2. If Å=8i - 8j, B = -i+ 6j and č=-25i. Find the constants a and b such that aA+ bB + Č = 0.
Given:
Å = 8i - 8j
B = -i + 6j
č = -25i
We want to find the constants a and b such that aÅ + bB + č = 0.
aÅ = a(8i - 8j) = 8ai - 8aj
bB = b(-i + 6j) = -bi + 6bj
Then, the equation becomes:
8ai - 8aj - bi + 6bj - 25i = 0
Now, let's equate the coefficients of the i and j terms separately.
For the i terms:
8a - b - 25 = 0 (equation 1)
For the j terms:
-8a + 6b = 0 (equation 2)
Solving equations 1 and 2 simultaneously, we get:
8a - b = 25
-8a + 6b = 0
Multiplying the second equation by 4:
-32a + 24b = 0
Adding this modified equation to the first equation:
(8a - b) + (-32a + 24b) = 25 + 0
-24a + 23b = 25
Now, we have a new equation:
-24a + 23b = 25 (equation 3)
To solve for a and b, we need another equation. We can choose any equation from equations 1 and 2. Let's use equation 1:
8a - b - 25 = 0
Solving equation 1 for b:
b = 8a - 25 (equation 4)
Substituting equation 4 into equation 3:
-24a + 23(8a - 25) = 25
-24a + 184a - 575 = 25
Collecting like terms:
160a = 600
Dividing both sides by 160:
a = 600/160
a = 3.75
Substituting the value of a back into equation 4 to find b:
b = 8(3.75) - 25
b = 30 - 25
b = 5
Therefore, the constants a = 3.75 and b = 5 will make aÅ + bB + č = 0.