Question

Find a vector equation and parametric equations for the line segment that joins p to q. p(1, −1, 7), q(7, 6, 1) vector equation r(t) = <1+6t,−1+7t,7−6t> parametric equations (x(t), y(t), z(t)) =

Answer 1

Answers: 

- vector equation: r(t) = <1 + 6t, -1 + 7t, 7 - 6t> 
- parametric equations:   
      x = 1 + 6t
      y = -1 + 7t
      z = 7 - 6t

Explanation:

To obtain the vector equation, we first get a vector v that is parallel to the line. To get the vector v, we subtract p from q. So,

v = q - p
   = (7,6,1) - (1,-1,7)
v  = (6, 7, -6)

The vector equation of the line is given by

$r(t) = v_0 + tv$

Where

$v_0$ = a point in the line (we choose point p(1,-1,7))

So, the equation of the line joining p and q is given by

$r(t) = v_0 + tv \\ \indent = \left \langle 1, -1, 7 \right \rangle + t\left \langle 6, 7, -6 \right \rangle \\ \indent = \left \langle 1, -1, 7 \right \rangle + \left \langle 6t, 7t, -6t \right \rangle \\ \indent \boxed{r(t) = \left \langle 1 + 6t, 1 + 7t, 7 - 6t \right \rangle }$

In the parametric equation of the line, we just need to get the x, y and z coordinates in the vector equation.

Since the vector equation is given by 

$r(t) = \left \langle 1 + 6t, 1 + 7t, 7 - 6t \right \rangle$

The parametric equations of the line are given by:

$x(t) = 1 + 6t \\y(t) = 1 + 7t \\z(t) = 7 - 6t$