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lockedup
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Homework Statement
Find the distance between (2,5,1) and the line 2i − 3j + 6k.
lockedup said:... and the line 2i − 3j +6k.
No, it isn't.Gunthi said:It's a plane.
2i - 3j + 6k isn't a line -- it's a vector. It has a certain length, while a line has infinite length. The problem is probably something more like this:lockedup said:Homework Statement
Find the distance between (2,5,1) and the line 2i − 3j + 6k.
Just as well. Given that you can't find a formula, how would you approach this problem? According to the forum rules, you have to give it a good shot before anyone can give you any help.lockedup said:Homework Equations
The Attempt at a Solution
I can't find a formula to figure this (or one that makes any sense)...
My assignment sheet says line...Mark44 said:2i - 3j + 6k isn't a line -- it's a vector. It has a certain length, while a line has infinite length. The problem is probably something more like this:
Find the distance between (2,5,1) and the line whose direction is given by the vector 2i − 3j + 6k.
Just as well. Given that you can't find a formula, how would you approach this problem? According to the forum rules, you have to give it a good shot before anyone can give you any help.
Mark44 said:No, it isn't.
pootette said:Gunthi,
I believe you are asking how to find the distance between a point in space, and a vector?
If so, start by looking at line-distance formulas and vector math.
I hope this gives you a jumping-off point.
Mark44 said:Gunthi,
If it were 2x - 3y + 6z = 0, you would be right
The formula for finding the distance between a point and a line in space is:
d = |ax0 + by0 + cz0 + d| / √(a² + b² + c²),
where (x0, y0, z0) is the coordinates of the point, and ax + by + cz + d = 0 is the equation of the line in standard form.
No, the distance between a point and a line cannot be negative. Distance is always a positive value, as it represents the shortest distance between the point and the line.
To find the coordinates of the closest point on a line to a given point in space, you can use the formula:
x = (b(bx0 - ay0) - acz0 - ad) / (a² + b²),
y = (a(-bx0 + ay0) - bcx0 - bd) / (a² + b²),
z = (a(ay0 + bx0) - bcx0 - bd) / (a² + b²),
where (x0, y0, z0) is the coordinates of the given point, and ax + by + cz + d = 0 is the equation of the line in standard form.
Yes, the distance between a point and a line can be calculated in any coordinate system as long as the coordinates of the point and the equation of the line are in the same coordinate system.
Yes, another way to find the distance between a point and a line is to use the perpendicular distance formula:
d = |ax0 + by0 + cz0 + d| / √(a² + b² + c²),
where (x0, y0, z0) is the coordinates of the point, and ax + by + cz + d = 0 is the equation of a line perpendicular to the given line that passes through the given point.