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Given two planes:

$\pi_1: 2x+4y-2z+1=0$

and

$\pi_2: \begin{cases} x(p,t)=2+p-t\\ y(p,t)=2p+t\\ z(p,t)=3-p \end{cases}$

Find the angle between the planes $\pi_1$ and $\pi_2$.

How to find the normal vector from the below plane equations?

$\pi_2: \begin{cases} x(p,t)=2+p-t\\ y(p,t)=2p+t\\ z(p,t)=3-p \end{cases}$

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    $\begingroup$ Write $\pi_2$ as $v+p \vec u + t\vec w$. Then $\vec n = \vec u\times\vec w$ $\endgroup$
    – David P
    Feb 9, 2014 at 17:27

1 Answer 1

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You need to find the normal vectors of these planes. Than the angle between those planes will be equal with the angle between their normal vectors. So you need to find the angle between vectors: $n_1=(2,4,-2)$ and $n_2=(1,1,3)$.

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  • $\begingroup$ Emin, Could you write to me, how to calculate vector $n_2$? $\endgroup$
    – Tomek
    Feb 9, 2014 at 17:35
  • $\begingroup$ from x=2+p−t and y=2p+t we have t=-x+2+p and t=y-2p. By equalizing right sides of the equations we have that -x+2+p=y-2p or p=(x+y-2)/3. Than by replacing that p in z=3−p we have x+y+3z-11=0. $\endgroup$
    – Emo
    Feb 9, 2014 at 17:39

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