Distance from a plane to a line

May 2010
7
0
Determine the distance from the line L1: r= [3,8,1] + t[-1,3,-2] to the plane II: 8x - 6y -13z - 12 = 0

1) Determine the equation of a line L2, perpendicular to II and passing through a point P on L1.

2) Determine point A; the point where L2 and II intersect.

3) Determine the distance from P to A. This distance represents the distance between the given line and plane.
 

Plato

MHF Helper
Aug 2006
22,507
8,664
Determine the distance from the line L1: r= [3,8,1] + t[-1,3,-2] to the plane II: 8x - 6y -13z - 12 = 0
1) Determine the equation of a line L2, perpendicular to II and passing through a point P on L1.
The equation for \(\displaystyle L_2\) is \(\displaystyle <3,8,1>+t<8,-6,-13>\).
You answer the other questions.
 
May 2010
7
0
I really do not know how to tackle this problem. There were actually 2 questions before the 3 i posted there and I answered them but I really needed help with those three.

I hope you can help me finish this problem so I can see how to do it in the future!
 

Plato

MHF Helper
Aug 2006
22,507
8,664
I hope you can help me finish this problem so I can see how to do it in the future!
I will give you further help only if you reply with an explanation of the equation for \(\displaystyle L_1\) that I gave you.
Otherwise, I am simply doing the problem for you.
That is against my principles: I what you to learn to do mathematics.
I do not think that people learn mathematics by watching it done or copying examples.
Please reply with your explanation.
 
May 2010
7
0
Since I already proved that the given line is parallel and distinct to the plane, then we know that we can take any point on the line to use as the closest distance from the plane.

So we can automatically use the point given to us in the equation of the original line, P(3,8,1), and then we need the line to be perpendicular to the plane. To do that we need the normal vector to the plane which is [8,-6,-13] (taken from the equation of the plane).

Is that what you wanted me to explain?
 

Plato

MHF Helper
Aug 2006
22,507
8,664
Yes it is. It shows me you have a grasp of what is going on.
Now write the line in parametric form:
\(\displaystyle L_2 = \left\{ {\begin{array}{*{20}c}
{x = 3 + 8t} \\
{y = 8 - 6t} \\
{z = 1 - 13t} \\ \end{array} } \right.\)
Put that in the equation of the plane and solve for t.
 
May 2010
7
0
Okay so I solved for t and found that equals 49/269. So I need to somehow use this to figure out the intersection point?

So far in my class we have done intersections of two planes and we ended up solving for the line of intersection which meant that the parametric equations were the final solutions. Since we are now looking for a point and not a line though, how do I use the value of t to get that point? We have not learned this yet so I am really confused right now.
 
May 2010
7
0
Nevermind, I figured it out.

Sub t into the parametric equations and solve for the values of x,y,z and that is the point of intersection of the perpendicular vector and the plane (Point A).

It then asked me to find the distance between point A and P (the original point on the line) so I used the distance formula: Sqrroot((X2-X1)^2 + (Y2-Y1)^2 + (Z2-Z1)^2) and that gave me the distance of the line between the plane and the original line.

Thanks for all of your help, now I have to figure out how to find the same solution using vector projections! To do that I have made a diagram so far where I found another point on the plane B(0,-2,0), but I am not sure what to do next. I don't know which vector to project on the other and don't really know how to use this to find the distance of the line AP.

I would appreciate some help.
 

Plato

MHF Helper
Aug 2006
22,507
8,664
If \(\displaystyle P:~(p,q,r)\) is point and \(\displaystyle \Pi:~<x-a,y-b,z-c>\cdot N=0\) is a plane
the distance from the point to the plane is \(\displaystyle \frac{|<p-a,q-b,r-c>\cdot N|}{||N||}\).