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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

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  1. Zenmo
    • one year ago
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    Parametric Equations of a Line in Space: \[x = x _{1}+at, y = y _{1}+bt, and z= z _{1}+ct\]

  2. Michele_Laino
    • one year ago
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    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

  3. Michele_Laino
    • one year ago
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    so we have: \[\Large x\left( t \right) = - 4 + \frac{t}{2}\] similarly for y(t) and z(t)

  4. Zenmo
    • one year ago
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    \[< -4, 1, 0 > + t <\frac{ 1 }{ 2 }, \frac{ 4 }{ 3 }, -1>\]

  5. Zenmo
    • one year ago
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    =\[<-4+\frac{ 1 }{ 2 }t, 1+\frac{ 4 }{ 3 }t, -t>\] Is that correct for parametric equations?

  6. Zenmo
    • one year ago
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    @Michele_Laino

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

  8. Michele_Laino
    • one year ago
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    your parametric equations are right!

  9. Michele_Laino
    • one year ago
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    other parametric equations, which are equivalent to the first ones, are: \[\Large \left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 4} \\ 1 \\ 0 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 3 \\ 8 \\ { - 6} \end{array}} \right)\]

  10. Michele_Laino
    • one year ago
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    so you are right!

  11. Zenmo
    • one year ago
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    Yea, I figured it out, the book just formatted the answer differently by multiplying 6 to each T to get rid of the fractions.

  12. Zenmo
    • one year ago
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    so its x= -4t +3, y= 1+8t, z=-6t

  13. Michele_Laino
    • one year ago
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    it suffice that you change your parameter, namely: \[\Large t \to 6\tau \] where \tau is the new parameter

  14. Michele_Laino
    • one year ago
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    \[\Large \left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 4} \\ 1 \\ 0 \end{array}} \right) + \tau \left( {\begin{array}{*{20}{c}} 3 \\ 8 \\ { - 6} \end{array}} \right)\]

  15. Zenmo
    • one year ago
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    I need a slight help on another problem on converting into symmetric equations, I already did the parametric part.

  16. Michele_Laino
    • one year ago
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    ok! I see your new problem

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