91Ó°ÊÓ

The cross product of two vectors is an essential operation in vector mathematics that is particularly useful in physics and engineering. It helps us find a vector that is perpendicular to two given vectors in three-dimensional space. To compute the cross product between two vectors \( \mathbf{A} \) and \( \mathbf{B} \), we use the determinant of a 3x3 matrix that includes the unit vectors \( \hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}} \), and the components of \( \mathbf{A} \) and \( \mathbf{B} \): \[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} \] This results in a new vector that has components:

• \( (a_2b_3 - a_3b_2) \) along \( \hat{\mathbf{i}} \)
• \( (a_3b_1 - a_1b_3) \) along \( \hat{\mathbf{j}} \)
• \( (a_1b_2 - a_2b_1) \) along \( \hat{\mathbf{k}} \)

The newly derived vector is orthogonal—meaning perpendicular—to both \( \mathbf{A} \) and \( \mathbf{B} \). This process was employed in our example to find the vector orthogonal to \( \mathbf{OA} \) and \( \mathbf{BC} \).

The magnitude of a vector is a measure of its length or size. It is an important concept because it helps us understand how large or small a vector is, irrespective of its direction. In three dimensions, the formula to calculate the magnitude of a vector \( \mathbf{V} = x \hat{\mathbf{i}} + y \hat{\mathbf{j}} + z \hat{\mathbf{k}} \) is: \[ |\mathbf{V}| = \sqrt{x^2 + y^2 + z^2} \] This equation comes from the Pythagorean theorem and essentially finds the Euclidean distance from the origin to the point \( (x, y, z) \) in space.

• If all the components of the vector are squared and then summed up, the square root of this sum will give the magnitude.
• In our example, calculating \( |\mathbf{OA} \times \mathbf{BC}| \) involved squaring the components from the cross product, adding them, and taking the square root to find \( \sqrt{2} \).

The shortest distance between two skew lines—which are lines that do not intersect and are not parallel—is a fascinating geometric problem often dealt with using vector algebra. To find this distance, you use the formula: \[ m = \frac{|\mathbf{OA} \times \mathbf{BC}|}{|\mathbf{BC}|} \] Here,

• \(|\mathbf{OA} \times \mathbf{BC}| \) represents the magnitude of the cross product of the direction vectors of the two lines.
• \(|\mathbf{BC}| \) is the magnitude of one of the line's direction vectors, which acts as the denominator.

This gives the perpendicular distance between the two lines, which corresponds to the minimum distance where a segment perpendicular to both lines can span across them. The exercise used this technique to determine the shortest distance \( m \) between the lines \( OA \) and \( BC \), resulting in a distance of 1.