[Linear algebra] [College] Finding points on a line when given a distance to a plane.

Hello everyone,

I'm studying for my linear algebra test tomorrow and came across the following problem:

Find all the points on e which are located on a distance sqrt26 from the plane a.

e: x/2 = (y-5)/3 = z-2

A is the plane through:

The point Q(0,10,1)

And the line (let's call it f) :

3x-4z+9 = 0

7z-3y+3 = 0

This was my idea:

- Find the direction vector for f by the cross product off the normal vectors of the planes which define f.

- Find the normal vector of a perpendicular to f.

- Define the plane with the point Q and the normal vector

- Determine the distance from the plane to the line a with the distance formula.

I already went wrong on the second step, i can't find the normal vector for the plane. Does anybody have any ideas for that?

Also, if there is another way of solving this problem, please tell me.

Any help at all is very much appreciated!

Re: [Linear algebra] [College] Finding points on a line when given a distance to a pl

Re: [Linear algebra] [College] Finding points on a line when given a distance to a pl

Quote:

Originally Posted by

**TheEmptySet** The distnace from a plane to a point can be found here

Point-Plane Distance -- from Wolfram MathWorld
I am not sure what you plane is. You only specify one point for the plane, we need more information. If you can put the equation of the plane in standard form

and write the equaiton of the line in parametric form

If you use the reasoning or the formula in the link above you will get

Now just sub into the equation and solve for t.

I'm sorry if this wasn't clear, the plane goes through the point Q(0,10,1) and the line f which is determined by two planes 3x-4z+9 = 0 and

7z-3y+3 = 0.

I'm not quite sure how to determine a plane with this information.

Re: [Linear algebra] [College] Finding points on a line when given a distance to a pl

Quote:

Originally Posted by

**tonykart44** I'm sorry if this wasn't clear, the plane goes through the point Q(0,10,1) and the line f which is determined by two planes 3x-4z+9 = 0 and

7z-3y+3 = 0.

You need a point in the intersection of the two planes: works.

is the direction vector of the line,

Consider that is the normal of plane you need.