91Ó°ÊÓ

The position vector is a crucial component when identifying a point on a line in three-dimensional space. It serves as a reference point and is often found in parametric equations of lines. In the context of the exercise, we are given the line equation \( \mathbf{r}(t) = \langle 5t+7, 2-t, 12t+4 \rangle \).
From this, we can ascertain that the position vector \( \mathbf{A} \), representing a fixed point on the line, is \( \langle 7, 2, 4 \rangle \). This vector encompasses the constant terms in the line's equation and gives us a starting point on the line.

• The position vector points to a specific location in space on the line.
• It usually includes the non-variable components of the parametric equations for a line.

Position vectors are foundational for establishing the line's presence in space and serve as a starting reference from any parametric point or vector reasoning.

The direction vector indicates the direction of a line. It is crucial because it defines the trajectory along which a line extends. In this example, the direction vector is extracted from the coefficients of \( t \) in the line equation:\( \mathbf{d} = \langle 5, -1, 12 \rangle \).
This vector determines the line's path through space, indicating "how much" the line moves in each coordinate direction as the parameter \( t \) changes.

• The direction vector is derived from the coefficients of the variable \( t \).
• It directly influences the line’s slope and orientation.
• Changes in the parameter \( t \) will shift the point along the direction dictated by the direction vector.

The direction vector is vital in specifying the line’s orientation and is an essential piece for understanding the line’s spatial behavior.

The cross product is a vector operation that is key when working with vectors that are perpendicular to each other. It creates a new vector that is perpendicular to both of the original vectors involved. In this exercise, we calculate the cross product of the line's direction vector \( \mathbf{d} = \langle 5, -1, 12 \rangle \) and vector \( \mathbf{AQ} = \langle -12, 0, 5 \rangle \).
Given by the determinant:\[ \mathbf{d} \times \mathbf{AQ} = \langle -5, -144, -12 \rangle \].

• The cross product yields a vector perpendicular to both \( \mathbf{d} \) and \( \mathbf{AQ} \).
• It is calculated using the determinant of a matrix involving the components of both vectors.

The cross product is especially valuable for determining areas or solving problems involving perpendicular vectors.

The magnitude of a vector is a measure of its length or size. It is calculated as the square root of the sum of the squares of its components. It gives a scalar value representing how "long" or "large" a vector is, independent of its direction. In this scenario, we need the magnitudes of two vectors for our distance calculation:

• For \( \mathbf{d} \times \mathbf{AQ} = \langle -5, -144, -12 \rangle \):\[ |\mathbf{d} \times \mathbf{AQ}| = \sqrt{(-5)^2 + (-144)^2 + (-12)^2} \approx 144.64 \]
• For \( \mathbf{d} \):\[ |\mathbf{d}| = \sqrt{5^2 + (-1)^2 + 12^2} \approx 13.04 \]

The magnitude is a key step for many vector calculations, such as determining distances, and is instrumental in ensuring the overall precision of vector mathematics.