Question

How do you find a vector parametric equation $r\left( t \right)$for the line through points $P = \left( { - 3, - 1,1} \right)$and $Q = \left( { - 8, - 4,5} \right)$. If $r\left( 6 \right) = P$ and $r\left( {10} \right) = Q$

Answer 1

Given the coordinates. We have to find the parametric equation for the line. First, we will determine the parametric equation at $x = 6$ and then at $x = 10$. Then, solve the equations for by eliminating one variable and solve for another variable. Later, substitute the value of the variable and solve for the variable. Then, determine the y-parametric equations by substituting the y-coordinates into the equation. Solve the equations to determine the z-parametric equations by substituting z-coordinates into the equation.

Formula used:
The general form of vector parametric equation is given by:
$r\left( t \right) = \left( {{x_p},{y_p},{z_p}} \right) + t\left( {{x_v},{y_v},{z_v}} \right)$
Where the parametric equations are
$x = t{x_v} + {x_p}$
$y = t{y_v} + {y_p}$
$z = t{z_v} + {z_p}$

Complete step by step answer:
We are given the points $P = \left( { - 3, - 1,1} \right)$ and $Q = \left( { - 8, - 4,5} \right)$. First, we will write the parametric equations for x-coordinate.
$\Rightarrow x = t{x_v} + {x_p}$
Now, substitute $t = 6$ and $x = - 3$ into the equation.
$\Rightarrow - 3 = 6{x_v} + {x_p}$ -----(1)
Then, substitute $t = 10$ and $x = - 8$ into the equation.
$\Rightarrow - 8 = 10{x_v} + {x_p}$ -----(2)

Subtract the equation (1) from equation (2) to eliminate ${x_p}$.
$\Rightarrow - 8 - \left( { - 3} \right) = 10{x_v} - 6{x_v} + {x_p} - {x_p}$
$\Rightarrow - 5 = 4{x_v}$
$\Rightarrow {x_v} = - \dfrac{5}{4}$

Now, substitute the value of ${x_v}$ into the equation (1), we get:
$\Rightarrow - 3 = 6\left( { - \dfrac{5}{4}} \right) + {x_p}$
$\Rightarrow - 3 = - \dfrac{{15}}{2} + {x_p}$
$\Rightarrow - 3 + \dfrac{{15}}{2} = {x_p}$
$\Rightarrow \dfrac{9}{2} = {x_p}$

Now, we will determine the y-parametric equation by substituting $t = 6$ and $y = - 1$ into the equation.
$\Rightarrow - 1 = 6{y_v} + {y_p}$ -----(3)
Then, substitute $t = 10$ and $y = - 4$ into the equation.
$\Rightarrow - 4 = 10{y_v} + {y_p}$ -----(4)

Subtract the equation (3) from equation (4) to eliminate ${y_p}$.
$\Rightarrow - 4 - \left( { - 1} \right) = 10{y_v} - 6{y_v} + {y_p} - {y_p}$
$\Rightarrow - 3 = 4{y_v}$
$\Rightarrow {y_v} = - \dfrac{3}{4}$

Now, substitute the value of ${y_v}$ into the equation (3), we get:
$\Rightarrow - 1 = 6\left( { - \dfrac{3}{4}} \right) + {y_p}$
$\Rightarrow - 1 = - \dfrac{9}{2} + {y_p}$
$\Rightarrow - 1 + \dfrac{9}{2} = {y_p}$
$\Rightarrow \dfrac{7}{2} = {y_p}$

Now, we will determine the z-parametric equation by substituting $t = 6$ and $z = 1$ into the equation.
$\Rightarrow 1 = 6{z_v} + {z_p}$ ----(5)
Then, substitute $t = 10$ and $z = 5$ into the equation.
$\Rightarrow 5 = 10{z_v} + {z_p}$ -----(6)

Subtract the equation (5) from equation (6) to eliminate ${z_p}$.
$\Rightarrow 5 - 1 = 10{z_v} - 6{z_v} + {z_p} - {z_p}$
$\Rightarrow 4 = 4{z_v}$
$\Rightarrow {z_v} = 1$

Now, substitute the value of ${z_v}$ into the equation (5), we get:
$\Rightarrow 1 = 6\left( 1 \right) + {z_p}$
$\Rightarrow 1 = 6 + {z_p}$
$\Rightarrow 1 - 6 = {z_p}$
$\Rightarrow - 5 = {z_p}$
Substitute the values into the general form of vector parametric equation $r\left( t \right) = \left( {{x_p},{y_p},{z_p}} \right) + t\left( {{x_v},{y_v},{z_v}} \right)$
$\Rightarrow r\left( t \right) = \left( {\dfrac{9}{2},\dfrac{7}{2}, - 5} \right) + t\left( { - \dfrac{5}{4}, - \dfrac{3}{4},1} \right)$

Hence, the vector parametric equation for the line is $r\left( t \right) = \left( {\dfrac{9}{2},\dfrac{7}{2}, - 5} \right) + t\left( { - \dfrac{5}{4}, - \dfrac{3}{4},1} \right)$

Note: The vector parametric equation of the line passing through the points is determined by writing the x, y and z coordinates of the parametric equation plus the coordinates of the vector equation.