Distance between point coordinates in a straight line

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SUMMARY

The discussion centers on determining a point P on the line segment AB, where A = (1,2,5) and B = (0,1,0), such that the distance ||\vec{PB}|| equals three times the distance ||\vec{PA}||. The parametric form of the line is established as r: (1, 2, 5) + λ(-1,-1,-5), leading to the equations x = 1 - λ, y = 2 - λ, and z = 5 - 5λ. The user struggles to reconcile their calculations with the expected solution from their textbook, particularly regarding the correct application of distance formulas and vector subtraction.

PREREQUISITES
  • Understanding of vector notation and operations
  • Familiarity with parametric equations of a line
  • Knowledge of distance formulas in three-dimensional space
  • Basic algebraic manipulation skills
NEXT STEPS
  • Review vector distance calculations in three-dimensional space
  • Study the derivation of parametric equations from two points
  • Practice solving problems involving ratios of distances in vector contexts
  • Explore examples of vector subtraction and its implications in geometry
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Students studying geometry, particularly those focusing on vector analysis and distance calculations, as well as educators seeking to clarify concepts related to parametric lines and vector operations.

Bunny-chan
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Homework Statement


Let A = (1,2,5) and B = (0,1,0). Determine a point P of the line AB such that ||\vec{PB}|| = 3||\vec{PA}||.

Homework Equations

The Attempt at a Solution


Initially, writing the line in parametric form\vec{AB} = B - A = (0-1,1-2,0-5) = (-1,-1,-5)\\<br /> \\ <br /> \Rightarrow \vec{v} = (-1,-1,-5)\\r: (1, 2, 5) + \lambda(-1,-1,-5)\\<br /> \\<br /> x = 1 - \lambda\\<br /> y = 2 - \lambda\\<br /> z = 5 - 5\lambdaI know that \text{dist}\{PB\} = ||\vec{PB}||, which in turn means||\vec{PB}|| = \sqrt{(0 - x)^2 + (1 - y)^2 + (0 - z)^2} = 3||\vec{PA}|| \\ \Rightarrow \sqrt{(0 - x)^2 + (1 - y)^2 + (0 - z)^2} = 3\left(\sqrt{(1 - x)^2 + (2 - y)^2 + (5 - z)^2}\right)And then I just replace the variables with their values from the parametric system of equations.

While this does seem correct to me, I can never get to the value of my book. I'd like to know what I'm doing wrong. I've checked a few solutions on the web, and sometimes they subtract P - B instead of B - P like I did, and it doesn't make sense to me... Any help would be greatly appreciated.
 
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Bunny-chan said:

Homework Statement


Let A = (1,2,5) and B = (0,1,0). Determine a point P of the line AB such that ||\vec{PB}|| = 3||\vec{PA}||.

Homework Equations

The Attempt at a Solution


Initially, writing the line in parametric form\vec{AB} = B - A = (0-1,1-2,0-5) = (-1,-1,-5)\\<br /> \\<br /> \Rightarrow \vec{v} = (-1,-1,-5)\\r: (1, 2, 5) + \lambda(-1,-1,-5)\\<br /> \\<br /> x = 1 - \lambda\\<br /> y = 2 - \lambda\\<br /> z = 5 - 5\lambdaI know that \text{dist}\{PB\} = ||\vec{PB}||, which in turn means||\vec{PB}|| = \sqrt{(0 - x)^2 + (1 - y)^2 + (0 - z)^2} = 3||\vec{PA}|| \\ \Rightarrow \sqrt{(0 - x)^2 + (1 - y)^2 + (0 - z)^2} = 3\left(\sqrt{(1 - x)^2 + (2 - y)^2 + (5 - z)^2}\right)And then I just replace the variables with their values from the parametric system of equations.

While this does seem correct to me, I can never get to the value of my book. I'd like to know what I'm doing wrong. I've checked a few solutions on the web, and sometimes they subtract P - B instead of B - P like I did, and it doesn't make sense to me... Any help would be greatly appreciated.
Your work is correct so far, show what you did further.
 

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