Three vectors extending from the origin are given as \(\mathbf{r}_{1}=(7,3,-2)\), \(\mathbf{r}_{2}=(-2,7,-3)\), and \(\mathbf{r}_{3}=(0,2,3)\). Find \((a)\) a unit vector perpendicular to both \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2} ;(b)\) a unit vector perpendicular to the vectors \(\mathbf{r}_{1}-\mathbf{r}_{2}\) and \(\mathbf{r}_{2}-\mathbf{r}_{3} ;\) (c) the area of the triangle defined by \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2} ;(d)\) the area of the triangle defined by the heads of \(\mathbf{r}_{1}, \mathbf{r}_{2}\), and \(\mathbf{r}_{3}\).

In vector mathematics, a unit vector is one that has a magnitude of 1. It essentially points in the same direction as the original vector but is scaled down to a unit length. Calculating a unit vector is a fundamental step in vector operations such as defining directions in space.

The process of obtaining a unit vector, denoted as \( \hat{v} \), involves dividing the vector \( \mathbf{v} \) by its magnitude \( ||\mathbf{v}|| \). This process is also known as normalization. To normalize a vector, we simply use the formula \( \hat{v} = \frac{\mathbf{v}}{||\mathbf{v}||} \), where \( ||\mathbf{v}|| \) represents the Euclidean norm (or length) of \( \mathbf{v} \).

Let's say vector \( \mathbf{v} \) has components \( (x, y, z) \), then its magnitude is \( ||\mathbf{v}|| = \sqrt{x^2 + y^2 + z^2} \). By dividing each component of \( \mathbf{v} \) by \( ||\mathbf{v}|| \), we obtain the corresponding unit vector. It's useful to remember that unit vectors are vital in expressing directions clearly in various fields of science and engineering.

Vector normalization is the process of converting a vector into a unit vector that maintains the same direction as the original. This is particularly important when you want to focus on the direction of the vector, rather than its magnitude.

The normalization is done by dividing each component of the vector by the vector's magnitude. If you have a vector \( \mathbf{A} \) with components \( (A_x, A_y, A_z) \), then the normalized vector \( \hat{A} \) can be calculated as \( \hat{A} = \frac{\mathbf{A}}{||\mathbf{A}||} = \left( \frac{A_x}{\sqrt{A_x^2 + A_y^2 + A_z^2}}, \frac{A_y}{\sqrt{A_x^2 + A_y^2 + A_z^2}}, \frac{A_z}{\sqrt{A_x^2 + A_y^2 + A_z^2}} \right) \).

In practice, normalization is used when vectors are needed for comparisons, physics simulations, or to represent rotations in 3D space without scaling effects.

To find the area of a triangle formed by vectors, we can utilize the cross product of two of its sides. The cross product \( \mathbf{A} \times \mathbf{B} \) of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) results in a vector that is perpendicular to both, with a length equal to the area of the parallelogram they span.

To get the area of the triangle, we need half of the parallelogram's area, so we calculate \( A_{\triangle} = \frac{1}{2} ||\mathbf{A} \times \mathbf{B}|| \). The magnitude of the cross product gives us this area directly. It's essential for geometry problems in vector calculus and physics where the determination of areas in 3D space is required.

Vector subtraction can be pictured as finding the difference between two points in space represented by vectors, which yields a new vector. For instance, if we have two vectors \( \mathbf{A} \) and \( \mathbf{B} \) with components \( \mathbf{A} = (A_x, A_y, A_z) \) and \( \mathbf{B} = (B_x, B_y, B_z) \) respectively, the resultant vector \( \mathbf{C} = \mathbf{A} - \mathbf{B} \) has components \( (A_x - B_x, A_y - B_y, A_z - B_z) \).

This operation is used for finding relative positions or the displacement vector between two points. For instance, it's essential in mechanics to determine the net displacement of an object, and in computer graphics, to calculate lighting and shading based on angles between vectors.