For p 0, a vector v2 = 2i + (p + 1)j is obtained by rotating the vector v1 = √3pi + j by an angle

Q: Wait, why do we reject the negative root $p = -1$?

The problem explicitly states $p > 0$. In JEE Advanced, always check the domain constraints! Ignoring this is a classic trap that can lead to an incorrect vector definition. Always validate your roots against the problem statement.

Q: Why use the cross product for $\sin \theta$ instead of $\sqrt{1 - \cos^2 \theta}$?

Using $\sqrt{1 - \cos^2 \theta}$ introduces a square root and potential sign ambiguity. The cross product gives the value directly and cleanly, avoiding the messy algebra of nested radicals.

Q: Can we use rotation matrices to solve this?

Yes, you absolutely can! Using the rotation matrix $\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$ is a valid approach. However, it is often more computationally intensive than the dot/cross product method. Choose the path that minimizes your chance of calculation errors.

Question

For $p > 0$, a vector $\vec{v}_2 = 2\hat{i} + (p + 1)\hat{j}$ is obtained by rotating the vector $\vec{v}_1 = \sqrt{3}p\hat{i} + \hat{j}$ by an angle $	heta$ about origin in counter clockwise direction. If $	an 	heta = \frac{(\alpha\sqrt{3}-2)}{(4\sqrt{3}+3)}$, then the value of $\alpha$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, holding a vector $\vec{v}_1 = \sqrt{3}p\hat{i} + \hat{j}$. You are tasked with rotating this vector counter-clockwise by an angle $	heta$ to reach a new position $\vec{v}_2 = 2\hat{i} + (p + 1)\hat{j}$.

The most profound realization you can have here is that rotation is an isometry. When you spin a vector around the origin, its length is preserved. It does not stretch, and it does not shrink.

This simple, elegant fact—that $|\vec{v}_1| = |\vec{v}_2|$—is the master key that unlocks the entire problem.

### The Algebraic Hunt

Since the magnitude is preserved, we can work with the squared magnitudes to avoid the headache of square roots: $|\vec{v}_1|^2 = |\vec{v}_2|^2$. Substituting our components, we get:

$$(\sqrt{3}p)^2 + 1^2 = 2^2 + (p + 1)^2$$

Expanding this, we arrive at $3p^2 + 1 = 4 + p^2 + 2p + 1$. Now, gather your terms. Bringing everything to one side gives us $2p^2 - 2p - 4 = 0$.

Dividing by $2$, we find the quadratic $p^2 - p - 2 = 0$. Factoring this, we get $(p - 2)(p + 1) = 0$. We have two candidates: $p = 2$ and $p = -1$.

But look closely at the problem statement—it demands $p > 0$. We discard the negative root and lock in **$p = 2$**. With $p$ determined, our vectors become concrete: $\vec{v}_1 = 2\sqrt{3}\hat{i} + \hat{j}$ and $\vec{v}_2 = 2\hat{i} + 3\hat{j}$.

### The Trigonometric Toolkit

Now, we need $	an 	heta$. We know that $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$. We use the two pillars of vector algebra: the dot product and the cross product.

The dot product gives us the cosine: $\vec{v}_1 \cdot \vec{v}_2 = |\vec{v}_1||\vec{v}_2| \cos 	heta$. Calculating the dot product, we get $(2\sqrt{3})(2) + (1)(3) = 4\sqrt{3} + 3$.

The magnitude of our vectors is $\sqrt{(2\sqrt{3})^2 + 1^2} = \sqrt{13}$. Thus:

$$\cos 	heta = \frac{4\sqrt{3} + 3}{13}$$

Next, the cross product gives us the sine: $|\vec{v}_1 	imes \vec{v}_2| = |\vec{v}_1||\vec{v}_2| \sin 	heta$. For 2D vectors, the magnitude of the cross product is $|x_1y_2 - x_2y_1| = |(2\sqrt{3})(3) - (2)(1)| = 6\sqrt{3} - 2$. Thus:

$$\sin 	heta = \frac{6\sqrt{3} - 2}{13}$$

### The Grand Finale

Finally, we divide:

$$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{\frac{6\sqrt{3} - 2}{13}}{\frac{4\sqrt{3} + 3}{13}}$$

The denominators of $13$ cancel out beautifully, leaving us with:

$$	an 	heta = \frac{6\sqrt{3} - 2}{4\sqrt{3} + 3}$$

Comparing this to the given form $	an 	heta = \frac{\alpha\sqrt{3} - 2}{4\sqrt{3} + 3}$, we see that **$\alpha = 6$**. You have successfully navigated the geometry, the algebra, and the trigonometry.

Enter Numerical Value:

Visualized Solution

Visualizing Vector Rotation

Setting up the Magnitude Equation

Expanding the Equation

Forming the Quadratic Equation

Solving for $p$

Finding the Specific Vectors

Strategy for Finding $tan θ$

Calculating $cos θ$ via Dot Product

Calculating $sin θ$ via Cross Product

Calculating $tan θ$

Comparing and Finding $α$

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Algebraic Hunt

The Trigonometric Toolkit

The Grand Finale

Similar Questions

The vector $\frac{1}{3} (2 \hat{i} - 2 \hat{j} + \hat{k})$ is

Let a unit vector $C$ which makes an angle of $6 0^{\circ}$ with $2 \hat{i} + 2 \hat{j} - \hat{k}$ and angle $4 5^{\circ}$ with $\hat{i} - \hat{k}$ be $C$ . Then $C + (- \frac{1}{2} \hat{i} + \frac{1}{3 2} \hat{j} - \frac{2}{3} \hat{k})$ is :

Let $a = \hat{i} + \hat{j} + \hat{k}$ , $b = \hat{i} - \hat{j} + \hat{k}$ and $c = \hat{i} - \hat{j} - \hat{k}$ be three vectors. A vector $v$ in the plane of $a$ and $b$ , whose projection on $c$ is $\frac{1}{3}$ , is given by

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Let $a, b, c \in R$ be such that $a^{2} + b^{2} + c^{2} = 1$ . If $a cos θ = b cos (θ + \frac{2 π}{3}) = c cos (θ + \frac{4 π}{3})$ , where $θ = \frac{π}{9}$ , then the angle between the vectors $a \hat{i} + b \hat{j} + c \hat{k}$ and $b \hat{i} + c \hat{j} + a \hat{k}$ is :

Let the angle $θ, 0 < θ < \frac{π}{2}$ between two unit vectors $\overset{a}{^}$ and $\hat{b}$ be $sin^{- 1} (\frac{65}{9})$ . If the vector $c = 3 \overset{a}{^} + 6 \hat{b} + 9 (\overset{a}{^} \times \hat{b})$ , then the value of $9 (c \cdot \overset{a}{^}) - 3 (c \cdot \hat{b})$ is

Enter Numerical Value:

Visualized Solution

Visualizing Vector Rotation

Setting up the Magnitude Equation

Expanding the Equation

Forming the Quadratic Equation

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

Finding the Specific Vectors

Strategy for Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tanθ

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cosθ via Dot Product

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sinθ via Cross Product

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

Comparing and Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α