91Ó°ÊÓ

In the context of lines in space, the direction vector plays a crucial role. Simply put, this mathematical entity indicates the direction in which the line extends. For a line to be defined clearly, it must have this vector associated with it. Think of the direction vector like a pointer, specifying not just any path but the specific way or route the line will follow.

Typically, a direction vector is expressed as a combination of unit vectors, such as:

• In the exercise, the direction vector is given by the expression \( a \vec{i} + b \vec{j} + c \vec{k} \).

Here, \( a \), \( b \), and \( c \) are the constants that scale the respective unit vectors \( \vec{i} \), \( \vec{j} \), and \( \vec{k} \).

It's essential to understand that the actual length of the direction vector doesn't change the line itself, only the speed at which a point moves along it. Any multiples of \( \langle a, b, c \rangle \) will still translate the same line.

To describe a line in space, parametric equations are incredibly useful. They allow for the representation of the line in terms of a parameter, commonly denoted as \( t \). This parameter helps trace all possible points on the line through continuous variation.

The parametric form of a line involves:

• A known point on the line, often called the initial point, represented as \( (x_0, y_0, z_0) \).
• A direction vector \( \langle a, b, c \rangle \) as discussed.

Using these, the line's equations are structured as follows:

• \( x = x_0 + at \)
• \( y = y_0 + bt \)
• \( z = z_0 + ct \)

The exercise applies these concepts. We take the point (2, 1, 3) with the direction vector \( \langle a, b, c \rangle \) to derive the parametric equations:

\( x = 2 + at \)
\( y = 1 + bt \)
\( z = 3 + ct \)

This approach allows for the easy specification of any point along the line by altering \( t \). Each value of \( t \) results in a different coordinate, thus mapping out the entire line.

It’s vital to understand the conditions necessary for a line to pass through a given point in space. This is particularly important when you need a line to intersect a specific point, such as the origin \( (0,0,0) \) in this case.

The exercise requires finding values of \( a, b, c \) so the line, originally passing through \( (2,1,3) \), also passes through the origin.
We set up conditions using the equation for a point on the line, substituting the origin coordinates into our parametric representation:

• \( 0 = 2 + at \)
• \( 0 = 1 + bt \)
• \( 0 = 3 + ct \)

From these equations, each solved for \( t \), you derive:

\[t = -\frac{2}{a}, \quad t = -\frac{1}{b}, \quad t = -\frac{3}{c}\]
All these must result in the same \( t \) for any real solution. Equating these gives relationships between \( a, b, \) and \( c \), specifically that:

• \( a = 2b \)
• \( b = \frac{1}{3}c \)

Ultimately, the direction numbers must maintain a ratio, \( a : b : c = 2 : 1 : \frac{2}{3} \). This ensures the line transitions through the desired point.