Find a set of (a) parametric equations and (b) symmetric equations for the line through the point and parallel to the specified vector or line. (For each line, write the direction numbers as integers.)
Point ( -4, 1, 0), Parallel to v = (1/2)i + (4/3)j - k

Question

zenmo · Answer

Parametric Equations of a Line in Space: $$x = x _{1}+at, y = y _{1}+bt, and z= z _{1}+ct$$

michele_laino · Answer

for part A, we have to apply this eqaution:
$$\Large  X = A + tv$$
whic, can be rewritten by components, like below:
$$\Large  \left( {\begin{array}{*{20}{c}}
  x \ 
  y \ 
  z 
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - 4} \ 
  1 \ 
  0 
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
  {1/2} \ 
  {4/3} \ 
  { - 1} 
\end{array}} \right)$$
where t is the parameter

michele_laino · Answer

so we have:
$$\Large  x\left( t \right) =  - 4 + \frac{t}{2}$$
similarly for y(t) and z(t)

zenmo · Answer

$$< -4, 1, 0 > + t <\frac{ 1 }{ 2 }, \frac{ 4 }{ 3 }, -1>$$

zenmo · Answer

=$$<-4+\frac{ 1 }{ 2 }t, 1+\frac{ 4 }{ 3 }t, -t>$$
Is that correct for parametric equations?

zenmo · Answer

@Michele_Laino

zenmo · Answer

The solution is: x = -4 + 3t, y= 1+8t, z = -6t.     It doesn't make sense?

michele_laino · Answer

your parametric equations are right!

michele_laino · Answer

other parametric equations, which are equivalent to the first ones, are:
$$\Large  \left( {\begin{array}{*{20}{c}}
  x \ 
  y \ 
  z 
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - 4} \ 
  1 \ 
  0 
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
  3 \ 
  8 \ 
  { - 6} 
\end{array}} \right)$$

michele_laino · Answer

so you are right!

zenmo · Answer

Yea, I figured it out, the book just formatted the answer differently by multiplying 6 to each T to get rid of the fractions.

zenmo · Answer

so its x= -4t +3, y= 1+8t, z=-6t

michele_laino · Answer

it suffice that you change your parameter, namely:
$$\Large t \to 6\tau $$
where \tau is the new parameter

michele_laino · Answer

$$\Large  \left( {\begin{array}{*{20}{c}}
  x \ 
  y \ 
  z 
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - 4} \ 
  1 \ 
  0 
\end{array}} \right) + \tau \left( {\begin{array}{*{20}{c}}
  3 \ 
  8 \ 
  { - 6} 
\end{array}} \right)$$

zenmo · Answer

I need a slight help on another problem on converting into symmetric equations, I already did the parametric part.

michele_laino · Answer

ok! I see your new problem