I need to find the value of a for which the line r which is passing through the point $Q=(0, 0, 0)$ and $P=(a, 1, 2)$ is parallel to the plane with equation
$$ \alpha: a(x+y)-z=0 $$
So, I know from where I could start but I get stuck.
First, I find the parametric equation of $r$, knowing that it's passing through $Q$ and $P$, I get a directional vector:
$$ OP = (a, 1, 2) $$
Using $Q$ as a starting system, I find its equation
$$ r: \begin{cases} x=at \\y=t \\z=2t \end{cases}$$
To see if the plane and the line are parallel, I'd do the cross product between two directional vectors and see if it's $0$. However, I don't know how I should do it considering I have a real parameter in both the equation of the line and the plane.
Best Answer
The easiest way here is to determine $a$ such that the line directional vector and the plane normal vector are perpendicular. For this, their dot product needs to be zero, i.e.
$$(a,1,2) \cdot (a,a,-1) = 0$$
(The coordinates of the plane normal vector are the coefficients next to $x$, $y$ and $z$ respectively in the implicit equation of the plane.)