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Math Help - Vector equation of intersecting lines. Answer check

  1. #1
    Member
    Joined
    Jun 2008
    Posts
    175

    Red face Vector equation of intersecting lines. Answer check

    The question asks for the vector equation for the line of intersection by the planes x+2y-2z=5 and 6x-3y+2z=8

    Answer is given as
    <br />
\vec r = \left\langle {\frac{{31}}{{15}},\frac{{22}}{{15}},0} \right\rangle  + t\left\langle {2,14,15} \right\rangle <br />

    But when I do the working

    Labeling x+2y-2z=5 as equation 1
    and 6x-3y+2z=8 as equation 2

    Multiply equation 1 by 6 to get 6x+12y-12z=30 call this equation 3

    equation 2 minus equation 3 to get

    <br />
\begin{array}{l}<br />
 15y = 22 - 14z \\ <br />
 y = \frac{{22}}{{15}} - \frac{{14}}{{15}}z \\ <br />
 \end{array}<br />

    Let z=t

    <br />
y = \frac{{22}}{{15}} - \frac{{14}}{{15}}t<br />

    Than from equation 1

    <br />
\begin{array}{l}<br />
 x + 2y - 2z = 5 \\ <br />
 x = 5 - 2\left( {\frac{{22}}{{15}} - \frac{{14}}{{15}}t} \right) - 2t \\ <br />
 x = 5 - \frac{{44}}{{15}} + \frac{{28}}{{15}}t - 2t \\ <br />
 x = \frac{{31}}{{15}} - \frac{2}{{15}}t \\ <br />
 \end{array}<br />

    and so

    <br />
\begin{array}{l}<br />
 \vec r = \left\langle {\frac{{31}}{{15}},\frac{{22}}{{15}},0} \right\rangle  + t\left\langle {\frac{{ - 2}}{{15}},\frac{{ - 14}}{{15}},1} \right\rangle  \\ <br />
 \vec r = \left\langle {\frac{{31}}{{15}},\frac{{22}}{{15}},0} \right\rangle  + t\left\langle { - 2. - 14,15} \right\rangle  \\ <br />
 \end{array}<br />

    Am I right with my negative signs in <-2,-14,15>

    and the answer in the text of <2,14,15> is wrong?
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  2. #2
    Newbie
    Joined
    Oct 2008
    From
    Poland
    Posts
    9
    Well, You've made two mistakes while 'rearranging' equations. There is:

    <br />
\begin{array}{l}<br />
 15y = 22 - 14z \\ <br />
 y = \frac{{22}}{{15}} - \frac{{14}}{{15}}z \\ <br />
 \end{array}<br />
    and should be:
    <br />
\begin{array}{l}<br />
 15y = 22 + 14z \\ <br />
 y = \frac{{22}}{{15}} + \frac{{14}}{{15}}z \\ <br />
 \end{array}<br />
    and the next minus is here:
    <br />
\begin{array}{l}<br />
 x + 2y - 2z = 5 \\ <br />
 x = 5 - 2\left( {\frac{{22}}{{15}} - \frac{{14}}{{15}}t} \right) - 2t \\ <br />
 x = 5 - \frac{{44}}{{15}} + \frac{{28}}{{15}}t - 2t \\ <br />
 x = \frac{{31}}{{15}} - \frac{2}{{15}}t \\ <br />
 \end{array}<br />
    whereas it should be
    <br />
 \begin{array}{l}<br />
  x + 2y - 2z = 5 \\ <br />
  x = 5 - 2\left( {\frac{{22}}{{15}} - \frac{{14}}{{15}}t} \right) + 2t \\ <br />
  x = 5 - \frac{{44}}{{15}} + \frac{{28}}{{15}}t + 2t \\ <br />
  x = \frac{{31}}{{15}} + \frac{2}{{15}}t \\ <br />
  \end{array}<br />
    So, if You take it into consideration, it occurs that the answer in the text is correct.

    best regards!
    katastrofa.nadfioletu
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