The resultant of forces P and Q is R. If Q is doubled then R is doubled. If the direction of Q is re

Q: Wait, why do we use $180^\circ - \theta$ when we reverse $\vec{Q}$?

Think of the vector $\vec{Q}$ as an arrow. If you reverse it, it points in the exact opposite direction. If the original angle between $\vec{P}$ and $\vec{Q}$ was $\theta$, the angle between $\vec{P}$ and the new vector $-\vec{Q}$ is the supplementary angle, which is $180^\circ - \theta$. Since $\cos(180^\circ - \theta) = -\cos \theta$, this naturally introduces the negative sign in our equation.

Q: Why can't we just use the triangle law of vector addition?

You absolutely can! The triangle law and the parallelogram law are geometrically equivalent. However, the Law of Cosines is simply the algebraic manifestation of the triangle law. Using the Law of Cosines allows us to work directly with magnitudes, which is much faster for this specific problem than drawing triangles and calculating side lengths.

Q: Is it mathematically sound to treat $P^2$ and $Q^2$ as variables?

Yes, it is a brilliant strategy! In JEE problems, we often look for symmetries. By treating $P^2$, $Q^2$, and $R^2$ as our primary variables, we transform a complex trigonometric problem into a simple system of linear equations. It is a powerful tool to keep in your arsenal.

Question

The resultant of forces $\vec{P}$ and $\vec{Q}$ is $\vec{R}$. If $\vec{Q}$ is doubled then $\vec{R}$ is doubled. If the direction of $\vec{Q}$ is reversed, then $\vec{R}$ is again doubled. Then $P^2 : Q^2 : R^2$ is

Options:

(A) 2 : 3 : 1

(B) 3 : 1 : 1

(C) 2 : 3 : 2

(D) 1 : 2 : 3

Sigma Percentile · Accepted Answer

### The Vector Dance: A Masterclass in Symmetry

My dear student, welcome to the arena. Today, we are not just solving a physics problem; we are performing a dance with vectors.

When you look at the resultant of two forces, $\vec{P}$ and $\vec{Q}$, do not see them as static arrows. See them as a dynamic system, a parallelogram that breathes and changes as we manipulate its sides.

The problem gives us three snapshots of this system, and our goal is to find the hidden ratio between their magnitudes.

### Phase 1: The Baseline

We begin with the fundamental Law of Cosines. If you have two vectors $\vec{P}$ and $\vec{Q}$ with an angle $	heta$ between them, their resultant $\vec{R}$ is governed by the equation:

$$R^2 = P^2 + Q^2 + 2PQ \cos 	heta$$

This is our baseline. It is the anchor for everything that follows.

Never underestimate the power of this single equation; it contains the entire geometry of the system.

### Phase 2: The Perturbations

Now, we introduce change. The problem presents two scenarios.

First, we double $\vec{Q}$. The new resultant is $2\vec{R}$. Substituting these into our baseline, we get:

$$(2R)^2 = P^2 + (2Q)^2 + 2P(2Q) \cos 	heta$$

Which simplifies to:

$$4R^2 = P^2 + 4Q^2 + 4PQ \cos 	heta$$

This is our second snapshot. Next, we reverse $\vec{Q}$.

The angle between $\vec{P}$ and $-\vec{Q}$ becomes $180^{\circ} - 	heta$. Since $\cos(180^{\circ} - 	heta) = -\cos 	heta$, our third snapshot becomes:

$$4R^2 = P^2 + Q^2 - 2PQ \cos 	heta$$

Notice the beauty here? The negative sign appears naturally, and the system is now ready to be solved.

### Phase 3: The Algebraic Symphony

This is where the magic happens. We have three equations, but we have an annoying term: $2PQ \cos 	heta$.

It is the obstacle preventing us from finding the ratio. But look at equations (1) and (3). If we add them, the cosine terms cancel out entirely!

$$(R^2) + (4R^2) = (P^2 + Q^2 + 2PQ \cos 	heta) + (P^2 + Q^2 - 2PQ \cos 	heta)$$

This yields:

$$5R^2 = 2P^2 + 2Q^2 \implies P^2 + Q^2 = 2.5R^2$$

We have successfully isolated a relationship between $P^2$, $Q^2$, and $R^2$. Now, we repeat the process to eliminate the cosine term using equation (2).

By subtracting equation (3) from equation (1), we find that $4PQ \cos 	heta = -3R^2$. Substituting this into equation (2) gives us:

$$4R^2 = P^2 + 4Q^2 - 3R^2 \implies P^2 + 4Q^2 = 7R^2$$

### The Final Victory

We now have a simple system of two linear equations: $P^2 + Q^2 = 2.5R^2$ and $P^2 + 4Q^2 = 7R^2$.

Subtracting the first from the second, we get $3Q^2 = 4.5R^2$, which means $Q^2 = 1.5R^2$. Substituting this back, we find $P^2 = R^2$.

The ratio $P^2 : Q^2 : R^2$ is $1 : 1.5 : 1$. Multiplying by $2$ to clear the decimal, we arrive at the elegant result:

**$2 : 3 : 2$**

You have conquered the problem not by brute force, but by understanding the symmetry of the vectors. Keep this mindset, and no problem will ever be too difficult for you.

Animated Solution for Mathematics - Vector Algebra: The resultant of forces $P$ and $Q$ is $R$ . If $Q$ is doubled then $R$ is doubled. If the direction of $Q$ is reversed, then $R$ is again doubled. Then $P^{2} : Q^{2} : R^{2}$ is

Select Answer:

Visualized Solution

Initial Vector Setup

Case 1: Doubling $Q$

Case 2: Reversing $Q$

Eliminating $cos θ$

Isolating $4 P Q cos θ$

Substituting into Equation (2)

Solving for $Q^{2}$

Solving for $P^{2}$

Final Ratio $P^{2} : Q^{2} : R^{2}$

The Sigma Insight: Addition of Vectors

The Vector Dance

Phase 1

Phase 2

Phase 3

The Final Victory

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