 # Find a vector equation and parametric equations for the line segment that joins P to Q. P(-1, 0, 1) and Q(-2.5, 0, 2.1). The question aims to find the vector equation and the parametric equations for the line that joins two points, P and Q. The points P and Q are given.

The question depends on the concepts of the vector equation of the line. The vector equation for a finite line with $r_0$ as the initial point of the line. The parametric equation of two vectors joined by a finite line is given as:

$r(t) = (1\ -\ t) r_0 + tr_1 \hspace{0.2in} where \hspace{0.2in} 0 \leq t \leq 1$

The vectors P and Q are given as:

$P = < -1, 0, 1 >$

$Q = < -2.5, 0, 2.1 >$

Here, taking P as the first vector as $r_0$ and Q as the second vector as$r_1$.

Substituting the values of both vectors in the parametric equation, we get:

$r(t) = ( 1\ -\ t) < -1, 0, 1 > + t < -2.5, 0, 2.1 >$

$r(t) = < -1 + t , 0, 1\ -\ t > + < -2.5t, 0, 2.1t >$

$r(t) = < -1 + t\ -\ 2.5t, 0 + 0, 1\ -\ t + 2.1t >$

$r(t) = < -1\ -\ 1.5t, 0, 1 + 1.1t >$

The corresponding parametric equations of the line are calculated to be:

$x = -1\ -\ 1.5t \hspace{0.2in} | \hspace{0.2in} y = 0 \hspace{0.2in} | \hspace{0.2in} z = 1 + 1.1t$

Where the value to t only ranges from [0, 1].

## Numerical Result

The parametric equation of the line joining P and Q is calculated to be:

$r(t) = < -1\ -\ 1.5t, 0, 1 + 1.1t >$

The corresponding parametric equations of the line are calculated to be:

$x = -1\ -\ 1.5t \hspace{0.2in} | \hspace{0.2in} y = 0 \hspace{0.2in} | \hspace{0.2in} z = 1 + 1.1t$

Where the value to t only ranges from [0, 1].

## Example

The vectors $r_0$ and v are given below. Find the vector equation of the line containing $r_0$ parallel to v.

$r_0 = < -1, 2, -1 >$

$v = < 1, -3, 0 >$

We can use the vector equation of the line, which is given as:

$r(t) = r_0 + tv$

Substituting the values, we get:

$r(t) = < -1, 2, -1 > + t < 1, -3, 0 >$

$r(t) = < -1, 2, -1 > + < t, -3t, 0 >$

$r(t) = < -1 + t, 2\ -\ 3t, -1 >$

The corresponding parametric equations are calculated to be:

$x = 1 + t \hspace{0.2in} | \hspace{0.2in} y = 2\ -\ 3t \hspace{0.2in} | \hspace{0.2in} z = -1$