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I have a question asking me to find a line in the symmetrical equation that passes through $(1,1,1)$ and also parallel to the following two planes

  • $\pi_1 = [x,y,z] = [1,-3,4]+r[-1,3,2]+s[2,1,2]$
  • $\pi_2 = [x,y,z] = [1,2,1]+r[1,-3,2]+s[1,1,2]$

I manage to find the normal equation of these two planes:

  • $\pi_1 = 4x+6y-7z=-42 $
  • $\pi_2 = -8x+4z=-4 $

Then I am not sure how to proceed with this question. However, we can see that these two planes are not even parallel to each other, so how can we find a line that is parallel two both of them?

The answer key provided the solution as: $\frac{x-1}{3}=\frac{y-1}{5}=\frac{z-1}{6}$, but I drew the line in Geogebra and I do not think the line is parallel to both planes. enter image description here

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  • $\begingroup$ Are you sure you calculated the correct normal form of the plane $\pi_2$? $\endgroup$
    – user289143
    Nov 28, 2019 at 20:44
  • $\begingroup$ Hmmm but I just tried with a calculator, it holds $4*1+6*-3-7*4 = -42$ $\endgroup$
    – Yan Zhuang
    Nov 28, 2019 at 20:48
  • $\begingroup$ Yes sorry, indeed it is correct! $\endgroup$
    – user
    Nov 28, 2019 at 20:49
  • $\begingroup$ Shouldn't it be $\pi_2 : -8x+4z=-4$? $\endgroup$
    – user289143
    Nov 28, 2019 at 20:50
  • $\begingroup$ Oh right, sorry for the mistake I made. I gonna change it now $\endgroup$
    – Yan Zhuang
    Nov 28, 2019 at 20:53

1 Answer 1

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To be parallel to both planes, the direction vector $\vec v$ of the line must be orthogonal to both normal vectors, then a direction vectors is given by

$$\vec v=\vec n_1\times \vec n_2=\begin{vmatrix}\hat i&\hat j&\hat k\\4&6&-7\\-8&0&-4\end{vmatrix}$$

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  • $\begingroup$ Thank you for your help, I got the right equation! However, do you have any idea how my book would got the direction vector of (3,5,6)? I understand they are parallel, but by doing what calculation would render such a result directly as it is not required to simplify direction vector. $\endgroup$
    – Yan Zhuang
    Nov 28, 2019 at 21:01
  • $\begingroup$ By cross product we obtain $(-24,-40,48)$ that is a multiple of $(3,5,-6)$. Therefore it seems there is a typo in the book. I think that using cross product is a very effective way in this case. $\endgroup$
    – user
    Nov 28, 2019 at 21:16
  • $\begingroup$ Yes indeed it is, and I can't think of any other ways to do this. Thank you very much! $\endgroup$
    – Yan Zhuang
    Nov 28, 2019 at 21:18
  • $\begingroup$ @YanZhuang You are welcome! Bye $\endgroup$
    – user
    Nov 28, 2019 at 21:19

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