Determine the equation in normal form that contains the line $\left\{\begin{matrix} x=1+t\\ y=3t\\ z=2t \end{matrix}\right.$ and is parallel to the intersection between the planes $x+z=1$ and $y+z=2$.
$$x+z=1 \Rightarrow \overrightarrow{n}_{1}=(1,0,1) \\y+z=2 \Rightarrow \overrightarrow{n}_{2}=(0,1,1)\\ (1,0,1)\times (0,1,1)=(-1,-1,1)\\$$
Now I need to figure out a normal vector that is parallel to (-1,-1,1). If I use dot product I get $(-1,-1,1)\cdot M_{1}=0 \Rightarrow M_{1}=0$ which doesn't seem right.
What's my next move?
$\endgroup$3 Answers
$\begingroup$Say equation of the plane $ax + by + cz = d$
Take points on the given line $(1, 0, 0), (2, 3, 2)$ for $t = 0, 1$. As the line is in the plane, these points are also in the plane.
From $(1, 0, 0)$, we get $d = a$ so equation of plane becomes $ax + by + cz - a = 0$.
From $(2, 3, 2)$, we get $a + 3b + 2c = 0$
From dot product of the normal vector with direction vector $(-1, -1, 1)$, we get $a + b - c = 0$.
So, $b = - \frac{3c}{2}, a = \frac{5c}{2}$
So equation of the plane is $5x - 3y + 2z = 5$.
$\endgroup$ $\begingroup$A normal vector to the plane has to be orthogonal to the directing vector to the line contained in the plane, $\vec u=(1,3,2)$, and to the directing vector of the intersection of the given planes, $\vec v=(1,1,-1)$.
An obvious candidate is $\:\vec u\times \vec v=(5,-3,2)$.
Furthermore, the plane contains the point $(1,0,0)$, so the equation is$$\color{red}{5x-3y+2z=}5\cdot 1-3\cdot 0+2\cdot 0=\color{red}5.$$
$\endgroup$ $\begingroup$We have that a normal vector for the plane $n=(a,b,c)$ is
- orthogonal to the direction vector for the line $(1,3,2)$
- orthogonal to the direction vector for the intersection of the planes $(-1,-1,1)$
that is
$a+3b+2c=0$
$-a-b+c=0$
which can be solved assuming a free parameter, e.g. taking $a=5$ we obtain $n=(5,-3,2)$ and the plane equation is
$$5x-3y+2z+d=0$$
and form the condition that $P=(1,0,0)$ belongs to the plane we find that $d=-5$.
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