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.