2. If Ā= 8i- 8j. B = -i +6j and č= -25. Find the constants a and b such that aÃ+ bB + Č = 0.

Answer 1

Given Ā = 8i - 8j, B = -i + 6j, and č = -25, we need to find the constants a and b such that aĀ + bB + č = 0.

Substituting the given values, we have:

a(8i - 8j) + b(-i + 6j) - 25 = 0

Expanding the equation, we get:

8ai - 8aj - bi + 6bj - 25 = 0

Rearranging the terms, we have:

(8a - b)i + (-8a + 6b - 25)j = 0

This equation should hold for all values of i and j. Therefore, the coefficients of i and j must equal zero:

8a - b = 0
-8a + 6b - 25 = 0

From the first equation, we can solve for b:

b = 8a

Substituting this value into the second equation, we get:

-8a + 6(8a) - 25 = 0

-8a + 48a - 25 = 0

40a - 25 = 0

40a = 25

a = 25/40

Simplifying the fraction, we get:

a = 5/8

Substituting this value back into the equation b = 8a, we find:

b = 8(5/8)

b = 5

Therefore, the constants a = 5/8 and b = 5 satisfy the equation aĀ + bB + č = 0.