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$l: x=0,y=t,z=t$ and $\pi:6x+2y-2z=3$ find if they are parallel and how is above the other.

So I took the dot product $(0,1,1)\cdot(6,2,-2)=0$ so they are parallel. To test how is above/below I have set $x=0$ and looked at the $z$ component.

for the line I got $z=-y$ and for the plane I got $z=y-\frac{3}{2}$, which has different values for positive and negative $y$ values.

Is there a different approach?

Christopher Carl Heckman
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gbox
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  • For the line you get $z=y$. – Intelligenti pauca May 14 '16 at 21:21
  • @Aretino why is that? – gbox May 14 '16 at 21:31
  • @ChristopherCarlHeckman the equation of the line is $\frac{x}{0}=\frac{y}{t}=\frac{z}{t}$? – gbox May 14 '16 at 21:40
  • Sort of. However, you're getting away from the original problem. Since the line is parallel to the plane, that means the line lies in a plane whose equation is of the form $6x+2y-2z=k$, for some $k$. (Find $k$.) Then solve for $z$ in both equations, and see which formula between those two is larger; that plane will be above the other one. (And the line will lie in its plane as well.) – Christopher Carl Heckman May 14 '16 at 21:43
  • @ChristopherCarlHeckman so $k=2y-2z$ but how can I solve for $z$ without taking $x=0$ in equation of the plane? – gbox May 14 '16 at 21:51
  • You know that the line $x=0, y=t, z=t$ lies in the plane, so $6\cdot 0 + 2 \cdot t - 2 \cdot t = k$ for all $t$. – Christopher Carl Heckman May 14 '16 at 22:07

2 Answers2

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Take any point $\boldsymbol{p}=(x_p,y_p,z_p)$ on the line and find it's signed distance to the plane

$$ d = \frac{6 x_p + 2 y_p +(-2) z_p - 3}{\sqrt{(6)^2+(2)^2+(-2)^2}} $$

if it is positive then the point is in the same side as the plane normal direction. If it is negative it is in the opposite side.

In your case, the line passes through the origin $(0,0,0)$ yielding a negative $d$.

John Alexiou
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Based on my comments ... The plane containing the line $x=0,y=t,z=t$ has an equation of $6x+2y-2z =0$, which means $z_2={3x+y}$ (the value of $z$ for plane 2). The original plane had an equation of $6x+2y-2z=3$, or $z_1={3x+y} -{3\over2}$. Since $z_2> z_1$, the plane containing the line is above the original plane.

Christopher Carl Heckman
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