- #1
Bunny-chan
- 105
- 4
Homework Statement
Let [itex]A = (1,2,5)[/itex] and [itex]B = (0,1,0)[/itex]. Determine a point [itex]P[/itex] of the line [itex]AB[/itex] such that [itex]||\vec{PB}|| = 3||\vec{PA}||[/itex].
Homework Equations
The Attempt at a Solution
Initially, writing the line in parametric form[tex]\vec{AB} = B - A = (0-1,1-2,0-5) = (-1,-1,-5)\\
\\
\Rightarrow \vec{v} = (-1,-1,-5)\\[/tex][tex]r: (1, 2, 5) + \lambda(-1,-1,-5)\\
\\
x = 1 - \lambda\\
y = 2 - \lambda\\
z = 5 - 5\lambda[/tex]I know that [itex]\text{dist}\{PB\} = ||\vec{PB}||[/itex], which in turn means[tex]||\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)[/tex]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 [itex]P - B[/itex] instead of [itex]B - P[/itex] like I did, and it doesn't make sense to me... Any help would be greatly appreciated.