91Ó°ÊÓ

The cross product is a vector operation that helps find a vector perpendicular to two given vectors. It is used here to determine the direction of the area parallelogram formed by two vectors. For vectors \(\mathbf{a}\) and \(\mathbf{b}\), the cross product \(\mathbf{a} \times \mathbf{b}\) results in a vector. This vector is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\) and has a magnitude equal to the area of the parallelogram formed by the two vectors.

• The cross product is calculated using a determinant involving the unit vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\).
• For our exercise, this is calculated as:
  \[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -1 & -2 & -1 \ 1 & -5 & -7 \end{vmatrix} = (9, -6, 7) \]

This result allows us to further calculate other necessary values like vector magnitudes and distances.

Understanding vector magnitude, or the length of a vector, is essential when working with vectors. It gives the "size" of the vector, crucial when calculating distances in coordinate geometry. The magnitude of a vector \(\mathbf{v} = (x, y, z)\) is found using the following formula:
\[|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}\]
For example, within our exercise, the magnitude of vector \(\mathbf{a}\), \((-1, -2, -1)\), is:
\[|\mathbf{a}| = \sqrt{(-1)^2 + (-2)^2 + (-1)^2} = \sqrt{6} \]
Similarly, we calculate the magnitude of the cross product \(\mathbf{a} \times \mathbf{b}\) using the same formula:\[|\mathbf{a} \times \mathbf{b}| = \sqrt{9^2 + (-6)^2 + 7^2} = \sqrt{166}\]
These calculations form the components of the distance formula.

The distance formula between a point and a line in three-dimensional space can be derived using vector operations such as cross product and magnitude. The formula used in this problem is:
\[d = \frac{ |\mathbf{a} \times \mathbf{b}| }{ |\mathbf{a}| }\]
This formula represents the shortest distance from a point \(P\) to a line defined by a vector \(\mathbf{a}\) with another point \(Q\) on the line from which vector \(\mathbf{b}\) is formed. This formula elegantly combines vector cross product and magnitude to project the distance onto the line's normal.
Applying in our example, with vectors already calculated:
\[d = \frac{ \sqrt{166} }{ \sqrt{6} }\approx 5.26\]

The calculations demonstrate how vector operations can simplify complex three-dimensional problems.

Coordinate geometry forms the backbone of solving problems related to points, lines, and planes in space. Using vectors, one can efficiently calculate distances and angles within coordinate systems. By defining vectors for line direction and points, we establish relationships between geometric entities.

• The process involves identifying point coordinates and transforming them into vectors.
• In our exercise, the vectors are formed by points \(Q\), \(R\), and \(P\) using their coordinates:
• \(\mathbf{a} = \vec{QR}\) and \(\mathbf{b} = \vec{QP}\).

Through vector transformations and operations, coordinate geometry enables us to solve spatial problems comprehensively. Utilizing operations such as cross products, you can check perpendicularity and areas, which are crucial in verifying correct arrangements or calculations in space.