QUESTIONS - LINEAR ALGEBRA - VECTORS

QUESTIONS / ANSWERS .ACF.LINEAR ALGEBRA
On 2/9/07, acf wrote:
> acf has left a new comment on your post "QUESTIONS - LINEAR ALGEBRA &
> VECTORS":
>
> Let B and C be non-parallel vectors in R superscript 3 and let A be an
> element in R superscript 3 orthgonal to both B and C. Use liner algebra to
> prove A is parallel to B cross C.
>
> Hint: Find a basis. Please do not use vector compounents or simple vector
> algebra. Must use linear algebra techniques.
>
>
>
> Posted by acf to Maths,management,counselling at 11:50 AM

R3 DIMENSION.
B AND C ARE NON PARALLEL.
HENCE THEY ARE INDEPENDENT VECTORS
A IS ORTHOGONAL TO BOTH B AND C.
HENCE A IS NEITHER DEPENDENT ON B NOR ON C.
HENCE A,B,C ARE 3 INDEPENDENT VECTORS IN R3
HENCE A,B,C FORM A BASIS IN R3
HENCE
B X C = xA+yB+zC
BUT
B X C IS PERPENDICULAR TO BOTH B AND C.
HENCE IT CAN NOT HAVE A COMPONENT ALONG B OR C...THAT IS ITS COMPONENTS ALONG B AND C ARE ZEROS
THAT IS
y=z=0
HENCE
B X C = xA
HENCE
B X C AND A ARE DEPENDENT ....COLLINEAR OR PARALLEL VECTORS.

QUESTIONS - LINEAR ALGEBRA & VECTORS

QUESTIONS/ANSWERS - ACF - LINEAR ALGEBRA & VECTORS

On 2/8/07, acf wrote:
acf has left a new comment on your post "QUESTIONS-ANSWERS-LINEAR ALGEBRA":

Let A and B be orthogonal unit vectors in R superscript 3 and let C=A cross
B. Show that {A,B,C} is an orhogonal basis for R superscript 3 and prove
that if A=B cross C then B=C cross A. Show that C,A,C coss A form a basis
for R superscript 3 and B = C cross A

Posted by acf to Maths,management,counselling at 2:55 PM


GIVEN
A AND B ARE ORTHOGONAL VECTORS IN R3....SO
A.B=0=|A||B|COS(T)=0,WHERE T IS ANGLE BETWEEN A & B
|A| = |B| =1....HENCE
COS(T)=0....SO .....T=90
A & B ARE PERPENDICULAR AND HENCE LINEARLY INDEPENDENT.
GIVEN
A X B = C IN R3
HENCE C IS PERPENDICULAR TO A AND B BY DEFINITION OF VECTOR CROSS PRODUCT.
FURTHER |C|=|A||B|SIN(T)=1*1*1...SINCE T=90 AND |A|=|B|=1..GIVEN
SO A,B,C ARE 3 MUTUALLY PERPENDICULAR UNIT VECTORS.
THEY ARE INDEPENDENT
THEY ARE IN R3 WHOSE DIMENSION IS 3
SINCE ANY SET OF 3 INDEPENDENT VECTORS (THAT IS EQUAL TO THE DIMENSION OF R3) IN R3 FORM A BASIS, AND THESE 3 ARE MUTUALLY PERPENDICULAR UNIT VECTORS , THEY FORM AN ORTHONORMAL BASIS FOR R3.
GIVEN
A=B X C.....................
CROSS BOTH SIDES WITH C ON LEFT
C X A = C X (B X C) = (C.C)B - (C.B)C.....
C.C=1....SINCE |C|=1.....C.B=0....SINCE C AND B ARE PERPENDICULAR TO EACH OTHER.HENCE
C X A = B......PROVED.
BY THE SAME REASONING AS GIVEN ABOVE , WE CAN PROVE THAT C,A,C X A FORM AN ORTHONORMAL BASIS FOR R3.

QUESTIONS-ANSWERS-LINEAR ALGEBRA

QUESTIONS/ANSWERS... BY ACF - LINEAR ALGEBRA

Feb.6 th.(12 hours ago)
acf has left a new comment on your post "QUESTIONS - CDOG - MATRICES":

If A1, A2,...AK which is an element of R to the nth dimenion are non-zero and mutually orthogonal show that they are linearly independent.

Posted by acf to Maths,management,counselling at 7:06 AM

FOR EASE OF NOMENCLATURE LET ME CHANGE THE PROBLEM AS FOLLOWS.
A,B,C,........K ARE NON ZERO,MUTUALLY ORTHOGONAL,N DIMENSIONAL VECTORS.
TST A,B,C ......K ARE INDEPENDENT.
LET US PROVE THIS BY REDUCTIO-AD-ABSURDUM.
THAT IS BY CONTRADICTION.
LET US ASSUME,A,B,C.....K ARE DEPENDENT.
THEN WITHOUT ANY LOSS OF GENERALITY, WE CAN TAKE THAT
A=X2*B+X3*C+X4*D+.............Xk*K......................1
DOTTING IT WITH B,C,ETC...AND NOTING THAT THE VECTORS ARE MUTUALLY
ORTHOGONAL THAT IS A.B=A.C=...=B.C=B.D=.....0...WE GET,
A.B=0=X2(B.B)+X3(C.B)+.......Xk(K.B)=X2(B.B).....2
A.C=0=X2(B.C)+X3(C.C)+.......Xk(K.C)=X3(C.C)......3
...........................................ETC
A.K=0=X2(B.K)+X3(C.K)+.......Xk(K.K)=Xk(K.K)......4
---------------------------------------------------------------------------
SINCE B,C.....K ARE NON ZERO VECTORS ,
B.B,C.C,........K.K CANNOT BE ZERO .
HENCE X2=X3=......=Xk=0
THAT IS
A=0...........FROM EQN.1
BUT THIS A CONTRADICTION SINCE A IS NON ZERO VECTOR.
SO OUR ASSUMPTION THAT
A,B,C,......K....ARE LINEARLY DEPENDENT IS WRONG.
HENCE A,B,C.....K ARE LINEARLY INDEPENDENT.