# Distance from a plane to a line

• May 25th 2010, 01:41 PM
Distance from a plane to a line
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.
• May 25th 2010, 02:13 PM
Plato
Quote:

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 $L_2$ is $<3,8,1>+t<8,-6,-13>$.
• May 25th 2010, 03:30 PM
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!
• May 25th 2010, 03:42 PM
Plato
Quote:

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 $L_1$ 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.
• May 25th 2010, 04:16 PM
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?
• May 25th 2010, 04:40 PM
Plato
Yes it is. It shows me you have a grasp of what is going on.
Now write the line in parametric form:
$L_2 = \left\{ {\begin{array}{*{20}c}
{x = 3 + 8t} \\
{y = 8 - 6t} \\
{z = 1 - 13t} \\ \end{array} } \right.$

Put that in the equation of the plane and solve for t.
• May 25th 2010, 07:05 PM
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 26th 2010, 07:38 AM
If $P:~(p,q,r)$ is point and $\Pi:~\cdot N=0$ is a plane
the distance from the point to the plane is $\frac{|\cdot N|}{||N||}$.