Question

The resultant of two vectors P and Q is R.If Q is doubled then the new resultant vector is perpendicular to P.Then magnitude of R is-

Answer 1

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

Answer:

$|\overrightarrow{\mathrm{R}}|=|\overrightarrow{\mathrm{Q}}|$

Explanation:

Step 1: Initial resultant between two vectors [ Fig. 1]

Resultant of $\vec{P}$and $\vec{Q}$,

$|\overrightarrow{\mathrm{R}}|^{2}=|\overrightarrow{\mathrm{P}}|^{2}+|\overrightarrow{\mathrm{Q}}|^{2}+2|\overrightarrow{\mathrm{P}}||\overrightarrow{\mathrm{Q}}| \cos \theta$

Step 2: Dot product between new resultant and $\vec{P}$ [Fig. 2]

When $\overrightarrow{\mathrm{Q}}$ is doubled, new resultant, $\overrightarrow{\mathrm{R}}_{1}=\overrightarrow{\mathrm{P}}+2 \overrightarrow{\mathrm{Q}}$ becomes perpendicular to$\overrightarrow{\mathrm{P}}$

$\therefore \overrightarrow{\mathrm{R}}_{1} \cdot \overrightarrow{\mathrm{P}}=0$

$\Rightarrow(\overrightarrow{\mathrm{P}}+2 \overrightarrow{\mathrm{Q}}) \cdot \overrightarrow{\mathrm{P}}=0$

$\Rightarrow|\overrightarrow{\mathrm{P}}|^{2}+2|\overrightarrow{\mathrm{P}}||\overrightarrow{\mathrm{Q}}| \cos \theta=0$

$\Rightarrow \cos \theta=\frac{-|\overrightarrow{\mathrm{P}}|}{2|\overrightarrow{\mathrm{Q}}|}$

Step 3: Solving equation

Putting value of $\cos \theta$ in equation (1)

$|\overrightarrow{\mathrm{R}}|^{2}=|\overrightarrow{\mathrm{P}}|^{2}+|\overrightarrow{\mathrm{Q}}|^{2}+2|\overrightarrow{\mathrm{P}}||\overrightarrow{\mathrm{Q}}|\left[\frac{-|\overrightarrow{\mathrm{P}}|}{2|\overrightarrow{\mathrm{Q}}|}\right]$

$\Rightarrow|\overrightarrow{\mathrm{R}}|^{2}=|\overrightarrow{\mathrm{Q}}|^{2}$

$\Rightarrow|\overrightarrow{\mathrm{R}}|=|\overrightarrow{\mathrm{Q}}|$

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