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Let $L_1$ and $L_2$ be two lines.

$L_1:r(t)=\langle1,-2,4\rangle+t\langle4,-3,-1\rangle$

$L_2:s(t)=\langle1,1,0\rangle+t\langle2,3,-2\rangle$

The plane vector $ax+by+cz=1$ contains the line $L_2$ and perpendicular to $n$, where n is the normal to the two lines. Determine $a,b,c$.

My way of doing this was to find the normal to the two lines, which is $\frac{1}{7}\langle3,2,6\rangle$. This is n.

Next I used this formula, $A(x-x_0)+B(y-y_0)+C(z-z_0)=0$, where $A=3$, $B=2$, $C=6$, and the point $\langle1,1,0\rangle$.

I got $3(x-1)+2(y-1)+6(z)=0$, after simplification, got me, $3x+2y+6z=5$

and so, $a=3/5$, $b=2/5$ and $c=6/5$.

My answer is really short though, but I cannot understand.

It says

The equation of the plane perpendicular to $7n=\langle 3,2,6\rangle $ and contains $R$ is $3x+2y+6z=3(1)+2(1)+6(0)=1$.

Hence $a=3/5$, $b=2/5$ and $c=6/5$.

Got not a single idea how it got the $5$ from.

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"The plane vector..."? I think you meant simply "The plane", as no "vector" contains a line, though the other way can be true. You also didn't tell us what is $\,n\,$... –  DonAntonio Feb 28 '13 at 12:15
    
Sorry don, n is the normal to the plane and the two lines. –  Yellow Skies Feb 28 '13 at 12:26

1 Answer 1

up vote 1 down vote accepted

The equation of the plane perpendicular to $7n=\langle 3,2,6\rangle $ and contains $R$ is $3x+2y+6z=3(1)+2(1)+6(0)=1$.

You say that you're not sure where the 5 came from. The answer is that there's a typo in the section you quoted. $3(1)+2(1)+6(0) = 5$ and there's your 5.

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My I definitely am too blind to see the typo... Thanks! –  Yellow Skies Feb 28 '13 at 15:05
    
Sorry for all the trouble stack :)# –  Yellow Skies Feb 28 '13 at 15:06

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