x 121 y 022 z 103 Find the set of all linear combinations

x = [1,2,1] y = [0,2,-2], z = [1,0,3] Find the set of all linear combinations of x, y, z.

Solution: Plane through (0,0,0) which is perpendicular to [3,-1,-1] or 3x-y-z = 0.

Solution

x=[1 2 1] y=[0 2 -2] z =[1 0 3]

linear combination of x y z is

ax+by+cz=d

this must satisfy above 3 points

so a+2b+c=d ---->eq1

2b-2c =d ------->eq2

a+3c=d ----->eq3

eq1 - eq2 = a+3c =0 ----->eq4

from eq3 and eq4 d=0;

so the linear combination is ax+by+cz=0 this sastisfies thr origin [0 0 0]

the normalized normal vector to this plane and passes through origin can be abtained by cross multiplying (x-y) and (y-z) vectors

x-y =[1 0 3] & y-z =[-1 2 -5]

so (x-y) * (y-z) = [3 -1 -1]-----> normalized coordinates

so it s plane which passes through origin[0 0 0] and perpendicular to [3 -1 -1]