Answer
A) $(x,y,z)= (1\dfrac{7}{8},0, \ln (1/2))$
B) $r(t)=\lt 1-3t, 1+2t, t \gt$ or, $x=1-3t; y=1+2t,z=t$
C) $3x-2y-z=1$
Work Step by Step
A) When a curve intersects the xz palne, $y=0$.
Here, we have $y=2t-1 =0 \implies t=\dfrac{1}{2}$
Thus, $x=2-(1/2)^3=1\dfrac{7}{8}$ and $z=\ln (1/2)$
Thus, $(x,y,z)= (1\dfrac{7}{8},0, \ln (1/2))$
B) Given: $x(t)=1=2-t^3; t=1$
Then $x'(t)=\lt 2-3t^2, 2, \dfrac{1}{t}$
At the point $(1,1,0)$, the parametric equations are:
$r'(1)=\lt -3,2,1 \gt$
$r(t)=\lt 1-3t, 1+2t, t \gt$ or, $x=1-3t; y=1+2t,z=t$
C) A plane normal to a vector equation at a point can be written as:
$ax+by+cz=d$
In this problem, $\lt a,b,c \gt =\lt -1, 2, 1 \gt$
Thus, we have $-3x+2y+z=-3(1)+2 (1) +1 \cdot (0)$
or, $-3x+2y+z=-1$
Hence, $3x-2y-z=1$
