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Find the equation of a plane that contains both r and s.

Let r be $x=3t, y=-t+1,z=t$ and $x=3t, y=-t+1,z=t-2$,

So far I found the cross product and it gave me $(3t-2)\hat{i}-6t\hat{j}.$

What do I do from here?

brujaja
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  • You can find answer from almost same question here http://math.stackexchange.com/questions/209579/find-the-equation-of-the-plane-that-contains – Ömer Sep 29 '13 at 20:06

1 Answers1

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Your lines are $${\bf l_1}(t)=t(3,-1,1)+(0,1,0)$$ $${\bf l_2}(t)=t(3,-1,1)+(0,1,-2)$$

These are parallel lines through different points. Thus, you may first find a vector $\bf v$ which starts at the first and ends at the second, and then take the cross product of this $\bf v$ with your direction ${\bf w}=(3,-1,1)$ to obtain a normal vector to the plane you seek.

Make a drawing if you cannot visualize what is going on, it usually helps.

Pedro
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