Problem 3 - A Complete Resource Book in Mathematics for JEE Main 2017

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

The coordinates of C are \(\left(2 + \frac{1}{\sqrt{2}}, \sqrt{\frac{3}{2}}\right)\), which corresponds to option (C).

Step 1: Translate point B relative to A

The coordinate of point A is \((2,0)\), and B is \((3,1)\). We first translate B such that A is the origin. The new coordinates of B with respect to A are: \[ (x_B', y_B') = (3 - 2, 1 - 0) = (1, 1). \]

Step 2: Apply rotation matrix

Since we are rotating point B around point A by an angle of \(15^{\circ}\), we use the rotation matrix:\[\begin{pmatrix}\cos(15^{\circ}) & -\sin(15^{\circ}) \\sin(15^{\circ}) & \cos(15^{\circ})\end{pmatrix}.\]Substitute \(\cos(15^{\circ}) = \frac{\sqrt{6} + \sqrt{2}}{4}\) and \(\sin(15^{\circ}) = \frac{\sqrt{6} - \sqrt{2}}{4}\) into the matrix, and apply it to B's translated coordinates:\[\begin{pmatrix}\frac{\sqrt{6} + \sqrt{2}}{4} & -\frac{\sqrt{6} - \sqrt{2}}{4} \\frac{\sqrt{6} - \sqrt{2}}{4} & \frac{\sqrt{6} + \sqrt{2}}{4}\end{pmatrix}\begin{pmatrix}1 \1\end{pmatrix} =\begin{pmatrix}x_c \y_c\end{pmatrix}\]which results in:\[\begin{pmatrix}\frac{\sqrt{6} + \sqrt{2} - \sqrt{6} + \sqrt{2}}{4} \\frac{\sqrt{6} - \sqrt{2} + \sqrt{6} + \sqrt{2}}{4}\end{pmatrix} = \begin{pmatrix}\frac{2\sqrt{2}}{4} \\frac{2\sqrt{6}}{4}\end{pmatrix} = \begin{pmatrix}\frac{1}{\sqrt{2}} \\sqrt{\frac{3}{2}}\end{pmatrix}.\]

Step 3: Translate back to the original coordinates

The final step is to translate the rotated point back from point A's origin, moving the result of the previous step back to the original coordinate system of point A. So, add A's original coordinates to the rotated coordinates of B:\[ (x_C, y_C) = \left(2 + \frac{1}{\sqrt{2}}, 0 + \sqrt{\frac{3}{2}}\right).\]

Step 4: Compare with given options

The calculated coordinates for point C are \(\left(2 + \frac{1}{\sqrt{2}}, \sqrt{\frac{3}{2}}\right)\). Comparing this with the options provided, it matches option (C).

Key Concepts

Rotation Matrix

In coordinate geometry, rotating a vector or point in a 2D plane requires the use of a rotation matrix. This magical-sounding matrix lets us rotate a point about an origin by a specific angle in a counterclockwise direction. The matrix for rotation by an angle \( \theta \) is represented as follows:
\[\begin{pmatrix} \cos(\theta) & -\sin(\theta) \ \sin(\theta) & \cos(\theta)\end{pmatrix}.\]
What this matrix does is re-align the vector by changing its coordinates while keeping the same relative direction to the origin. When applied to the coordinates of point \( B \) translated relative to point \( A \), it calculates the new position after rotation. By substituting the cosine and sine of the rotation angle, you can determine the transformed coordinates. Using the correct trigonometric values, like those provided for \( 15^{\circ} \), ensures the point is accurately placed after the rotation.

Translation of Coordinates

Translation in coordinate geometry refers to shifting a point or figure across the plane. To rotate point \( B \) around \( A \), it is easier if \( A \) is made the origin. This involves translating \( B \) relative to \( A \).

• Translation results in new coordinates \( (x_B', y_B') = (x_B - x_A, y_B - y_A) \).
• For our example, \( A(2, 0) \) and \( B(3, 1) \) translate to \( B'(1, 1) \).

This translation simplifies calculations because rotating around the origin using a rotation matrix is straightforward. After rotation, the inverse operation returns to the original coordinate system, effectively reversing the initial translation, thus placing the rotated point \( C \) back in the right context of the plane.

Angle of Rotation

The angle of rotation defines how much a point or line should be turned around a fixed point. In the problem, we use an angle of \( 15^{\circ} \). This angle is measured in an anti-clockwise direction but can similarly be handled for clockwise rotations.

• The unit of measure here is degrees, a common ancient yet effective way to determine angles.
• The measure of \( 15^{\circ} \) is critical since it directly affects the sine and cosine values used in the rotation matrix.

Remembering the exact trigonometric values for certain angles, like \( \cos(15^{\circ}) = \frac{\sqrt{6} + \sqrt{2}}{4} \) and \( \sin(15^{\circ}) = \frac{\sqrt{6} - \sqrt{2}}{4} \), is crucial for accurate rotations. These exceptions ensure that transformations are precise.

2D Vector Transformation

2D vector transformation involves changing the position or orientation of a vector, often involving rotations and translations. In this exercise, the transformation is achieved by translating, rotating, and then translating back a vector.
This procedure modifies the initial vector \( B \) relative to \( A \) to a new position as \( C \).

• Initially, it transforms the vector from Cartesian coordinates to a simpler local system (translation step).
• The rotation matrix then alters its orientation while maintaining a specified pivot point (usually the origin after translation).
• The transformation concludes by reverting the vector back to its initial reference framework, ensuring that the vector's new coordinates are appropriately aligned on the plane.

These vector transformations are essential for numerous applications in physics, engineering, and computer graphics, where spatial awareness and coordinate precision are vital.