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.