = B = Cũ U2, 43) Then we can write is matrix foom As [ūa, us] A = 0 Viz V12 1. Leviz 0812 12 Now find Deteominant 1 Ala Viz (12V12 + 12V12) - 12 (0-12.) +12 10+12) TA = 24 V12XV12 + 144 +14 4 Tale 24x12 + 288 JA = 2888 +288 | | |- 576 # o => Abowe Bet B= cū, va, ū) coe linearly 1 independent set 5 B is not dependent set . singu r (viz, 0, -V12) lui 1 = V1U2)+02 +(-V12)2 = V12+0+2 ū l = 174 # 1 · Smily lūIE V 1447 12 444 = V30071 143) = V144 +12+144 = V100 #1 ü,, az and us are not amit vector th

sinu ūs, te (VT2,0,-V12) (12,812, +2) [3= (12,082, 12) clearly from aboue u. 470 t its cu feo upu= <(V12,0, -V12) (12,0;V12 ) = 12 +12= 24 # 1 S ie üzeug +1 isy: aloo üzeug #o Paj and uiue 1 + ºHJ. so only one option is toue : : 8 = cū, we, us) are linemly independent set . B is not orthonal basis, because B is not osthonomof inaboree pooceso