Straight line perpendicular, parallel, or intersects a plane.

Question

Can someone guide me how to solve this.

Find valuea of $a$ so the line $\ ε: x-2=y=4-z $

is perpendicular, parallel or just intersects the plane $\ π: ax+y-z=2 $

Excuse me if I dont use the proper terminology.

Answer 1

I would approach this using vectors, but I don't know how much you know about them. The line is given by $\vec r(t) = \vec{v}t + \vec{b}$, where $\vec{v}=(1,1,-1)$ and $\vec b = (-2, 0, 4)$ The plane $\pi$ has equation $\vec n \cdot \vec r = 2$, where $\vec n = (a, 1, -1)$ is a vector normal to the plane.

For $\varepsilon$ to be perpendicular to $\pi$, its direction vector $\vec v$ has to be a multiple of the normal vector $\vec n$.

For $\varepsilon$ to be parallel to $\pi$, its direction vector $\vec v$ has to be perpendicular to $\vec n$, which is equivalent to saying that their inner product is zero: $\vec v \cdot \vec n = 0$.

The line and plane will intersect unless the line is parallel to the plane and does not lie in the plane. So we want to check if, when the line is parallel to the plane, it actually lies within it. That will occur if any point on the parallel line lies in the plane, and fortunately, we happen to know a point on that line, namely $\vec b$. So we check if $\vec b$ satifies the equation for $\pi$: $\vec n \cdot \vec b = 2$, where $\vec n$ has the value of $a$ found for the parallel case. If $\vec b$ satisfies this equation, then the line will always intersect the plane. If not, the line will intersect the plane for all values of $a$ except for the value in the parallel case.