Inproduct and norm Search for a representation of a line that goes through a point $p$ and that is orthogonal with the plane $\alpha$.
$p=\begin{bmatrix} 3 \\ 0 \\ 5\end{bmatrix}, \alpha =\begin{bmatrix} x_1 \\ x_2 \\ x_3\end{bmatrix}= \begin{bmatrix} 0 \\ -1 \\ -1\end{bmatrix} + \rho\begin{bmatrix} 1 \\ 2 \\ 2\end{bmatrix} + \tau \begin{bmatrix} 5 \\ 5 \\ 6\end{bmatrix}$
How do I this? I tried to convert $\alpha$ to a cartesian equation and then reverse the signs of each coefficient of that equation and cycle the variable with those coefficients but to no avail. Help would be appreciated. 
 A: The vectors $$\mathbf{v}_1=\left(\begin{matrix}1\\2\\2\end{matrix}\right)\;\;\;\;\mathbf{v}_2=\left(\begin{matrix}5\\5\\6\end{matrix}\right)$$ are the generator of the plane. To be perpendicular to this plane you have to find a third vector perpendicular to $\mathbf{v}_1$ and $\mathbf{v}_2$. Obviously if you take the vector product of this to you are assured to find a perpendicular vector, so $$\mathbf{v}_1\times\mathbf{v}_2 = \det\left(\begin{matrix}\hat{i}&\hat{j}&\hat{k}\\1&2&2\\5&5&6\end{matrix}\right) = 2\hat{i}+4\hat{j} -5\hat{k}$$ The third perpendicular vector is, then $$\mathbf{v}_3 = \left(\begin{matrix}2\\4\\-5\end{matrix}\right)$$
Which is the characteristic vector of the plane. This will be the generator of our line. The parametric equation of a line with a certain generator vector passing through a point $P(x_p,y_p,z_p)$ is $$\left\{\begin{matrix}x = x_p+v_1t\\y=y_p+v_2t\\z=z_p+v_3t\end{matrix}\right.$$
so, in our case:
$$\left\{\begin{matrix}x = 3+2t\\y=4t\\z=5-5t\end{matrix}\right.$$
A: The most straightforward solution is to simply construct the vector perpendicular to the plane as the cross product of your two planar vectors $(1,2,2)^t$ and $(5,5,6)^t$ (possible only in 3D). The line $l$ you're looking for would then be simply be represented by the point $p$ plus the orthogonal vector you just constructed.
Spelled out: 
$$l=\begin{bmatrix} 3 \\ 0 \\ 5\end{bmatrix}+\lambda \left( \begin{bmatrix} 1 \\ 2 \\ 2\end{bmatrix} \times \begin{bmatrix} 5 \\ 5 \\ 6\end{bmatrix}\right)=\begin{bmatrix} 3 \\ 0 \\ 5\end{bmatrix}+\lambda \begin{bmatrix} 2 \\ 4 \\ -5\end{bmatrix}$$
with $\lambda \in \mathbb{R}$.
A: Obviously the characteristic vector of the plane is the cross product of $\begin{bmatrix} 1 \\ 2 \\ 2\end{bmatrix}$ and $\begin{bmatrix} 5\\ 5 \\ 6\end{bmatrix}$ (this is because any point of the plane is a linear combination of these two vectors) which is $$\begin{bmatrix} 2\\4\\-5\\\end{bmatrix}$$therefore the line is$$\dfrac{x-3}{2}=\dfrac{y}{4}=\dfrac{5-z}{5}$$
