# Distance Between Origin and Lin

1. Sep 17, 2011

### TrueStar

1. The problem statement, all variables and given/known data
Find the distance from the origin to the line
x=1+t
y=2-t
z=-1+2t

2. Relevant equations

Possibly d=|aXb| / |a|

3. The attempt at a solution

I think I need to use the formula provided above, but I'm not sure how. A different distance formula was covered in class, but I think there are different formulas to use based on what you're trying to find the distance between.

There is an example in the textbook like this that asks to use this formula, but it gives out three points and tells assigns vectors for a and b.

I'm not sure how to approach this.

2. Sep 17, 2011

### HallsofIvy

Staff Emeritus
Don't just memorize formulas without learning when and how they apply.

Find the plane perpendicular to the given line containin (0, 0, 0). The shortest distance from the origin to the line must be along a line in that plane. Where does the line cross that plane?

3. Sep 17, 2011

### TrueStar

I tried to find an equation of a plane with the parametric equation given to me. I came up with:

x-y+2z=0 - I plugged in the origin points, which makes d=0.

Using the origin as a point, I applied the distance formula (the absolute value of the equation of a plane with the point plugged in divided by the square root of the sum of the normal vector squared), but I'm getting zero for an answer.

I don't know what I'm doing wrong. Since the origin is a 0,0,0, the equation ofthe line will always cancel out.

4. Sep 17, 2011

### ehild

The distance of a point P from a line is defined as the minimum of the distance of any point on the line from the given point P. Write the distance from the origin of a point (x,y,z) on the line in terms of t and find t where it is minimal.

ehild

5. Sep 17, 2011

### HallsofIvy

Staff Emeritus
I said before, "Where does the line cross that plane?"

6. Sep 17, 2011

### TrueStar

I think I got the equation of a plane wrong...I think it should be -x-y-3z=0

I used this and the origin points to put into a vector equation:

(0,0,0) + t(-1,-1,-3) - This gives me position vectors of

x=0-t
y=0-t
z=0-3t

I plug this into my equation of a plane which was -x-y-3z=0 so that it looks like:

-1(0-t)-1(0-t)-3(0-3t)=0

This mean t=0. I use this to plug back into the three equations above and I still get (0,0,0). I don't feel this is correct.

7. Sep 17, 2011

### ehild

it should be r˙t=0 where t is the direction vector of the line, (1,-1,2), so x-y+2z=0.
Plug in x=1+t, y=2-t and z=-1+2t, and find t.

ehild

8. Sep 17, 2011

### TrueStar

OK, I think I see what I did wrong with the equation of a line. I think the multiples of t in the parametric equations are the scalers for the equation of a line. I may not have my terminology right here.

So I plugged in the parametric equations for x, y, and z and solved for t. It was 1/2. I understand that this would be a minimum distance.

The points are (3/2, 3/2, 0) and the actual distance is the sqrt of 4.5.

Hopefully this is the correct answer.