Resultant of 2P and Q with Q

❓ Question

The sum of two forces:

$$\vec P \text{ and } \vec Q$$

is:

$$\vec R$$

such that:

$$|\vec R|=|\vec P|$$

The angle $\theta$ (in degrees) that the resultant of:

$$2\vec P \text{ and } \vec Q$$

will make with:

$$\vec Q$$

is ______.

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🖼 Question Image

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✍️ Short Explanation

This is a vector addition problem.

👉 Use magnitude condition first
👉 Find angle between $P$ and $Q$
👉 Then calculate direction of new resultant.

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🔷 Step 1 — Given Vector Relation 💯

Given:

$$\vec R=\vec P+\vec Q$$

and:

$$|\vec R|=|\vec P|$$

Square both sides:

$$|\vec P+\vec Q|^2=|\vec P|^2$$

Using vector formula:

$$P^2+Q^2+2PQ\cos\phi=P^2$$

So:

$$Q^2+2PQ\cos\phi=0$$
$$Q+2P\cos\phi=0$$
$$\boxed{{\displaystyle Q=-2P\cos \varphi }}$$

where $\phi$ is angle between $P$ and $Q$.

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🔷 Step 2 — Resultant of $2\vec P$ and $\vec{Q}$

Let new resultant be:

$$\vec S=2\vec P+\vec Q$$

We need angle made by $\vec S$ with $\vec Q$.

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🔷 Step 3 — Use Direction Formula

Angle $\theta$ with $\vec Q$:

$$\tan\theta = \frac{2P\sin\phi}{Q+2P\cos\phi}$$

But from earlier:

$$Q=-2P\cos\phi$$

So denominator:

$$Q+2P\cos\phi=0$$

Hence:

$$\tan\theta\to\infty$$

Therefore:

$$\theta=90^\circ$$

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🔷 Step 4 — Physical Interpretation

Since denominator becomes zero:

$$\vec S \perp \vec Q$$

So resultant is perpendicular to $Q$.

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🔷 Step 5 — JEE Trap Alert 🚨

❌ Direct triangle law apply kar dena

❌ Magnitude equation square na karna

❌ Direction formula mein denominator sign mistake kar dena

Remember:

$$|\vec A+\vec B|^2=A^2+B^2+2AB\cos\theta$$

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✅ Final Answer

$$\boxed{90^\circ}$$

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