Answer
a) $(x,y,z)= (1\dfrac{7}{8},0, \ln (\dfrac{1}{2}))$
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) Here, the curve intersects at the xz palne; this means that we have $y=0$.
when $y=2t-1 =0 $, then we have $t=\dfrac{1}{2}$
and $x=2-(0.5)^3=1\dfrac{7}{8}$ and $z=\ln (0.5)$
Thus, $(x,y,z)= (1\dfrac{7}{8},0, \ln (1/2))$
2) Here, we have $x'(t)=\lt 2-3t^2, 2, \dfrac{1}{t}$
The parametric equations at the point $(1,1,0)$:
$r'(1)=\lt -3,2,1 \gt$
and the vector form is: $r(t)=\lt 1-3t, 1+2t, t \gt$ or, $x=1-3t; y=1+2t,z=t$
C) A plane normal to a vector equation is given as follows:
$ax+by+cz=d$
Here, we have $\lt a,b,c \gt =\lt -1, 2, 1 \gt$
$-3x+2y+z=-3(1)+(2) \times (1) +(1) \times (0)$
This implies that $-3x+2y+z=-1$
Hence, $3x-2y-z=1$
