Statement I Two forces (P+Q) and (P-Q) where P perpendicular to Q, when act at an angle theta_1 to e

Q: Why is the magnitude of P+Q and P-Q the same?

Because P and Q are perpendicular ($\cos 90^\circ = 0$), the cross term $2PQ \cos \theta$ vanishes in both $|P+Q|^2$ and $|P-Q|^2$, leaving just $P^2+Q^2$ for both.

Q: Can we assume P and Q are equal in magnitude?

No, the problem doesn't state that. However, the resultants depend only on the combination $(P^2+Q^2)$, so we can treat $\sqrt{P^2+Q^2}$ as a single variable $F$ without needing individual values of P and Q.

Q: What if P and Q were not perpendicular?

If they weren't perpendicular, the magnitudes of P+Q and P-Q would be different, and the algebra would be much more complex, involving the angle between P and Q.

Q: Why did we square the resultant?

Squaring the resultant removes the square root from the vector addition formula, making it much easier to isolate and solve for the cosine of the angle.

Q: Is it always true that a larger resultant means a smaller angle?

Yes! For two vectors of fixed magnitude, their resultant is maximum when the angle between them is $0^\circ$ and minimum when it's $180^\circ$. As the angle increases, the cosine decreases, reducing the resultant.

Question

Statement I Two forces $(\mathbf{P}+\mathbf{Q})$ and $(\mathbf{P}-\mathbf{Q})$ where $\mathbf{P} \perp \mathbf{Q}$, when act at an angle $	heta_1$ to each other, the magnitude of their resultant is $\sqrt{3(P^2+Q^2)}$, when they act at an angle $	heta_2$, the magnitude of their resultant becomes $\sqrt{2(P^2+Q^2)}$. This is possible only when $	heta_1 < 	heta_2$.
Statement II In the situation given above. $	heta_1 = 60^\circ$ and $	heta_2 = 90^\circ$.
In the light of the above statements, choose the most appropriate answer from the options given below.

Options:

(A) Statement I is false but statement II is true.

(B) Both statement I and statement II are true.

(C) Statement I is true but statement II is false.

(D) Both statement I and statement II are false.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you have two force vectors, $\mathbf{P}$ and $\mathbf{Q}$, that are perfectly perpendicular to each other. The problem introduces two new forces derived from these: $\mathbf{F}_1 = \mathbf{P} + \mathbf{Q}$ and $\mathbf{F}_2 = \mathbf{P} - \mathbf{Q}$.

Before we even think about angles and resultants, we need to understand the nature of these two new forces. Because $\mathbf{P}$ and $\mathbf{Q}$ are at $90^\circ$ to each other, the cross term in their magnitude calculation vanishes.

$$|\mathbf{F}_1| = \sqrt{P^2 + Q^2 + 2PQ \cos 90^\circ} = \sqrt{P^2 + Q^2}$$

$$|\mathbf{F}_2| = \sqrt{P^2 + Q^2 - 2PQ \cos 90^\circ} = \sqrt{P^2 + Q^2}$$

This is a beautiful symmetry! Both of our new forces have the exact same magnitude. Let's call this common magnitude $F$, where $F = \sqrt{P^2 + Q^2}$.

### The Master Equation

Now, we are told that these two forces, each of magnitude $F$, act on a point at an angle $	heta_1$, producing a resultant $R_1 = \sqrt{3(P^2+Q^2)} = \sqrt{3}F$.

We can use the fundamental law of vector addition:

$$R_1^2 = F_1^2 + F_2^2 + 2F_1 F_2 \cos 	heta_1$$

Substituting our known values:

$$(\sqrt{3}F)^2 = F^2 + F^2 + 2(F)(F) \cos 	heta_1$$

$$3F^2 = 2F^2 + 2F^2 \cos 	heta_1$$

### Final Calculation

Let's solve for $	heta_1$. Subtracting $2F^2$ from both sides gives:

$$F^2 = 2F^2 \cos 	heta_1$$

Dividing by $2F^2$, we get:

$$\cos 	heta_1 = \frac{1}{2}$$

This immediately tells us that $	heta_1 = 60^\circ$.

Now, let's repeat the exact same process for the second scenario, where the angle is $	heta_2$ and the resultant is $R_2 = \sqrt{2(P^2+Q^2)} = \sqrt{2}F$.

$$(\sqrt{2}F)^2 = F^2 + F^2 + 2(F)(F) \cos 	heta_2$$

$$2F^2 = 2F^2 + 2F^2 \cos 	heta_2$$

Subtracting $2F^2$ from both sides yields:

$$0 = 2F^2 \cos 	heta_2 \implies \cos 	heta_2 = 0$$

This means $	heta_2 = 90^\circ$.

Comparing the two angles, we clearly see that $60^\circ < 90^\circ$, meaning $	heta_1 < 	heta_2$.

Looking at the given statements: Statement I claims this scenario is possible only when $	heta_1 < 	heta_2$, which aligns perfectly with our finding. Statement II explicitly states $	heta_1 = 60^\circ$ and $	heta_2 = 90^\circ$, which is exactly what we calculated. Therefore, **both statements are undeniably true!**

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Visualized Solution

$Magnitudes of P + Q and P - Q$

$Resultant at angle θ_{1}$

$Calculating θ_{1}$

$Resultant at angle θ_{2}$

$Calculating θ_{2}$

$Final Conclusion$

The Sigma Insight: Vector Addition, Subtraction, and Resolution

Analyzing the Setup

The Master Equation

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnitudes of P+Q and P−Q

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Resultant at angle θ1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating θ1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Resultant at angle θ2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating θ2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Conclusion