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You are given two planes in parametric form :

\begin{eqnarray*} S_{1} : \begin{pmatrix} x_{1}\\ x_{2}\\ x_{3} \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} + u_1 \begin{pmatrix} 1\\ 2\\ 0 \end{pmatrix} + v_1 \begin{pmatrix} 0\\ 6\\ 1 \end{pmatrix} \end{eqnarray*}

\begin{eqnarray*} S_{2} : \begin{pmatrix} x_{1}\\ x_{2}\\ x_{3} \end{pmatrix} = \begin{pmatrix} 0\\ -1\\ -1 \end{pmatrix} + u_2 \begin{pmatrix} 1\\ 0\\ 3 \end{pmatrix} + v_2 \begin{pmatrix} 0\\ 2\\ 1 \end{pmatrix} \end{eqnarray*}

Substitute expressions for $x_1, x_2$ and $x_3$ from the parametric form of $S_{2}$ into Cartesian equation for $S_1$ and hence find a parametric vector form of the line of intersection $L$.

I found Cartesian equation for $S_1$ which is $2x_1-x_2+6x_3=0$. But when I substitute expressions from $S_2$, they come with $u_2$ and $v_2$ vectors. How can I eliminate them?

begbeg
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  • $u_1$ and $v_1$ aren’t vectors. They are scalar parameters. After substitution you’ve got a linear equation in these parameters, so solve for one in terms of the other. – amd Jul 09 '19 at 00:45
  • Sorry I wanted to write real numbers for $u_2$ and $v_2$. But I substitute the following expressions: \begin{eqnarray} x_1 = u_2\ x_2 = -1+2v_2\ x_3 = -1 + 3u_2 +v_2 \end{eqnarray} into $S_1$. \begin{equation} 2(u_2)-(-1+2v_2)+6(-1+3u_2+v_2)=0\ 20u_2+4v_2=5 \end{equation} I don't know what I will do after solving one in terms of others.. – begbeg Jul 09 '19 at 10:58
  • Substitute back into $S_2$. – amd Jul 09 '19 at 19:02
  • Thank you I solved it! – begbeg Jul 10 '19 at 10:39
  • You should write your solution up as an answer to this question, then! – amd Jul 10 '19 at 19:45

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Given plane $$S : \vec r = \vec a + u_1 \vec b + u_2 \vec c $$ Alternate way of interpreting this equation as the equation of plane spanned by the vectors $( \vec b - \vec a) = \vec r_1 $ and $(\vec c - \vec a) = \vec r_2$. The normal of the plane can be computed, as $\vec n = \vec r_1 \times \vec r_2 $ and other thing you need is a point which lies on the plane, which here is $\vec a$. Equation of the plane can be written as $ \vec r \cdot \vec n = \vec a \cdot \vec n$. If you will do so, you will obtain the equation of $S_2$ as $10x_1 + 2 x_2 - 3 x_3 = 1$

xrfxlp
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  • If I understood the question correctly, they want me to do a transition between parametric form of $S_2$ and Cartesian form of $S_1$. And finally find the intersection line in parametric form. – begbeg Jul 09 '19 at 11:41
  • At last you have to find the parametric form of intersection of line, right? So what is the problem with my solution, you've got the equation of plane, now carry on. – xrfxlp Jul 09 '19 at 11:45
  • @begbeg Do you have any problem with my approach? – xrfxlp Jul 09 '19 at 12:00
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    Actually I solved before as you said for another part of the question. I mean I've already found the Cartesian equations for both planes and then find the parametric form of the line. But now they asked me specially solve in this way as a second method and compare with what I've found before. There is not any problem with your solution. But I don't think yours is what they asked. – begbeg Jul 09 '19 at 12:41