## University Calculus: Early Transcendentals (3rd Edition)

$x=1,y=2-t, z=-t$
The parametric equations of a straight line can be found by knowing the value of a vector, such as $v=v_1i+v_2j+v_3k$, passing through a point $P(x_0,y_0,z_0)$ as follows: $x=x_0+t v_1,y=y_0+t v_2; z=z_0+t v_3$ Here, we have $P(1,2,0)$ and $v=\lt 1-1,1-2,-1-0 \gt =\lt 0,-1,-1 \gt$ Thus, we get the parametric equations: $x=1,y=2-t, z=-t$