Question

The resultant of two vectors P and Q is R. If the magnitude of Q is doubled, the new resultant becomes perpendicular to P. Then the magnitude of R is ?

Answer 1

Answer:

esultant of two forces P and Q is R

R = P + Q

|R| = sqrt(P^2 + Q^2 + 2PQCosθ)

When Q is doubled new resultant R’ is

R’ = P + 2Q

Since it is perpendicular to P, dot product of R’ and P is zero

(R’).(P) = 0

(P + 2Q).(P) = 0

P^2 + 2PQCosθ = 0

(Dot product of P and P is P^2 and that of P and 2Q is 2PQCosθ)

Cosθ = -P/(2Q)

R = sqrt(P^2 + Q^2 + 2PQCosθ)

R = sqrt(P^2 + Q^2 + 2PQ*(-P/(2Q))

R = sqrt(P^2 + Q^2 - P^2)

R = Q

Answer 2

Explanation:

resultant of two forces P and Q is R

R = P + Q

Square of Resultant R is R^2 = P^2 + Q^2 + 2PQ cos α,

α is the angle between P and Q.

tan θ = Q sin α / ( P + Q cosα), θ is the angle between R and P.

If Q is doubled , and if the angle between P and the new resultant S is 90 °,

tan 90 = 2Q sin α / ( P + 2Q cosα),

Therefore, P = - 2Q cosα.

But R^2 = P^2 + Q^2 + 2PQ cos α,

R^2 = P^2 + Q^2 - P^2

R^2 = Q^2

R = P ( in magnitude)