Chapter 16: Integrals and Vector Fields - Section 16.1 - Line Integrals - Exercises 16.1 - Page 943: 6

Graph(b)

Given: $r(t)=tj+(2-2t)k$ Here, $x=0; y=t,z=2-2t$ This is a line $z=2-2y$ along the plane $x=0$ whose intercept has the form of $y+\dfrac{z}{2}=1$. Thus, $y$ intercept is $=1$, and $z$-intercept is $=2$ The answer is: Graph(b).