How to find intersection point of lines $p... \frac{x + 1}{0} = \frac{y - 2}{-1} = \frac{z}{1}$ and $q... \frac{x - 1}{-1} = \frac{y + 6}{3} = \frac{z + 6}{4}$?
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3(1) What have you tried? (2) What's that division by 0 supposed to mean? – Kyle Miller Nov 18 '21 at 18:54
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1I don't even know what that means. It's just a notation trick I guess. So, $p$ has always x-coordinate $-1$. – 1b3b Nov 18 '21 at 18:56
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Have you tried solving this system of equations for $x$, $y$, and $z$? Those are intersection points (if they exist). – Kyle Miller Nov 18 '21 at 18:58
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Let $\frac{x + 1}{0} = \frac{y - 2}{-1} = \frac{z}{1}=t\in\mathbb{R}$. Then $x=-1, y=2-t, z=t$. So, the parametric form of the first line is $c_1(t)=(-1,2-t,t),\ t\in\mathbb{R}$. If we do the same for the second line, we get it's parametric form $c_2(t)=(1-t, 3t-6,4t-6),\ t\in\mathbb{R}$. Thus, $$c_1(t)=c_2(t) \iff t=2. $$ Consequently, the intersection point is $A(-1,0,2)$.
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