A vector a has components 2p and 1 with respect to a rectangular cartesian system. This system is ro

Q: Wait, why does the y-component stay 1 in the new system? Does rotation not affect both components?

Great observation! In this specific problem, the problem statement explicitly gives us the new components as $(p+1, 1)$. We aren't calculating the rotation matrix transformation here; we are using the *given* information about the new state. The fact that the y-component remains 1 is a constraint provided by the problem, not a general rule of rotation.

Q: Can't we just use rotation matrices to find the angle $\theta$ first?

You absolutely could! You could use the rotation matrix $R(\theta)$ to express the new components in terms of the old ones and $\theta$. However, that introduces a trigonometric variable $\theta$ which you would then need to eliminate. Using the invariance of magnitude is a much more elegant 'JEE-style' shortcut that bypasses the need for trigonometry entirely.

Q: Why do we square the magnitude instead of just taking the square root?

Squaring is a tactical choice! It eliminates the square root sign, making the algebra cleaner and preventing potential errors with signs. In physics, we often work with squared magnitudes (like in energy or power) to keep expressions polynomial.

Question

A vector $\vec{a}$ has components $2p$ and 1 with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If, with respect to the new system, $\vec{a}$ has components $p + 1$ and 1, then

Options:

(A) $p = 0$

(B) $p = 1$ or $p = -1/3$

(C) $p = -1$ or $p = 1/3$

(D) $p = 1$ or $p = -1$

(E) none of these.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We begin with our vector $\vec{a}$ in a standard Cartesian system $(x, y)$. Its components are given as $2p$ and $1$. Mathematically, we write this as $\vec{a} = 2p\hat{i} + 1\hat{j}$.

When we rotate our coordinate system by an angle $	heta$, we enter a new frame $(x', y')$. The vector itself is anchored; it does not move. However, its projection onto the new axes changes.

We are told the new components are $p+1$ and $1$. This is our bridge. Because the vector's length is an intrinsic property, the squared magnitude in the initial system must equal the squared magnitude in the rotated system: $|\vec{a}|^2_{	ext{initial}} = |\vec{a}|^2_{	ext{rotated}}$.

### The Algebraic Dance

Let us calculate the squared magnitude in the initial system using the Pythagorean theorem:
$$|\vec{a}|^2_{	ext{initial}} = (2p)^2 + 1^2$$

Now, we do the same for the rotated system:
$$|\vec{a}|^2_{	ext{rotated}} = (p+1)^2 + 1^2$$

Since these two quantities are equal, we set them against each other:
$$(2p)^2 + 1^2 = (p+1)^2 + 1^2$$

Notice the elegance here? The $1^2$ term appears on both sides. We can subtract it immediately, simplifying our equation to $(2p)^2 = (p+1)^2$.

This is where many students rush and make a sign error. We expand both sides carefully:
$$4p^2 = p^2 + 2p + 1$$

Bringing all terms to one side gives us the quadratic equation:
$$3p^2 - 2p - 1 = 0$$

### The Final Resolution

We are now in familiar territory. We need to factorize $3p^2 - 2p - 1 = 0$. We look for two numbers that multiply to $-3$ and add to $-2$. Those numbers are $-3$ and $1$.

So, we rewrite the middle term:
$$3p^2 - 3p + p - 1 = 0$$

Grouping the terms, we get $3p(p - 1) + 1(p - 1) = 0$, which factors beautifully into:
$$(3p + 1)(p - 1) = 0$$

Setting each factor to zero, we find our solutions:
**$p = 1$** or **$p = -\frac{1}{3}$**

This result is not just a set of numbers; it is the mathematical confirmation that there are two distinct physical configurations of the coordinate system that satisfy the conditions given in the problem. By focusing on the invariance of the vector's magnitude, we bypassed the complex trigonometry of rotation matrices and arrived at the solution with clarity and confidence.

Select Answer:

Visualized Solution

Initial Vector $a$

Rotated Coordinate System

Invariance of Magnitude

Magnitude in Initial System

Magnitude in Rotated System

Equating the Magnitudes

Simplifying Constants

Expanding the Terms

Standard Quadratic Form

Factorizing the Equation

Solving for $p$

The Sigma Insight: Components of a Vector

Analyzing the Setup

The Algebraic Dance

The Final Resolution

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