A particle has two velocities of equal magnitude inclined to each other at an angle θ. If one of the

Q: Wait, why does the resultant of two equal vectors always bisect the angle?

Think of it as a parallelogram. When two adjacent sides are equal in length, the parallelogram becomes a rhombus. The diagonal of a rhombus always bisects the vertex angle. Since the resultant vector is the diagonal of the parallelogram formed by the two velocity vectors, it must bisect the angle between them.

Q: Can we use the Law of Sines instead of the tangent formula?

Absolutely! You could form a triangle with sides $u$, $v'$, and $R'$, and use the Law of Sines to relate the angles. However, the tangent formula is often faster for JEE problems because it directly relates the resultant's direction to the magnitudes and the angle between the components. Both methods will lead you to the same elegant result.

Q: Why did we assume $\sin x \neq 0$ when canceling in the final steps?

Great catch on the rigor! We assumed $\sin x \neq 0$ because $x = \frac{\theta}{4}$. If $\sin x = 0$, then $x = 0$, which would imply $\theta = 0$. If $\theta = 0$, the vectors are parallel, and the concept of 'bisecting the angle' becomes trivial or undefined in this context. Since the problem implies a non-trivial geometric configuration, we can safely proceed with the cancellation.

Question

A particle has two velocities of equal magnitude inclined to each other at an angle $	heta$. If one of them is halved, the angle between the other and the original resultant velocity is bisected by the new resultant. Then $	heta$ is

Options:

(A) $90^\circ$

(B) $120^\circ$

(C) $45^\circ$

(D) $60^\circ$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, open field, watching two objects move away from a single point. Each object has a velocity of magnitude $V$, and they are moving at an angle $	heta$ relative to each other.

In the world of vectors, symmetry is our greatest ally. When two vectors have the same magnitude, their resultant vector $\vec{R}$ is not just a random line; it is the perfect mirror, the angle bisector that splits $	heta$ into two equal halves of $\frac{	heta}{2}$.

This is the geometric reality we must hold in our minds before we even touch a pen to paper.

### The Perturbation: A Shift in the Balance

Now, the problem introduces a twist. We take one of these velocities, let's call it $\vec{v}$, and we cut its magnitude in half. It becomes $\vec{v}' = \frac{1}{2}\vec{v}$, with a magnitude of $\frac{V}{2}$.

Suddenly, the perfect symmetry of our rhombus is broken. The new resultant $\vec{R}'$, formed by the original $\vec{u}$ and the new $\vec{v}'$, will no longer bisect the original angle $	heta$.

Instead, the problem gives us a specific condition: this new resultant $\vec{R}'$ bisects the angle between the original resultant $\vec{R}$ and the vector $\vec{u}$. Since the original resultant $\vec{R}$ was at an angle of $\frac{	heta}{2}$ from $\vec{u}$, the new resultant $\vec{R}'$ must be at an angle of $\alpha = \frac{1}{2} 	imes \frac{	heta}{2} = \frac{	heta}{4}$ from $\vec{u}$.

### The Mathematical Bridge

To solve this, we need a tool that connects the magnitudes of the vectors to the angle of the resultant. That tool is the classic direction formula for vector addition:

$$	an \alpha = \frac{B \sin 	heta}{A + B \cos 	heta}$$

Here, our base vector $\vec{A}$ is $\vec{u}$ with magnitude $V$, and our second vector $\vec{B}$ is $\vec{v}'$ with magnitude $\frac{V}{2}$. The angle $\alpha$ is $\frac{	heta}{4}$. Substituting these into our formula, we get:

$$	an\left(\frac{	heta}{4}ight) = \frac{\frac{V}{2} \sin 	heta}{V + \frac{V}{2} \cos 	heta}$$

Notice how the magnitude $V$ appears in every term? We can cancel it out immediately, simplifying our expression to:

$$	an\left(\frac{	heta}{4}ight) = \frac{\sin 	heta}{2 + \cos 	heta}$$

This is the heart of the problem. We have reduced a complex vector scenario into a single, elegant trigonometric equation.

### The Trigonometric Dance

Now, let us simplify the algebra. Let $x = \frac{	heta}{4}$, which means $	heta = 4x$. Our equation becomes $	an x = \frac{\sin 4x}{2 + \cos 4x}$.

This looks intimidating, but remember the power of double-angle identities. We know that $\sin 4x = 2 \sin 2x \cos 2x$ and $\cos 4x = 2 \cos^2 2x - 1$. Substituting these in, we get:

$$\frac{\sin x}{\cos x} = \frac{2 \sin 2x \cos 2x}{2 + (2 \cos^2 2x - 1)} = \frac{2 \sin 2x \cos 2x}{1 + 2 \cos^2 2x}$$

Expanding $\sin 2x$ as $2 \sin x \cos x$, we can cancel $\sin x$ from both sides (assuming $\sin x 
eq 0$). This leaves us with:

$$\frac{1}{\cos x} = \frac{4 \cos x \cos 2x}{1 + 2 \cos^2 2x}$$

Cross-multiplying gives us $1 + 2 \cos^2 2x = 4 \cos^2 x \cos 2x$. Using the identity $2 \cos^2 x = 1 + \cos 2x$, the right side becomes $2(1 + \cos 2x) \cos 2x = 2 \cos 2x + 2 \cos^2 2x$.

The $2 \cos^2 2x$ terms cancel out perfectly, leaving us with the stunningly simple $1 = 2 \cos 2x$, or $\cos 2x = \frac{1}{2}$.

### The Final Revelation

If $\cos 2x = \frac{1}{2}$, then $2x = 60^\circ$. Since $x = \frac{	heta}{4}$, we have $2x = \frac{	heta}{2} = 60^\circ$.

This leads us directly to the final result:

$$	heta = 120^\circ$$

We started with a complex vector problem, navigated through the geometry of bisectors, and arrived at a beautiful trigonometric conclusion. This is the essence of JEE Advanced physics: finding the hidden simplicity within the complexity.

Animated Solution for Mathematics - Vector Algebra: A particle has two velocities of equal magnitude inclined to each other at an angle $θ$ . If one of them is halved, the angle between the other and the original resultant velocity is bisected by the new resultant. Then $θ$ is

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Visualized Solution

Initial Setup: Two Equal Velocities

The Original Resultant $R$

Modifying the System

The New Resultant $R^{'}$

Formula for Resultant Direction

Substituting the Values

Simplifying the Equation

Variable Substitution for Simplicity

Expanding the Multiple Angles

Simplifying the Denominator

Canceling and Cross-Multiplying

Using Half-Angle Identity

Solving for $x$

Finding the Final Angle $θ$

The Sigma Insight: Addition of Vectors

Analyzing the Setup

The Perturbation

The Mathematical Bridge

The Trigonometric Dance

The Final Revelation

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