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There are 2 planes:

  • $Q: 2x-y=2$

  • $R: x-y-z=1$

How do I find the line $p=(1,2,-1)$, which intersects $Q$ and $R$ plane.

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Well, what have you tried? –  espen180 Nov 8 '12 at 6:57
    
I highlight letters and then i don´t know how to continued? –  Luka Toni Nov 8 '12 at 7:08
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1 Answer 1

up vote 2 down vote accepted

Hint:

The cross product of two normal vectors gives a vector which is perpendicular to both of the planes given in your question and is therefore parallel to the line of intersection of the two planes. You can either use the cross product to find the intersection of the two planes or find the parametric representation.

$Q$ X $R$ = $\left(\begin{array}{ccc} i& j& k \\ 2 & -1 & 0 \\ 1 & -1 & -1 \end{array}\right)$ = i$\left(\begin{array}{ccc} -1 & 0 \\ -1 & -1 \end{array}\right)$ - j$\left(\begin{array}{ccc} 2 & 0 \\ 1 & -1 \end{array}\right)$ + k$\left(\begin{array}{ccc} 2 & -1 \\ 1 & -1 \end{array}\right)$ = $ i + 2j - k$

Then we use the line P to find the intersection.

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I use the cross product to find the intersection Q (2, -1, 0) and R (1, -1, -1) = QxR (1,2,-1) and what can i do with this? THANKS –  Luka Toni Nov 8 '12 at 8:21
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