If A=6i-8j,B=-8i+3j andC=26i+19j.find a and b such that aA+bB+C=0.
To find values of a and b such that aA + bB + C = 0, we need to set up two equations by equating the coefficients of i and j separately.
Let's first equate the coefficient of i:
a * (6i) + b * (-8i) + (26i) = 0
This simplifies to:
(6a - 8b + 26) * i = 0
Since the coefficients of i must be equal to 0, we get the equation:
6a - 8b + 26 = 0 ---------------------- (Equation 1)
Now, let's equate the coefficient of j:
a * (-8j) + b * (3j) + (19j) = 0
This simplifies to:
(-8a + 3b + 19) * j = 0
Since the coefficients of j must be equal to 0, we get the equation:
-8a + 3b + 19 = 0 ---------------------- (Equation 2)
Now we have two equations (Equation 1 and Equation 2) that need to be solved simultaneously to find the values of a and b.
Solving Equation 1 and Equation 2:
Equation 1: 6a - 8b + 26 = 0
Equation 2: -8a + 3b + 19 = 0
To solve this system of linear equations, we can use any method such as substitution or elimination.
Let's use the elimination method to solve the equations:
Multiply Equation 1 by 3 and Equation 2 by 8 to make the coefficients of 'b' cancel:
18a - 24b + 78 = 0
-64a + 24b + 152 = 0
Add the two equations together:
18a - 64a - 24b + 24b + 78 + 152 = 0
-46a + 230 = 0
Solving for 'a':
-46a = -230
a = -230 / -46
a = 5
Substitute the value of a = 5 into Equation 1:
6(5) - 8b + 26 = 0
30 - 8b + 26 = 0
-8b + 56 = 0
-8b = -56
b = -56 / -8
b = 7
Therefore, we have found that a = 5 and b = 7 such that aA + bB + C = 0.