Question

If A = (6i-8j) units, B = (-8i-3j) units, and C = (26i-19j) units, determine a and b
such that aA + bB + C = 0

Answer 1

Explanation:

see , it may be helpful for you

Answer 2

Answer:

a=37/41

b=–161/41

Explanation:

aA+bB+c = 0 then a =? b= ?

assuming that is the vector sum has to be zero...

x,y components have to add to zero

x:

a(6) + b(8) + 26 = 0

y:

a(–8) + b(3) + 19 = 0

two equations, 2 unknowns

6a + 8b + 26 = 0

–8a + 3b + 19 = 0

3a + 4b + 13 = 0

–8a + 3b + 19 = 0

multiply first by 3 and second by –4 and add

9a + 12b + 39 = 0

32a – 12b – 76 = 0

41a – 37 = 0

a = 37/41

6a + 8b + 26 = 0

6(37/41) + 8b + 26 = 0

8b = –26 – (222/41) = –1288/41

b = –161/41

check

6a + 8b + 26 = 0

–8a + 3b + 19 = 0

6(37/41) + 8(–161/41) + 26 = 0

–8(37/41) + 3(–161/41) + 19 = 0

222 – 1288 + 1066 = 0

–296 – 483 + 779 = 0