91Ó°ÊÓ

The cross product is an important operation in vector algebra, especially in 3D geometry. It allows us to find a vector that is perpendicular to two given vectors. The cross product of two vectors \(\mathbf{A} = a_1 \hat{\mathbf{i}} + a_2 \hat{\mathbf{j}} + a_3 \hat{\mathbf{k}}\) and \(\mathbf{B} = b_1 \hat{\mathbf{i}} + b_2 \hat{\mathbf{j}} + b_3 \hat{\mathbf{k}}\) is given by the determinant:

• \(\mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}\).

In this process, \(\hat{\mathbf{i}}\), \(\hat{\mathbf{j}}\), and \(\hat{\mathbf{k}}\) are unit vectors along the x, y, and z axes respectively. The result, \(\mathbf{A} \times \mathbf{B}\), will be a vector that is perpendicular to both \(\mathbf{A}\) and \(\mathbf{B}\). This property is crucial when dealing with perpendicular lines in space. When applied in our exercise, this operation enabled us to find the direction vector for the new line that is perpendicular to the two given lines.

Direction vectors are vectors that give us information about the direction of a line in 3D space. A direction vector can be any scalar multiple of \(\mathbf{v}\), the basic direction vector, ensuring the line moves along \(\mathbf{v}\)'s direction. Direction vectors are often written in component form, such as \(\mathbf{v} = v_1 \hat{\mathbf{i}} + v_2 \hat{\mathbf{j}} + v_3 \hat{\mathbf{k}}\).

• \(v_1\), \(v_2\), and \(v_3\) are the components along the x, y, and z axes respectively.

In the given exercise, we started with the direction vectors \(\mathbf{d}_1 = 2 \hat{\mathbf{i}} - 2 \hat{\mathbf{j}} + \hat{\mathbf{k}}\) and \(\mathbf{d}_2 = \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + 2 \hat{\mathbf{k}}\) of the two lines. These vectors tell us in which direction each line proceeds in space. Understanding these vectors allowed for the subsequent computation of the cross product to find a vector perpendicular to both.

Perpendicular lines in 3D geometry are lines that intersect at a 90-degree angle. For two vectors \(\mathbf{A}\) and \(\mathbf{B}\), if their dot product is zero, the vectors are perpendicular. Mathematically, \(\mathbf{A} \cdot \mathbf{B} = a_1 b_1 + a_2 b_2 + a_3 b_3 = 0\). In the exercise, the key challenge was to find a line that is perpendicular to two existing lines. Here, the cross product came in handy since it always results in a vector perpendicular to the two initial vectors.

• This perpendicular vector then defined the direction vector for the new line.

Thus, the line we needed was based on this direction vector, which perfectly meets the perpendicular condition given in the problem.

3D geometry deals with solid figures like cubes, spheres, and other forms in three-dimensional space, characterized by three axes: x, y, and z. Each point, line, or shape is represented with coordinates that specify their positions and movements within this space.A foundational concept in 3D geometry is the vector equation of a line:

• This is expressed in the form \(\mathbf{r} = \mathbf{a} + t\mathbf{b}\).

Here, \(\mathbf{r}\) is any point on the line, \(\mathbf{a}\) is a fixed point on the line, t is a parameter (real number), and \(\mathbf{b}\) is the direction vector, showing how the line propagates through space. This formula allows for all possible points to be traced along the line based upon t values.Conclusively, 3D geometry allowed us to see how these vectors and lines relate to each other spatially. The ability to visualize geometric relationships in three-dimensional space aids in accomplishing tasks such as finding perpendicular lines, as seen in our exercise.