Vector Transformation: Rotation And Reflection Explained

When we talk about vectors in mathematics, we're often interested in how they change when subjected to certain operations. Two fundamental transformations that vectors can undergo are rotation and reflection. Understanding these transformations is crucial in various fields, from computer graphics and physics to engineering and linear algebra. In this article, we'll delve into how to perform these operations on a specific vector, $\langle 3, -7\rangle$, by first rotating it by $\frac{3\pi}{4}$ radians and then reflecting the result across the $y$-axis. We'll break down each step to make the process clear and accessible.

Understanding Vector Rotations

Rotations are a type of geometric transformation that turns a figure or a vector around a fixed point, called the center of rotation. In a 2D plane, when we rotate a vector, we're essentially changing its direction without altering its magnitude (length). The angle of rotation determines how much the vector is turned, and its direction (clockwise or counterclockwise) is also important. A positive angle, like the $\frac{3\pi}{4}$ radians we'll be using, typically signifies a counterclockwise rotation. To rotate a vector $\langle x, y\rangle$ by an angle $\theta$ counterclockwise, we can use a rotation matrix. The rotation matrix for an angle $\theta$ is given by:

$$R(\theta) = \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}$$

To find the rotated vector, we multiply this matrix by the column vector representing our original vector:

$$\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$$

Let's apply this to our specific vector $\langle 3, -7\rangle$ and the angle $\theta = \frac{3\pi}{4}$. First, we need to find the values of $\cos(\frac{3\pi}{4})$ and $\sin(\frac{3\pi}{4})$. The angle $\frac{3\pi}{4}$ is in the second quadrant, where cosine is negative and sine is positive. The reference angle is $\frac{\pi}{4}$.

$\cos(\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2}$ $\sin(\frac{3\pi}{4}) = \frac{\sqrt{2}}{2}$

Now, let's set up the multiplication:

$$\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} -\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \end{bmatrix} \begin{bmatrix} 3 \\ -7 \end{bmatrix}$$

Performing the matrix multiplication:

$x' = (-\frac{\sqrt{2}}{2})(3) + (-\frac{\sqrt{2}}{2})(-7) = -\frac{3\sqrt{2}}{2} + \frac{7\sqrt{2}}{2} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$

$y' = (\frac{\sqrt{2}}{2})(3) + (-\frac{\sqrt{2}}{2})(-7) = \frac{3\sqrt{2}}{2} + \frac{7\sqrt{2}}{2} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$

So, after rotation, our vector becomes $\langle 2\sqrt{2}, 5\sqrt{2}\rangle$. This new vector has the same length as the original but is now pointing in a new direction, specifically $\frac{3\pi}{4}$ radians counterclockwise from its original orientation.

Understanding Vector Reflections

Reflections, also known as flips, are another type of geometric transformation. When a vector is reflected across a line, it creates a mirror image of itself on the opposite side of that line. The distance from the line to the original vector is the same as the distance from the line to the reflected vector. In our case, we need to reflect the rotated vector $\langle 2\sqrt{2}, 5\sqrt{2}\rangle$ across the $y$-axis. Reflecting a point $(x, y)$ across the $y$-axis results in a new point $(-x, y)$. The $x$-coordinate changes its sign, while the $y$-coordinate remains the same. This is because the $y$-axis acts as the mirror. Any point on the $y$-axis has an $x$-coordinate of 0. When you reflect a point, you're essentially moving it horizontally to the other side of the $y$-axis, maintaining its vertical position.

Applying this reflection rule to our rotated vector $\langle 2\sqrt{2}, 5\sqrt{2}\rangle$:

Let the rotated vector be $\mathbf{v'} = \langle x', y'\rangle = \langle 2\sqrt{2}, 5\sqrt{2}\rangle$.

When we reflect $\mathbf{v'}$ across the $y$-axis, the new vector, let's call it $\mathbf{v''}$, will have coordinates $(-x', y')$.

So, the new $x$-coordinate will be $-(2\sqrt{2}) = -2\sqrt{2}$. The new $y$-coordinate will remain $5\sqrt{2}$.

