Find the distance from the origin to the line x=1+y, y=2-t, z=-1+2t

[goomer]

Question: Find the distance from the origin to the line x=1+t, y=2-t, z=-1+2t

Equations: r = r0 +tv

Attempt:

I think I solved for the vector line equation correctly:

r = < 1, 2, -1> + t< 0, -1, 2>

But I don't know where to go from there. Help please!

 

[flyingpig]

It's asking for distance...

 

[Dick]

What kind of a approach is your book, notes or lectures using? You could differentiate |r|^2 with respect to t and find the minimum if you are doing calculus. Or you could do vector operations like dot and cross product to find it. You've got to have some clue.

 

[goomer]

Well, we've learned that the distance between a point and a plane is

| ax0 + by0 + cz0 + d | / √ ( a^2 + b^2 + c^2)

but this is finding the distance between a point and a line, so I don't see how it would help, or if it's even relevant.

 

[flyingpig]

What is the distance formula?

 

[Dick]

Well, we've learned that the distance between a point and a plane is

| ax0 + by0 + cz0 + d | / √ ( a^2 + b^2 + c^2)

but this is finding the distance between a point and a line, so I don't see how it would help, or if it's even relevant.

No, probably not really relevant. Do you know the vector dot product? Stuff like that? If your line is L(t)=r0+tv, you want to find a point on your line where L(t)-<0,0,0> is orthogonal to the direction vector of your line, v. Do you see why?