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
[tex]r(t) = v_0 + tv[/tex]
Where
[tex]v_0[/tex] = 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
[tex]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 }[/tex]
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
[tex]r(t) = \left \langle 1 + 6t, 1 + 7t, 7 - 6t \right \rangle[/tex]
The parametric equations of the line are given by:
[tex]x(t) = 1 + 6t
\\y(t) = 1 + 7t
\\z(t) = 7 - 6t[/tex]
