distance between line and point, having trouble understanding formula provided

[Drunvalo]

Hi there, I'm just doing some revision by going over some mock exam questions and I'm having trouble understanding what some of these values represent in the question.

So the task is the compute the distance between a line and a point in 3D.

The point is < 1, 3, 2 >, I understand this.

The line is:

2x + y + 3z = 3
x + 2y + z = 4

My understanding of this that these are 2 vectors of the 2 points that form the line, < 2, 1, 3 > and < 1, 2, 1>

I was wondering what = 3 and = 4 represent?

Thanks and I really hope I explained this well. God bless <3

 

[pka]

So the task is the compute the distance between a line and a point in 3D.

The point is < 1, 3, 2 >, I understand this.

The line is:
2x + y + 3z = 3
x + 2y + z = 4

My understanding of this that these are 2 vectors of the 2 points that form the line, < 2, 1, 3 > and < 1, 2, 1>

\(\displaystyle 2x+y+3z=3\\x+2y+z=4\) is not a line, they are two planes.
The intersection of two planes is the line in question.

First find a point \(\displaystyle P\) on both planes. Then find \(\displaystyle \vec{D}=<2,1,3>\times<1.2.1>\)

Then the line is \(\displaystyle P+t\;\vec{D}\).

 

[Drunvalo]

Okay, I see. Thanks :smile: Sorry I took so long to reply, I just wanted to make sure I understood how to check whether the planes were perpendicular before actually reading about how to get the line which is just the cross product of the two planes? I hope I got that right, but I still don't understand where = 3 and = 4 tie into all this. Just figuring out how to plot a point along the line. I don't think I quite get it yet, but I'm getting there.

2x+y+3z=3

x+2y+z=4

<2, 1, 3> * <1, 2, 1> = <-5, -1, 3>

Not sure where to go from here, am I right to say that because plane a and the cross product both have a z axis of 3 that I should find a point along it?

 

[Drunvalo]

Sorry for the double post but I hoped to bump the thread and get some clarification.

So in my last post I figured out the directional line and I'd just like some feedback on whether I'm going in the right direction toward getting the point vertex.

I set my x parameter to 0 on both plane vectors and change Y and Z with the sum of the vectors, so I turned = 4 to = 3 and turn 2 to 1.5 and y to 1.5 and z to 0.75, is this acceptable?

This gives me

2.5y = 3.75z

y = 3.75 / 2.5 * z
z = y * 3.75 / 2.5

is this right?

 

[HallsofIvy]

I'm not sure what you are doing with your last post but you are correct that <-5, -1, -3> is a vector perpendicular to the given plane. I presume you know that the "distance from a point to a plane" will be measured perpendicular to the plane. So construct a line passing through the point (1, 3, 2) parallel to <-5, -1, -3> and determine where the line goes through the given plane.