# Thread: Distence bewteen a pt. and plane

1. ## Distence bewteen a pt. and plane

Find the distance between the point M(-2,3,4) and the line
L: x-3/1 = y+1/2 = z-1/1.
I was wondering if someone can start me off with this problem.
I wanted to take the Line: x-3/1 = y+1/2 = z-1/1 and turn it to a "Parametric Equation"
This is what I have for the PE: x= -2 +t, y = 3+2t, z= 4+t

2. I think the parametric equation should be $x=3+t,y=2t-1,z=t+1$ (from $\frac{(x-3)}{1}=\frac{(y+1)}{2}=\frac{(z-1)}{1}=t$, if so the point (3,-1,1) (call it $\vec{OA}$) is on the line. So the $\vec{AM}$ is (-5,4,3).

if the vector parallel to the line L is $\vec{PQ}$

Now you'll have two vectors that cross each other and one vector starts from point A to point M

take the definition of $\vec{PQ}\times\vec{AM}=|PQ||AM|\sin \theta$

if you draw the vector's you'll find $|AM|\sin \theta$ is the distance of from the point to the required line

3. May I ask what is the AM.

IF you can, can you use the following and Ill take from there and get back you. I appreciate help!!!

U = (1, 2, 1), which is the direction vector for the line, correct???
P = (3, -1, 1) point on the line, correct???

and Ill take P and U to get PQ, correct???

You dont seem 100% sure on the parametrics equation that you have. LOL

I use this formula correct???

D= ll PQ x u ll / ll u ll

4. U = (1, 2, 1), which is the direction vector for the line, correct???
P = (3, -1, 1) point on the line, correct???
correct, here the P is the equal to the point A that i've used

and Ill take P and U to get PQ, correct???
wrong, what is Q here it, shouldn't it be M? So use P and M to get PM i

I use this formula correct???

D= ll PQ x u ll / ll u ll
yes but it has to be M

5. What I would do is this: the plane x+ 2y+ z= C is perpendicular to the line for all C. Find the value of C such that the plane contains the point (-2, 3, 4) and then determine where the line intersects that plane. The distance from the point to the line is the distance between those two points.

6. where is the x+ 2y+ z= C coming from??? I dont see that. It seems there may be more than one way to solve this problem.

7. Nevermind

8. Question:

Instead of using a paramtric equation, what are the other ways to solve this problem.

now I am thinking of using the Standard Equation of a Plane in space.

Now I am confusing my self.

Im lost.

9. Given a plane $Ax+By+Cz+D=0$ and a point $(p,q,r)$ the distance between the two is:
$\dfrac{|Ap+Bq+Cr+D|}{\sqrt{A^2+B^2+C^2}}$.

This is what I got. By the way I didnt get your image, error???
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Parametric Eqaution: x = 3 + t, y = -1 + 2t, z = 1 + t

Direction Vector: u = (1,2,1)

To find the point, let t = 0 and get

P = (3,-1,1)

So,

PQ= (-2 - (3), 3 - (-1), 4 - 1) = (-5, 4, 3)

PQ x u =
i j k
-5 4 3
1 2 1

= (-2, -8, -14)

D= ll PQ x ull / ll u ll

finale answer = sqrt(264) / sqrt(6)

11. Would my answer below be correct if I do it that way???