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The angle between two intersecting planes is defined to be the angle between their normal vectors. Find the angle between the planes $x – 2y + z = 0$ and $2x + 3y – 2z = 0$. Find the parametric equations of the line of intersection of the two planes above.

For the first plane I said $\overrightarrow n_0 =\langle 1, -2, 1 \rangle$ and for the second plane $\overrightarrow n_1 = \langle 2, 3, -2 \rangle$

Then using $\cos(\theta)= \frac{\mathbf A \bullet \mathbf B}{|\mathbf A||\mathbf B|}$ so $\theta = \cos^{-1}\left(\frac{\mathbf A \bullet \mathbf B}{|\mathbf A||\mathbf B|}\right)$

$\mathbf A \bullet \mathbf B = -6$

$|\mathbf A||\mathbf B|= \sqrt{102}$

$\theta = \cos^{-1}\left(\frac{-6}{\sqrt{102}}\right)$

Assuming this is correct so far, how do I find the parametric equations from here?

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  • $\begingroup$ Hint: You have normals to both planes. The line of intersection lies in both planes. Can you think of a way to compute that line from the two normals? (BTW, as an unrelated note, to really complete the first part, what is $\theta$, explicitly?) $\endgroup$ – rogerl Nov 14 '15 at 18:23
  • $\begingroup$ @rogerl $\frac{3\pi}{4}$? $\endgroup$ – hax0r_n_code Nov 14 '15 at 18:24
  • $\begingroup$ Yes, that's right. $\endgroup$ – rogerl Nov 14 '15 at 18:25
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    $\begingroup$ How do you get $|n_0||n_1|$ to be $6\sqrt2$, though? I get $\sqrt 6\cdot\sqrt{17}$. $\endgroup$ – Henning Makholm Nov 14 '15 at 18:27
  • $\begingroup$ @HenningMakholm Yup I just caught that computational error. $\endgroup$ – hax0r_n_code Nov 14 '15 at 18:28
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I don't think calculating the angle between two lines will help you to find the equation of the line of intersection of two lines.

It should be clear that the line of intersection is the line which is perpendicular to the normal of both the given planes.

Then the line will be along the cross product of the normal vector of both the planes.

$\therefore$ doing simple calculation gives us $$ \vec{n_{0}}\times\vec{n_{1}} = \langle1,4,7\rangle $$ So now we know the direction in which the required line is pointing.

Also it is not hard to guess that $(0,0,0)$ lies in both the plane. Thus the equation of line should pass through it.

Hence the parametric equation of the line will become:- $$ t = \dfrac{t}{4} = \dfrac{t}{7} $$ with $t$ as a parameter.

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  • $\begingroup$ Thanks for the the help. Are you saying I calculated the angle between the two planes incorrectly? $\endgroup$ – hax0r_n_code Nov 14 '15 at 18:41
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    $\begingroup$ No. Your method and accuracy are correct. Only that it won't help. There is no point in calculating angle to find the equation of the line of intersection of lines. $\endgroup$ – Prakhar Gupta Nov 14 '15 at 18:45
  • $\begingroup$ Oh, ok. You're saying this won't help me with finding the parametric equations? So that's why I need to use the cross product? The cross product produces the line and direction that intersects both planes. I then use that to produce the parametric equation and determine the parameter t? $\endgroup$ – hax0r_n_code Nov 14 '15 at 18:48
  • $\begingroup$ I understand what you're saying, but finding the angle is also a part of the question. So I guess I'm confused as to if I needed to do this or not. $\endgroup$ – hax0r_n_code Nov 14 '15 at 18:49
  • $\begingroup$ $r(t)=\langle 1, -2, 1 \rangle + t\langle 1,4,7>$ $\endgroup$ – hax0r_n_code Nov 14 '15 at 18:55
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You can remember that $\vec a\times\vec b$ is perpendicular to both $\vec a$ and $\vec b$, so that $\vec n_0\times\vec n_1$ gives the direction of the intersection line. In addition, you know that both planes pass through $(0,0,0)$ and so does the intersection line.

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