# Determine if plane perpendicular to line

## Homework Statement

Find cartesian equations of the line L containing P(2, 0, -3) and Q(1, -1, 6) and determine if plane (8x + y - z = -1) is perpendicular to L

## The Attempt at a Solution

PQ = (1-2)i + (-1-0)j + (6+3)k = -i -j +9k

(x-2)i + (y-0)j + (z+3)k = -ti - tj +9tk

so, cartesian equation of L is:
$$\frac{x-2}{-1} = \frac{y}{-1} = \frac{z+3}{9}$$

The plane normal n = (8, 1, -1). If dot product n and L = 0 then they are perpendicular.

How do I take the dot product of L and n in that form?

I don't understand the general equation of the line, I just followed an example and plugged in some numbers. This question should take about 5 minutes but has taken me about 2 hours and I still have no clue.

L is already perpendicular to the plane.

And what is the characterisitc vector of the line. it is u=<-1,-1,9> right

and of the plane n=<8,1,-9>.

If the plane and the line were perpendicular it means that the characteristic vector of the line, namely u, and that of the plance, namely n, have to be parallel, dont they?

so one can be written as a linear combination of the other.

L is already perpendicular to the plane.

how?

And what is the characterisitc vector of the line. it is u=<-1,-1,9> right

So a characteristic vector is the denominators in the cartesian equation? And a characteristic vector is perpendicular to the line?

so one can be written as a linear combination of the other.

Is linear combination the same as scalar multiple?

how??

I meant the vector n. not L. my bad.
So a characteristic vector is the denominators in the cartesian equation? And a characteristic vector is perpendicular to the line?
yes. and no.

The line is in the same direction as the vector that i wrote in my previous post, it is not perpendicular to it.
the characterisitc vector of a plane is perpendicular to it, while that of a line is parallel to the line itself.

Is linear combination the same as scalar multiple?

in the case of two vectors, yes.