Answer
(a) $Q = \left( {34,18,0} \right)$
(b) ${\bf{r}}\left( t \right) = \left( {2 + \frac{{32}}{{\sqrt {21} }}t,2 + \frac{{16}}{{\sqrt {21} }}t,8 - \frac{8}{{\sqrt {21} }}t} \right)$, ${\ \ }$ for $t>0$
Time it will take to reach $Q$:
$t \simeq 4.58$ seconds
Work Step by Step
(a) From Exercise 58 we obtain the normal vector ${\bf{n}}$ to the surface:
${\bf{n}} = \left( {\frac{4}{{\sqrt {21} }},\frac{2}{{\sqrt {21} }}, - \frac{1}{{\sqrt {21} }}} \right)$
The parametrization of the line starting from $P = \left( {2,2,8} \right)$ in the direction of ${\bf{n}}$ can be written as
${\bf{r}}\left( t \right) = \left( {2,2,8} \right) + \left( {\frac{4}{{\sqrt {21} }},\frac{2}{{\sqrt {21} }}, - \frac{1}{{\sqrt {21} }}} \right)t$, ${\ \ }$ for $t>0$
${\bf{r}}\left( t \right) = \left( {2 + \frac{4}{{\sqrt {21} }}t,2 + \frac{2}{{\sqrt {21} }}t,8 - \frac{1}{{\sqrt {21} }}t} \right)$
A particle traveling in the path of the line will pass the point $Q$ on the $xy$-plane if the $z$-component of ${\bf{r}}\left( t \right)$ is equal to zero. Thus,
$8 - \frac{1}{{\sqrt {21} }}{t_Q} = 0$
${t_Q} = 8\sqrt {21} $
Substituting ${t_Q}$ in ${\bf{r}}\left( t \right)$ gives the point $Q$:
${\bf{r}}\left( {8\sqrt {21} } \right) = \left( {2 + \frac{4}{{\sqrt {21} }}\cdot8\sqrt {21} ,2 + \frac{2}{{\sqrt {21} }}\cdot8\sqrt {21} ,8 - \frac{1}{{\sqrt {21} }}\cdot8\sqrt {21} } \right)$
${\bf{r}}\left( {8\sqrt {21} } \right) = \left( {34,18,0} \right)$
So, $Q = \left( {34,18,0} \right)$.
(b) We are given the constant speed of the particle $8$ cm/s. So, in this case, the direction vector of the path is ${\bf{v}} = 8{\bf{n}}$, where ${\bf{n}} = \left( {\frac{4}{{\sqrt {21} }},\frac{2}{{\sqrt {21} }}, - \frac{1}{{\sqrt {21} }}} \right)$.
The path ${\bf{r}}\left( t \right)$ of the particle becomes
${\bf{r}}\left( t \right) = \left( {2,2,8} \right) + 8\left( {\frac{4}{{\sqrt {21} }},\frac{2}{{\sqrt {21} }}, - \frac{1}{{\sqrt {21} }}} \right)t$, ${\ \ }$ for $t>0$
${\bf{r}}\left( t \right) = \left( {2 + \frac{{32}}{{\sqrt {21} }}t,2 + \frac{{16}}{{\sqrt {21} }}t,8 - \frac{8}{{\sqrt {21} }}t} \right)$
The distance from $P$ to $Q$ is
$s = ||PQ|| = \sqrt {{{\left( {34 - 2} \right)}^2} + {{\left( {18 - 2} \right)}^2} + {{\left( {0 - 8} \right)}^2}} = 8\sqrt {21} $
So, the time it takes for the particle to reach $Q$ is
$t = \frac{{8\sqrt {21} }}{8} = \sqrt {21} \simeq 4.58$ seconds
