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**[Solved]Two planes->x+2y-2z=2 & 2x-3y+6z=3 intersect in straight line l**

## Homework Statement

Hi.I've two planes' equation x+2y-2z=2 & 2x-3y+6z=3.The planes intersect in the straight line l. The question:Find a vector equation for the line l.

## Homework Equations

a.n=0 if they're both perpendicular 2 ea other

r.n=a.n ,where n=perpendicular to both r and a

3 dimensional vector equation formula:{A position vector of line}+t{direction vector of line},where t is a variable.

## The Attempt at a Solution

1(x)+2(y)-2(z)=0

1(2)+2(1)-2(2)=0

[tex]\left(\begin{array}{cc}2\\1\\2\end{array}\right)[/tex]

x+2y-2z=2

(2)+2(1)-2(1)=2

[tex]\left(\begin{array}{cc}2\\1\\1\end{array}\right)[/tex]

[tex]r_{1}[/tex]=[tex]\left(\begin{array}{cc}2\\1\\1\end{array}\right)[/tex]+t[tex]\left(\begin{array}{cc}2\\1\\2\end{array}\right)[/tex]

2x-3y+6z=0

2(1)+3(8/3)+6(1)=0

3[tex]\left(\begin{array}{cc}1\\\frac{8}{3}\\1\end{array}\right)[/tex]=[tex]\left(\begin{array}{cc}3\\8\\3\end{array}\right)[/tex]

2x-3y+6z=3

2(3)-3(7)+6(3)=3

[tex]\left(\begin{array}{cc}3\\7\\3\end{array}\right)[/tex]

[tex]r_{2}[/tex]=[tex]\left(\begin{array}{cc}3\\7\\3\end{array}\right)[/tex]+m[tex]\left(\begin{array}{cc}3\\8\\3\end{array}\right)[/tex]

Putting [tex]r_{2}[/tex] in 1st plane cartesian equation getting:

(3+3m)+2(7+8m)-2(3+3m)=2

m=[tex]\left(\begin{array}{cc}\frac{-9}{13}\\\end{array}\right)[/tex]

[tex]r_{2}[/tex]=[tex]\left(\begin{array}{cc}3\\7\\3\end{array}\right)[/tex]+m[tex]\left(\begin{array}{cc}3\\8\\3\end{array}\right)[/tex]

=[tex]\left(\begin{array}{cc}3\\7\\3\end{array}\right)[/tex]+[tex]\left(\begin{array}{cc}\frac{-9}{13}\\\end{array}\right)[/tex][tex]\left(\begin{array}{cc}3\\8\\3\end{array}\right)[/tex]

=[tex]\left(\begin{array}{cc}\frac{12}{13}\\\frac{19}{13}\\\frac{12}{13}\end{array}\right)[/tex]...[tex]\alpha[/tex]

Putting [tex]r_{1}[/tex] into 2nd cartesian equation getting:

2(2+2t)-3(1+t)+6(1+2t)=3

t=[tex]\frac{-4}{13}[/tex]

[tex]r_{1}[/tex]=[tex]\left(\begin{array}{cc}2\\1\\1\end{array}\right)[/tex]+[tex]\left(\begin{array}{cc}\frac{-4}{13}\\\end{array}\right)[/tex][tex]\left(\begin{array}{cc}2\\1\\2\end{array}\right)[/tex]

=[tex]\left(\begin{array}{cc}\frac{18}{13}\\\frac{9}{13}\\\frac{5}{13}\end{array}\right)[/tex]...[tex]\beta[/tex]

[tex]\beta[/tex]-[tex]\alpha[/tex]=[tex]\left(\begin{array}{cc}\frac{18}{13}\\\frac{9}{13}\\\frac{5}{13}\end{array}\right)[/tex]-[tex]\left(\begin{array}{cc}\frac{12}{13}\\\frac{19}{13}\\\frac{12}{13}\end{array}\right)[/tex]=[tex]\frac{1}{13}[/tex][tex]\left(\begin{array}{cc}6\\-10\\-7\end{array}\right)[/tex]

l's vector equation=[tex]\left(\begin{array}{cc}\frac{18}{13}\\\frac{9}{13}\\\frac{5}{13}\end{array}\right)[/tex]+w[tex]\left(\begin{array}{cc}6\\-10\\-7\end{array}\right)[/tex]

Anything wrong thus?

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