# Equation of a Line in three space question

1. Jan 13, 2006

### thomasrules

This question is really pissing me off because I don't understand it. I know how to do it but don't understand the reasoning:

Find an equation of the line through the point (4,5,5) that meets the line

(x-11)/3=(y+8)/-1=(z-4)/1 at right angles.

I found out that you have to do the dot product of (4,5,5) and that equation but I don't understand why you don't need the direction vector for the other equation to find the answer. Also why couldn't you just do (4,5,5) dot (11,-8,4)?????????

2. Jan 13, 2006

### matt grime

You don't dot product a vector and an equation.

How're you reading off the direction vector of the line from its equation?

3. Jan 13, 2006

### thomasrules

direction vector=(3,-1,1)

4. Jan 13, 2006

### matt grime

Still not entirely sure I see what the difficulty is: you know that the point where the line from (4,5,5) hits the given line perpendicularly is also the closest point to (4,5,5) on the line.

5. Jan 13, 2006

### HallsofIvy

Staff Emeritus
No, you didn't find that out- it doesn't make sense. (4,5,5) is a point not a vector and you can't take a dot product of anything with an equation! That's one reason why its not a good idea to write a vector in the form (a,b,c). I prefer either ai+ bj+ ck or <a, b, c>.

Let (x,y,z) be the point on the given line where that perpendicular crosses. Then (x-4)i+ (y-5)j+ (z-5)k is a vector in the direction of that line. Yes, 3i- j+ 3k is a vector in the direction of the given line. Since they are perpendicular, their dot product: 3(x-4)- (y-5)+ 3(z-5)= 0.
Multiplying that out, 3x- y+ 3z= 12- 5+ 15= 22 (that may be where you got the idea that you were doing "the dot product of (4,5,5) and that equation"). That equation, together with the equations (x-11)/3=(y+8)/-1=(z-4)/1 is enough to solve for x, y, and z and then the equation of the line.

6. Jan 13, 2006

### thomasrules

Ok let me show you how i think it's to be done because I don't get your part of

"That equation, together with the equations (x-11)/3=(y+8)/-1=(z-4)/1 is enough to solve for x, y, and z and then the equation of the line."

r=(11,-8,4)+t(3,-1,1)

r1=(4,5,5)+t(x,y,z)

r dot r1= (11+3t,-8-t,4+1)dot(4+x,5+y,5+z)=0

IS THAT RIGHT?

And how did you get "Yes, 3i- j+ 3k" isn't it 3i-j+k?????

7. Jan 13, 2006

### neurocomp2003

yes the vector is 3,-1,1, best is to draw a diagram with the point/Line and new point and use the projection formula to understand the above.

Last edited: Jan 13, 2006
8. Jan 14, 2006

### thomasrules

anyone help

9. Jan 14, 2006

### HallsofIvy

Staff Emeritus
That was a "miscopy".
Rewrite:
Let (x,y,z) be the point on the given line where that perpendicular crosses. Then (x-4)i+ (y-5)j+ (z-5)k is a vector in the direction of that line. Yes, 3i- j+ k is a vector in the direction of the given line. Since they are perpendicular, their dot product: 3(x-4)- (y-5)+ (z-5)= 0.
Multiplying that out, 3x- y+ z= 12- 5+ 5= 12

Now you also know (x-11)/3=(y+8)/-1=(z-4)/1 since (x, y, z) lies on that line.
From 3x- y+ z= 12, we can write z= -3x+ y+ 12 and then we have, from the equations for the line, (x-11)/3= (-3x+y+ 8) and (y+8)/-1= (-3x+y+8), two equations to solve for x and y.

10. Jan 14, 2006

### matt grime

Plenty of help has been given, but it still isn't clear what *you* are doing i.e. your explanations of what you are doing do not make sense, we think of your claim that you know to take the dot product of sometihng with an equation.

The simplest thing to do is to draw a picture to see how to use the projection formula to get a vector parallel to the required line, and then you know a vector parallel to the desired line and a point on the line hence you konw the equation of the line.

So, draw a diagram: a long line represents the given one; mark a point not on the line to represent (4,5,5); draw a line from the point to the given line, and draw one the meets perpendicularly; think about projections using dot products etc, it is quite clear, and you shuold also attempt to bear in mind the limitations of the web as a place for explaining diagrams: it is up to you to draw one not us.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook