Question

If the resultant of three forces F1 = pi +3j-k; F2= 5i+j +2k and F3= 6i - k acting on a
particle has magnitude equal to 5 units, then the value (s) of P is​

Answer 1

given three forces F1 = P i + 3j - k , F2 = 5i + j + 2k and F3 = 6i - k are acting on a particle.

so, resultant force acting on particle , F = F1 + F2 + F3

= (p i + 3j - k) + (5i + j + 2k) + (6i - k)

= (p + 5 + 6)i + (3 + 1)j + (-1 + 2 - 1)k

= (p + 11)i + 4j

now, magnitude of resultant , |F| = √{(p + 11)² + 4²}

but it is given that, |F| = 5

so, 5 = √{(p + 11)² + 4²}

squaring both sides,

5² = (p + 11)² + 4²

⇒5² - 4² = (p + 11)²

⇒ 25 - 16 = 9 = (p + 11)²

⇒ 3² = (p + 11)²

⇒ ±3 = p + 11

⇒p = -11 ± 3

so, P = (-11 - 3) = -14

and P = (-11 + 3) = -8

therefore values of p = -14, -8