Therefore, the resulting vector after reflection is $\langle -2\sqrt{2}, 5\sqrt{2}\rangle$. This vector is the mirror image of the rotated vector across the $y$-axis.

Combining Transformations and Finding the Resulting Vector

We've now completed both steps: rotation and reflection. The problem states that the resulting vector is $\left[\begin{array}{l}a \ b\end{array}\right]$. Based on our calculations, the vector after rotation and then reflection is $\langle -2\sqrt{2}, 5\sqrt{2}\rangle$. Therefore, we can equate this to the given form:

$\left[\begin{array}{l}a \ b\end{array}\right] = \left\langle -2\sqrt{2}, 5\sqrt{2}\right\rangle$

This means:

$a = -2\sqrt{2}$ $b = 5\sqrt{2}$

So, the values are $a = -2\sqrt{2}$ and $b = 5\sqrt{2}$. These values represent the components of the final vector after undergoing both transformations. It's important to perform these transformations in the specified order, as the order of operations can significantly change the final outcome in vector transformations.

Step-by-Step Summary

1. Original Vector: Start with the vector $\langle 3, -7\rangle$. This can be written as a column matrix $\begin{bmatrix} 3 \ -7 \end{bmatrix}$.
2. Rotation: Rotate the vector by $\frac{3\pi}{4}$ radians counterclockwise. We use the rotation matrix:

   $$R(\frac{3\pi}{4}) = \begin{bmatrix} -\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \end{bmatrix}$$

   Multiply the rotation matrix by the original vector:

   $$\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} -\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \end{bmatrix} \begin{bmatrix} 3 \\ -7 \end{bmatrix} = \begin{bmatrix} 2\sqrt{2} \\ 5\sqrt{2} \end{bmatrix}$$

   The rotated vector is $\langle 2\sqrt{2}, 5\sqrt{2}\rangle$.
3. Reflection: Reflect the rotated vector $\langle 2\sqrt{2}, 5\sqrt{2}\rangle$ across the $y$-axis. The rule for reflection across the $y$-axis is $(x, y) \rightarrow (-x, y)$. Applying this rule gives: $x'' = -(2\sqrt{2}) = -2\sqrt{2}$ $y'' = 5\sqrt{2}$ The resulting vector is $\langle -2\sqrt{2}, 5\sqrt{2}\rangle$.
4. Final Result: The problem states the resulting vector is $\left[\begin{array}{l}a \ b\end{array}\right]$. Therefore, $a = -2\sqrt{2}$ and $b = 5\sqrt{2}$.

This detailed breakdown illustrates how vector transformations, such as rotation and reflection, are applied step-by-step. Each operation changes the vector's position or orientation in the plane, and understanding the underlying mathematical principles allows us to predict and analyze these changes accurately. These concepts are fundamental in many areas of applied mathematics and science.

For further exploration into the fascinating world of linear algebra and vector transformations, you can check out Khan Academy's comprehensive resources on the subject.

Conclusion

We've successfully navigated the process of rotating a vector and then reflecting it across the $y$-axis. Starting with $\langle 3, -7\rangle$, we first applied a rotation of $\frac{3\pi}{4}$ radians, transforming it into $\langle 2\sqrt{2}, 5\sqrt{2}\rangle$. Subsequently, reflecting this intermediate vector across the $y$-axis yielded the final vector $\langle -2\sqrt{2}, 5\sqrt{2}\rangle$. This means that for the resulting vector $\left[\begin{array}{l}a \ b\end{array}\right]$, we have $a = -2\sqrt{2}$ and $b = 5\sqrt{2}$. The journey through these transformations highlights the systematic nature of vector operations and their importance in various mathematical and scientific applications. Mastering these techniques provides a solid foundation for understanding more complex concepts in geometry, physics, and engineering.

To deepen your understanding of vector spaces and transformations, I highly recommend exploring the resources available on Brilliant.org which offers interactive learning experiences in mathematics and computer science.