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

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The discussion focuses on finding the distance from the origin to the line defined by the parametric equations x=1+t, y=2-t, z=-1+2t. The vector line equation is established as r = <1, 2, -1> + t<0, -1, 2>. Participants suggest using calculus to differentiate |r|^2 with respect to t to find the minimum distance or employing vector operations such as dot and cross products. The relevance of the distance formula for a point to a plane is acknowledged, but its application to a line is debated.

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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!
 
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It's asking for distance...
 
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.
 
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.
 
What is the distance formula?
 
goomer said:
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?
 

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