If the equation of my plane is $(x−1,y−5,z−6)\cdot\vec{N}=0$. What can I do to convert it to parametric equation and vector equation?
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Similar question. P.S. I actually like Ivo's answer better than mine. – Mar 18 '15 at 01:41
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i thought the general formula was like this: Ax + By + Cz + D = 0, if N⃗ =(-8,-14,16) – Julio Mar 18 '15 at 01:44
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If for example the general equation of the plane is: $3x + 4y - z = 5$, then let:
$x = s, y = t$, then solve for $z = 3s + 4t-5 \to (x,y,z) = (s,t,3s+4t-5) = s(1,0,3) + t(0,1,4) + (0,0,-5)$ is the parametric equation.
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But my general equation is (x-1,y-5,z-6)⋅N⃗ =0 N⃗ = (-8,-14,16) How do I convert it to the normal general equation – Julio Mar 18 '15 at 01:49
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What I don't understand is that in your example you have 3x+4y−z=5, that is easy to follow, but how can I simplify this: (x−1,y−5,z−6)⋅(-8,-14,16)= 0 – Julio Mar 18 '15 at 01:52
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@Julio That's correct. So as you can see $-8x-14y+16z=18$ is exactly the same equation as $(x-1,y-5,z-6)\cdot(-8,-14,16)=0$. – Mar 18 '15 at 02:29
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Also, if you want to comment a specific person you should add @[username] so that they get messaged. – Mar 18 '15 at 02:31
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You have $$ \eqalign{ & 0 = \left( {x - 1,y - 5,z - 6} \right) \cdot \mathop N\limits^ \to \quad \Rightarrow \cr & \Rightarrow \quad 0 = \left( {x,y,z} \right) \cdot \mathop N\limits^ \to - \left( {1,5,6} \right) \cdot \mathop N\limits^ \to \quad \Rightarrow \cr & \Rightarrow \quad \left( {x,y,z} \right) \cdot \mathop N\limits^ \to = \left( {1,5,6} \right) \cdot \mathop N\limits^ \to \quad \Rightarrow \cr & \Rightarrow \quad N_x \,x + N_y \,y + N_z \,z = N_x + 5N_y \, + 6N_z \cr} $$ and that's the "standard" equation of the plane $ax+by+cz=d$.
For $$ \mathop N\limits^ \to = \left( {\matrix{ { - 8} \cr { - 14} \cr {16} \cr } } \right) $$ you get $$ \eqalign{ & \Rightarrow \quad - 8\,x - 14\,y + 16\,z = - 8 - 70\, + 96 = 18 \cr & \Rightarrow \quad - 4\,x - 7\,y + 8\,z = 9 \cr} $$
For the parametric equation, you have two approaches: I don't know which one you are considering.
The linear system approach is to consider that the "standard" equation is one linear
equation in $3$ unknowns.
Its solution will depend on two "free" parameters.
In this case you can choose for instance to put
$$
\left\{ \matrix{
x = t \hfill \cr
y = s \hfill \cr
z = {{9 + 4t + 7s} \over 8} \hfill \cr} \right.
$$
Aside, be advised that when you have more that two equations (line) then
you shall take care of the rank of the coefficient- and total- matrix.
The vectorial approach consists in that we can choose two independent vectors $\bf u , \bf v$ normal to $\vec{N}$ and put $$ {\bf x} = \lambda \,{\bf u} + \mu \,{\bf v} $$ For instance $$ {\bf u} = \left( {\matrix{ 7 \cr { - 4} \cr 0 \cr } } \right)\quad {\bf v} = \left( {\matrix{ 2 \cr 0 \cr 1 \cr } } \right)\quad \left\{ \matrix{ x = 7\lambda + 2\mu \hfill \cr y = - 4\lambda \hfill \cr z = \mu \hfill \cr} \right. $$
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