Answer
Formula used:
$\mathbf{r} = \mathbf{r_0} + t\mathbf{v}$
Vector Equation:
$r=(4+2𝑡)𝑖+(2−𝑡)𝑗+(−3+6𝑡)𝑘$
Parametric Equations:
$𝑥=4+2𝑡, 𝑦=2−𝑡, 𝑧=−3+6𝑡$
Work Step by Step
Solution:
We know the general form of the vector equation of a line:
$r=r_0+tv$
where:
$r_0=(𝑥_0,𝑦_0,𝑧_0)$ is a point on the line , $v=(𝑎,𝑏,𝑐)$ is the direction vector, and $t$ is the parameter.
Here,
$r_0=(4,2,−3), v=(2,−1,6)$
Substituting the values:
$r=(4,2,−3)+t(2,−1,6)$
$r=(4+2t,2−t,−3+6t).$
Thus, the vector equation of the line is:
$r=(4+2t)i+(2−t)j+(−3+6t)k$.
To find the parametric equations of the line, compare the vector equation with the coordinates $(𝑥,𝑦,𝑧)$
$(x,y,z): 𝑥=4+2𝑡 ,𝑦=2−𝑡, 𝑧=−3+6𝑡$.
Thus, the parametric equations are:
$𝑥=4+2𝑡, 𝑦=2−𝑡, 𝑧=−3+6𝑡.$
Final Answer:
Vector Equation:
$r=(4+2𝑡)𝑖+(2−𝑡)𝑗+(−3+6𝑡)𝑘.$
Parametric Equations:
$𝑥=4+2𝑡, 𝑦=2−𝑡, 𝑧=−3+6𝑡.$
