A velocity 1/4 m/s is resolved into two components along OA and OB making angles 30° and 45° respect

Q: Wait, why can't I just use $v \cos 45^\circ$ for the component along OB?

That is the most common trap in JEE vector problems! The formula $v \cos \theta$ is only valid when the axes are orthogonal (at $90^\circ$ to each other). Here, the total angle between OA and OB is $30^\circ + 45^\circ = 75^\circ$. Because the axes are not perpendicular, the projection of the vector onto one axis is influenced by the existence of the other. You must use the Sine Rule to account for this non-orthogonality.

Q: How do I remember which angle goes in the numerator of the Sine Rule?

Think of it as a cross-multiplication logic. If you want the component along axis OB, you use the sine of the angle that the velocity vector makes with the *other* axis (OA). It is a beautiful symmetry: to find the component along one path, you look at the angle relative to the path you are ignoring.

Q: Is there a way to solve this using standard x-y components?

Absolutely! You could set up a standard Cartesian coordinate system, resolve the velocity vector into $v_x$ and $v_y$, and then solve a system of two linear equations with two variables (the magnitudes along OA and OB). However, that is a much longer path. The Sine Rule is the 'elegant' shortcut that saves you precious time in the exam hall.

Question

A velocity $\frac{1}{4}$ m/s is resolved into two components along OA and OB making angles $30^\circ$ and $45^\circ$ respectively with the given velocity. Then the component along OB is

Options:

(A) $\frac{1}{8}(\sqrt{6} - \sqrt{2})$ m/s

(B) $\frac{1}{4}(\sqrt{3} - 1)$ m/s

(C) $\frac{1}{4}$ m/s

(D) $\frac{1}{8}$ m/s

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a flat plane, watching a particle zip past you with a velocity of $v = \frac{1}{4}$ m/s. In your mind, you naturally want to break this motion down into simple horizontal and vertical components.

It is the instinct of every physics student—the comfort of the $x$ and $y$ axes. But today, we are going to challenge that instinct. We are resolving this velocity along two arbitrary lines, $OA$ and $OB$, which are not perpendicular to each other.

This is where the beauty of vector geometry truly shines.

### The Parallelogram of Reality

When we resolve a vector, we are essentially constructing a parallelogram where the original vector is the diagonal. If the axes were perpendicular, this parallelogram would be a rectangle, and our standard trigonometric ratios would hold perfectly.

But here, the angle between $OA$ and the velocity is $\alpha = 30^\circ$, and the angle between $OB$ and the velocity is $\beta = 45^\circ$. Because the axes are not at $90^\circ$, our parallelogram is skewed.

To find the component along $OB$, we cannot simply multiply by a cosine. Instead, we must look at the triangle formed by the velocity vector and the components. By applying the Sine Rule, we find that the component along $OB$, which we will call $v_{OB}$, is given by the elegant relation:

$$v_{OB} = \frac{v \sin \alpha}{\sin(\alpha + \beta)}$$

This formula is a powerful tool in your JEE arsenal. It accounts for the 'skew' of the axes by using the sine of the total angle between them, $\alpha + \beta = 75^\circ$, in the denominator.

### The Calculation: A Dance of Fractions

Now, let us breathe and execute the math. We have $v = \frac{1}{4}$ m/s, $\alpha = 30^\circ$, and $\beta = 45^\circ$. Substituting these into our formula, we get:

$$v_{OB} = \frac{\frac{1}{4} \sin 30^\circ}{\sin(30^\circ + 45^\circ)}$$

We know that $\sin 30^\circ = \frac{1}{2}$, so our numerator becomes $\frac{1}{4} 	imes \frac{1}{2} = \frac{1}{8}$. Now, for the denominator, we face $\sin 75^\circ$. Using the compound angle identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$, we calculate:

$$\sin 75^\circ = \sin(45^\circ + 30^\circ) = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}$$

Now, we combine these:

$$v_{OB} = \frac{\frac{1}{8}}{\frac{\sqrt{3}+1}{2\sqrt{2}}} = \frac{1}{8} 	imes \frac{2\sqrt{2}}{\sqrt{3}+1} = \frac{\sqrt{2}}{4(\sqrt{3}+1)}$$

### The Final Polish

We are left with an irrational denominator. To reach the final answer, we rationalize it by multiplying the numerator and denominator by the conjugate, $(\sqrt{3}-1)$:

$$v_{OB} = \frac{\sqrt{2}(\sqrt{3}-1)}{4(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{\sqrt{6}-\sqrt{2}}{4(3-1)} = \frac{\sqrt{6}-\sqrt{2}}{8}$$

And there it is! The component of the velocity along $OB$ is **$\frac{1}{8}(\sqrt{6} - \sqrt{2})$ m/s**.

It is a result that feels earned. By resisting the urge to force an orthogonal solution and instead embracing the geometry of the situation, we have navigated the problem with precision. Remember, in JEE, the most elegant path is often the one that respects the geometry of the problem.

Animated Solution for Mathematics - Vector Algebra: A velocity $\frac{1}{4}$ m/s is resolved into two components along OA and OB making angles $3 0^{\circ}$ and $4 5^{\circ}$ respectively with the given velocity. Then the component along OB is

Select Answer:

Visualized Solution

Visualizing the Velocity Vector

Non-Orthogonal Resolution

Substituting the Values

Evaluating the Denominator

Evaluating the Numerator

Combining the Terms

Simplifying the Fraction

Rationalizing the Denominator

Final Algebraic Steps

Final Answer

The Sigma Insight: Components of a Vector

Analyzing the Setup

The Parallelogram of Reality

The Calculation

The Final Polish

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