Let if possible
4¯i+13¯j−18¯k = x(¯i−2¯i+3¯k)+y(2¯i+3¯j−4¯k)
=(x+2y)¯i+(−2x+3y)¯j+(3x−4y)¯k
This is possible if the following equations are consistent.
x+2y=4 .....(1)
−2x+3y=13...(2)
3x+4y=−18....(3)
Solving equations (1) and (2) , we get
∴x=−2,y=3
These values satisfy equation (3) also
∴4¯i+13¯j−18¯k=−2(¯i−2¯j+3¯k+3(2¯i+3¯j−4¯k))
∴ The vectors 4¯i+13¯j−18¯k is a linear combination of the vectors ¯i−2¯j+3¯k and 2¯i=3¯j−4¯k.