Question

10. Three forces F1 =a(i – j+k), F2 = 2î – 3j +4k
and F3=8i – 7j+6k act simultaneously on a
particle. If the particle is in equilibrium, the
value of a is
1) 10 2 )-10 3)8 4)2
.​

Answer 1

Answer:

a= - 10

Explanation:

If the body is in equilibrium F1+F2+F3=0

Substitute and find the value of a

If a = -10, then

-10i+10j-10k +

2i -3j +4k +

8i -7j +6k

Then only F1+F2+F3=0

Answer 2

Answer:

answer is -10

Explanation:

if the body is in equilibrium then F1+F2+F3=0

(2i-3j+4k)+(8i-7j+6k)=-10i+10j+10k

then a value is -10