Show that a line does not intersect a plane (vectors)

[Jon.G]

Homework Statement

A plane is given by the equation: $4x + 5y + 7z = 21$
and a line by the equation $r = (1,2,3) + \lambda (1,2,-2)$ where λ is real.

Show that the line does not intersect the plane.


The attempt at a solution
So if I remember correctly, if $n . a = 0$, they do not intersect, where n is the normal vector and a is the direction of the line, ie. n = (4,5,7) and a = (1,2,-2)
n . a gives 4 + 10 - 14 which is 0.

However I'm more confused by the theory. If n . a = 0, does this not mean that they are perpendicular? So why wouldn't they intersect?
I know I must be looking at this the wrong way, but I can't see where :/
Thanks

 

[Simon Bridge]

However I'm more confused by the theory. If n . a = 0, does this not mean that they are perpendicular? So why wouldn't they intersect?

The n vector is perpendicular to the plane.
If the line is perpendicular to the n vector then...

Note: it is not good enough just to show that n.a=0, you also have to show the line is not in the plane.

 

[HallsofIvy]

Homework Statement

A plane is given by the equation: $4x + 5y + 7z = 21$
and a line by the equation $r = (1,2,3) + \lambda (1,2,-2)$ where λ is real.

Show that the line does not intersect the plane.


The attempt at a solution
So if I remember correctly, if $n . a = 0$, they do not intersect, where n is the normal vector and a is the direction of the line, ie. n = (4,5,7) and a = (1,2,-2)
n . a gives 4 + 10 - 14 which is 0.

However I'm more confused by the theory. If n . a = 0, does this not mean that they are perpendicular? So why wouldn't they intersect?

Your confusion is in the word "they". In your first sentence "they" refers to the two vectors. In the second sentence "they" refers to the line and plane.

I know I must be looking at this the wrong way, but I can't see where :/
Thanks

 

[Jon.G]

Yeah, that's a very simple mistake I made there :/
Thank for pointing that out