problem of geometry in the space

I have a plane $p: 2x+2y+z=0$ and the straight line $\left\{ \begin{array}{c} 2y-z-2=0\\ 4x-3z-8=0 \\ \end{array} \right.$

How can I determine the plane $\Pi$ such that is passing through r and perpendicular to $p$?

• You mean the line. Commented Jul 19, 2018 at 13:01
• Sorry, already fixed... Commented Jul 19, 2018 at 13:01

The two defining equations for the line are equations of planes whose intersection is that line. Every plane that contains this line can be expressed as a linear combination of these two equations: $$\lambda(2y-z-2)+\mu(4x-3z-8) = 4\mu x+2\lambda y-(\lambda+3\mu)z-(2\lambda+8\mu) = 0.$$ Two planes are perpendicular iff their normals are. Compute the dot product of the normals of this plane and $p$, both of which you can read directly from the equations. Setting the resulting expression equal to zero gives you a linear equation in $\lambda$ and $\mu$. By inspection, neither of the defining planes is perpendicular to $p$, so you can set either $\lambda$ or $\mu$ to $1$ and solve for the other variable.

Rewriting the line $r$ in vector (parametric) form, you have $r : P+\lambda D$ with $P$ a point on $r$ and $D$ a direction vector.

The plane $\Pi$ is uniquely determined by a point and two (non-parallel) direction vectors. Taking the point $P$ and direction vector $D$ from $r$ ensures $r$ is contained in $\Pi$. If we name the second direction vector $E$, then you have $\Pi : P+\lambda D+\mu E$.

If you want $\Pi$ to be perpendicular to the given plane $p$, use a normal vector of $p$ as the second direction vector $E$ of $\Pi$. Since $p$ is given in cartesian form, this normal vector is easy to find: $$p: 2x+2y+z=0 \implies \vec n_p = (2,2,1)$$

Two planes are perpendicular iff their normal vectors are perpendicular. So if the desired plane is $ax+by+cz+d=0$, then $2a+2b+c=0$. The points $(2,1,0), (5,3,4)$ are on the line, hence they are on the desired plane. This yields $2a+b+d=0$ and $5a+3b+4c+d=0$.

Solve this system of linear equations to find $a,b,c,d$. There are infinitely many solutions, but only because any multiple of a solution yields the same plane.