QUESIONS
Identifying the Equation of a Line Segment in 3D
Choose one $\cdot 5$ points
Identify the equation of aline-segment with origin $\vec{S}_{0}-(3,4,1)$ and direction $\vec{v}-(1,2,1)$
\[
\left\{\begin{array}{l}
z=3+t \\
y=4+2 t, t \in \mathbb{R}^{*} \\
z=1+t
\end{array}\right.
\]
\[
\left\{\begin{array}{l}
z=3+t \\
y=4+2 t, 0 \leq t \leq 1 \\
z=1+t
\end{array}\right.
\]
\[
\left\{\begin{array}{l}
z=3+t \\
y=4+2 t, t \geq 0 \\
z=1+t
\end{array}\right.
\]

Step 1 ：The question is asking for the equation of a line segment in 3D space. The line segment has an origin at the point (3,4,1) and a direction vector of (1,2,1).

Step 2 ：The general equation for a line in 3D space is given by: \(\vec{r} = \vec{a} + t\vec{b}\) where \(\vec{r}\) is the position vector of a point on the line, \(\vec{a}\) is the position vector of a known point on the line, \(\vec{b}\) is the direction vector of the line, and t is a scalar parameter.

Step 3 ：In this case, \(\vec{a} = (3,4,1)\) and \(\vec{b} = (1,2,1)\). So, the equation of the line is: \(\vec{r} = (3,4,1) + t(1,2,1)\)

Step 4 ：This can be written in component form as: \(x = 3 + t\), \(y = 4 + 2t\), \(z = 1 + t\)

Step 5 ：Since the question is asking for the equation of a line segment, not a line, we need to specify the range of t. The line segment starts at the origin when t=0 and follows the direction vector. Therefore, t should be greater than or equal to 0. The end of the line segment is not specified in the question, so we cannot determine the upper limit of t.

Step 6 ：Therefore, the equation of the line segment is: \(\left\{\begin{array}{l} x=3+t \\ y=4+2 t, t \geq 0 \\ z=1+t \end{array}\right.\)

Step 7 ：Final Answer: \(\boxed{\left\{\begin{array}{l} x=3+t \\ y=4+2 t, t \geq 0 \\ z=1+t \end{array}\right.}\)