Finding a Perpendicular Line in Analytical Geometry

[Odzy]

Homework Statement

Find the line that passes through point M(-5,2,-1) and cuts the line:
(x-1)/3 = y/5 = (z+2)/-2 at right angle (two lines are perpendicular)

Homework Equations

r = a + \lambdab

The Attempt at a Solution

i used the scalar product a*b = |a| *|b| * cos(a,b) , but i don't seem to have enough knowns to solve the problem, or should i try something else


Also, if someone would like to give me a hint on understanding and approaching these problems
(i.e. lines and planes meeting each other... ) in most universal sense. I try to picture things but problem appears when something has to be written down. Can't see analogy of any kind.
Help please

 

[dynamicsolo]

I think it'll make your life easier to start from the parametric equations for the line, rather than the symmetric equation. Pick a parameter name and express the coordinates for points on that line in terms of that parameter.

You would now want to consider a vector from a general point on the line to some reference point (of your choosing) on that line and a second vector from that general point to the point M. What must be true about these two vectors? The answer to that question would let you solve for a value of the parameter which gives you a specific point on the line. The coordinates of that point and of M now give you the information you need to find the desired line.

 

[Odzy]

ok, then I get x = 3t + 1
y = 5t
z = -2t - 2 --> parametric form

my other line should look like : (x+5)/a = (x-2)/b = (z+1)/c

Now, what do you mean by : from general point on the line to some ref. point??
Sorry, but it is just so abstract that I can't find the edge.

If we call these lines p and q than p*q = 0 should be fulfilled ?

 

[HallsofIvy]

He meant pick an arbitrary point on the line. Actually, I would be inclined to use the general (3t+1, 5t, -2t-2). Find the vector from M to that (the way I am doing it, it would depend on t). Find the scalar projection of that vector onto your line. The "shortest distance" is always measured along a line perpendicular to the given line. you want to find t so that the scalar projection is 0.

 

[Odzy]

Thanks a lot.

 

[dynamicsolo]

Thanks a lot.

I'll warn you, having worked this through, that the coordinates of the point on the line, which will be the second point on the line you seek, are not pretty, but they are at least rational...