If A = 8i-8j, B = -i+6j and C = -25i find the conistant a and b such that aA+bB+C=0

Answer 1

To find the constants a and b such that aA + bB + C = 0, we need to equate the real and imaginary parts of the equation.

Given:
A = 8i - 8j
B = -i + 6j
C = -25i

Equating the real parts:
8a - b + 0 = 0
8a - b = 0 ----(1)

Equating the imaginary parts:
-8a + 6b - 25 = 0 ----(2)

From equation (1), we can express b in terms of a:
b = 8a

Substituting this into equation (2):
-8a + 6(8a) - 25 = 0
-8a + 48a - 25 = 0
40a = 25
a = 25/40
a = 5/8

Substituting this value of a back into equation (1):
8(5/8) - b = 0
5 - b = 0
b = 5

Therefore, the constants a and b such that aA + bB + C = 0 are:
a = 5/8
b = 5