## Calculus 8th Edition

(a) $P(2,0,2)$ (b) $x+y=2$
(a) From the given points in the problem, we have $x_1=1+t, y_1=1-t,z_1=2t$ and $x_2=2-s, y_2=s,z_2=2$ Now, arranging $z's$ and $y's$, we have $t=1$ and $s=0$ Putting these values in either parametric equation. $x_1=1+1, y_1=1-1,z_1=2(1)$ Point of intersection: $P(2,0,2)$ (b) Equation of plane is defined as: $a(x-x_0)+b(y-y_0)+c(z-z_0)$ Here, $\lt a,b,c\gt=\lt-2,-2,0\gt$ and $(x_0,y_0,z_0)=(2,0,2)$ $-2(x-2)+(-2)(y-0)+0(z-2)$ $-2x-4-2y=0$ $2x+2y=4$ Hence, $x+y=2$