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Zenmo
 one year ago
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
Zenmo
 one year ago
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

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Zenmo
 one year ago
Best ResponseYou've already chosen the best response.0Parametric Equations of a Line in Space: \[x = x _{1}+at, y = y _{1}+bt, and z= z _{1}+ct\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1for 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
 one year ago
Best ResponseYou've already chosen the best response.1so we have: \[\Large x\left( t \right) =  4 + \frac{t}{2}\] similarly for y(t) and z(t)

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.0\[< 4, 1, 0 > + t <\frac{ 1 }{ 2 }, \frac{ 4 }{ 3 }, 1>\]

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.0=\[<4+\frac{ 1 }{ 2 }t, 1+\frac{ 4 }{ 3 }t, t>\] Is that correct for parametric equations?

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.0The solution is: x = 4 + 3t, y= 1+8t, z = 6t. It doesn't make sense?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1your parametric equations are right!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1other 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
 one year ago
Best ResponseYou've already chosen the best response.1so you are right!

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.0Yea, I figured it out, the book just formatted the answer differently by multiplying 6 to each T to get rid of the fractions.

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.0so its x= 4t +3, y= 1+8t, z=6t

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1it suffice that you change your parameter, namely: \[\Large t \to 6\tau \] where \tau is the new parameter

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large \left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {  4} \\ 1 \\ 0 \end{array}} \right) + \tau \left( {\begin{array}{*{20}{c}} 3 \\ 8 \\ {  6} \end{array}} \right)\]

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.0I need a slight help on another problem on converting into symmetric equations, I already did the parametric part.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1ok! I see your new problem
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