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zordoloom Group Title

Find a vector equation and parametric equations for the line segment that joins P(2, 0, 0) Q(6, 2, 2)

  • one year ago
  • one year ago

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  1. No-data Group Title
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    Remember that to get vector equation you need any point that lies on to the line and a vector that goes in the same direction.

    • one year ago
  2. No-data Group Title
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    You could pick point P or point Q.

    • one year ago
  3. across Group Title
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    You could try\[\mathbf{r}(t)=\mathbf{P}+t(\mathbf{Q}-\mathbf{P}),\qquad0\leq t\leq1.\]Notice that if \(t=0\), then \(\mathbf{r}(t)=\mathbf{P}\), and if \(t=1\), then \(\mathbf{r}(t)=\mathbf{Q}\).

    • one year ago
  4. No-data Group Title
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    How do you make the letters bold @across ?

    • one year ago
  5. zordoloom Group Title
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    @across Do I just plug in the values P and Q for that formula?

    • one year ago
  6. across Group Title
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    \mathbf{}

    • one year ago
  7. sami-21 Group Title
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    just write Large before you type

    • one year ago
  8. No-data Group Title
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    \[\vec{r}=\large{P}\]

    • one year ago
  9. sami-21 Group Title
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    @zordoloom did you get it now ?

    • one year ago
  10. zordoloom Group Title
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    So the formula that across provided, Can I use that to find the equations?

    • one year ago
  11. sami-21 Group Title
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    first find the required parallel vector to the line PQ=<6-2,2-0,2-0)= <4,2,2>

    • one year ago
  12. sami-21 Group Title
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    yes you can use that .

    • one year ago
  13. zordoloom Group Title
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    so, i end up with r(t)=(2,0,0)+t(4,2,-2). Is this correct?

    • one year ago
  14. sami-21 Group Title
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    r(t)=(2,0,0)+t(4,2,2 ) not -2 !!!

    • one year ago
  15. zordoloom Group Title
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    ok

    • one year ago
  16. zordoloom Group Title
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    Thanks!!

    • one year ago
  17. sami-21 Group Title
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    welcome :)

    • one year ago
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