Question

r1=1,4,7 r2=2,5,8 r3=3,6,9 (vectors)
Linear algebra says that these vectors must also lie in a plane. there must be many combinations with y1r1+y2r2+y3r3=0 find two sets of y's
please help!!

Answer 1

For linear dependence \[a(1,4,7)+b(2,5,8)+c(3,6,9)=(0,0,0)\]
For any non zero linear combination of a,b,c.
Solving we get,
\[a+2b+3c=0\]
\[4a+5b+6c=0\]
\[7a+8b+9c=0\]
Subtracting 3rd from 2nd
\[3(a+b+c)=0\]
\[b+c=-a\]
from 1st \[a=-2b-3c\]
\[2b+3c=b+c\]
\[b=-2c\]
\[a=-b-c=-b+2c=c\]
Thus linear combination \[c(1,-2,1)\] for all c in real numbers

Answer 2

thank you a lot!!!