Problem 3 - A Complete Resource Book in Mathematics for JEE Main

Question

A line joining two points \(A(2,0)\) and \(B(3,1)\) is rotated about \(A\) in anti- clockwise direction through an angle \(15^{\circ} .\) If \(B\) goes to \(C\) in the new position, then the coordinates of \(C\) are (A) \(\left(2, \sqrt{\frac{3}{2}}\right)\) (B) \(\left(2,-\sqrt{\frac{3}{2}}\right)\) (C) \(\left(2+\frac{1}{\sqrt{2}}, \sqrt{\frac{3}{2}}\right)\) (D) none of these

Step-by-Step Solution

Answer

(D) None of these

Step 1: Identify the Initial Vector

First, find the vector representing the line segment from point \( A(2, 0) \) to point \( B(3, 1) \). This vector is obtained by subtracting coordinates of \( A \) from \( B \): \( \vec{AB} = (3 - 2, 1 - 0) = (1, 1) \).

Step 2: Convert Angle from Degrees to Radians

To perform a rotation, we need the angle in radians. The given angle is \( 15^{\circ} \). Convert it to radians using the formula: \( \text{Radians} = \text{Degrees} \times \frac{\pi}{180} \). Thus, \( 15^{\circ} = \frac{15 \times \pi}{180} = \frac{\pi}{12} \) radians.

Step 3: Apply the Rotation Matrix

To rotate a point around another point in the plane, use the rotation matrix: \[\begin{bmatrix}\cos\theta & -\sin\theta \\sin\theta & \cos\theta\end{bmatrix}\] Multiply the rotation matrix by the vector \( \vec{AB} \):\[\begin{bmatrix}\cos\frac{\pi}{12} & -\sin\frac{\pi}{12} \\sin\frac{\pi}{12} & \cos\frac{\pi}{12}\end{bmatrix}\begin{bmatrix}1 \1\end{bmatrix}\]

Step 4: Calculate Trigonometric Values

Calculate \( \cos\left(\frac{\pi}{12}\right) \) and \( \sin\left(\frac{\pi}{12}\right) \) using trigonometric identities:\[\cos\left(\frac{\pi}{12}\right) = \sqrt{\frac{2 + \sqrt{3}}{2}} \]\[\sin\left(\frac{\pi}{12}\right) = \sqrt{\frac{2 - \sqrt{3}}{2}}\]

Step 5: Compute the New Coordinates

Substitute the values from the previous step into the rotation matrix and multiply:\[ \begin{bmatrix}\sqrt{\frac{2 + \sqrt{3}}{2}} & -\sqrt{\frac{2 - \sqrt{3}}{2}} \\sqrt{\frac{2 - \sqrt{3}}{2}} & \sqrt{\frac{2 + \sqrt{3}}{2}}\end{bmatrix}\begin{bmatrix}1 \1\end{bmatrix}= \]\[\begin{bmatrix}\sqrt{\frac{2 + \sqrt{3}}{2}} - \sqrt{\frac{2 - \sqrt{3}}{2}} \\sqrt{\frac{2 - \sqrt{3}}{2}} + \sqrt{\frac{2 + \sqrt{3}}{2}}\end{bmatrix}\approx \begin{bmatrix} 1.07 \ 0.26 \end{bmatrix}\]

Step 6: Find the Coordinates of Point C

Add the rotated vector (found as \(\begin{bmatrix} 1.07 \ 0.26 \end{bmatrix}\)) to the coordinates of point \(A\): \[ C(x, y) = \begin{bmatrix} 2 \ 0 \end{bmatrix} + \begin{bmatrix} 1.07 \ 0.26 \end{bmatrix} = \begin{bmatrix} 3.07 \ 0.26 \end{bmatrix}\]Therefore, the coordinates of point \(C\) can be approximated as \(C(3.07, 0.26)\).

Step 7: Compare with Answer Choices

None of the provided answer options match our calculated position for point \( C(3.07, 0.26) \). Therefore, the correct answer from the given choices is (D) none of these.

Key Concepts

Vector Rotation

Vector rotation in coordinate geometry focuses on changing the orientation of a vector while maintaining its magnitude. When you rotate a vector, you're essentially rotating the entire coordinate system around a fixed point. This process uses a specific angle for rotation which is measured counterclockwise. Rotation can be visualized as spinning a vector around one of its endpoints or any other point.

To rotate a vector, the initial vector forms the base, and the angle of rotation adjusts how the vector sits relative to its start point. It's crucial to remember that the magnitude or length of the vector remains unchanged in this rotation process, only the direction changes. This is quite analogous to rotating a pencil without changing its length, simply altering the direction it points.

Rotating a vector using a rotation matrix simplifies the procedure, ensuring that rotations are precise and mathematically sound. This technique is fundamental in computer graphics, robotics, and physics simulations.

Rotation Matrix

A rotation matrix is a powerful tool in mathematics, used to perform rotations in a plane. It allows for a systematic way to rotate points around a given center. In a two-dimensional coordinate system, the rotation matrix is given by:

• \[\begin{bmatrix}\cos\theta & -\sin\theta \\sin\theta & \cos\theta\end{bmatrix}\]

Here, \( \theta \) is the rotation angle. This matrix applies trigonometric functions to rotate coordinates precisely.

For example, if you have a vector or point, and you want to rotate it through an angle \( \theta \), you multiply the rotation matrix by the vector. This is done using matrix-vector multiplication, which results in a new vector indicating the rotated position. The beauty of this process is that it allows us to keep transformations smooth and consistent, important in applications like animation or simulations.

The rotation matrix is central to handling transformations in a structured way, offering both versatility and elegance in solving complex geometric problems.

Trigonometric Identities

Trigonometric identities are essential in dealing with angles and rotations in mathematics. They provide relationships between trigonometric functions like sine, cosine, and tangent, which are crucial for solving geometry problems involving angles.

In the context of a rotation, you often need precise values for functions like \( \cos(\theta) \) and \( \sin(\theta) \) to use in the rotation matrix. Certain angles have known exact values through identities or are calculated using formulas derived from those identities.

For example, when working with the angle \( \frac{\pi}{12} \), which is equivalent to 15 degrees, you can use trigonometric identities to express \( \cos(\frac{\pi}{12}) \) and \( \sin(\frac{\pi}{12}) \), as shown:

• \[\cos\left(\frac{\pi}{12}\right) = \sqrt{\frac{2 + \sqrt{3}}{2}}\]
• \[\sin\left(\frac{\pi}{12}\right) = \sqrt{\frac{2 - \sqrt{3}}{2}}\]

These identities simplify the process of plugging in values into the rotation matrix, allowing for calculations that keep the rotated vector precise. They act as mathematical shortcuts to solve trigonometric equations efficiently.