Need some help with this explanation

[braincell]

Hi, I stumbled across this formula but I'm a bit confused.
http://mathworld.wolfram.com/Point-LineDistance3-Dimensional.html

From (9)-(11), do the X1's and X2's and X0's (in bold text) in the formula represent the x-coordinates of 3 points X1, X2, X0 respectively? Or do I have to do the calculation for the y-coordinates and z-coordinates to solve for d?

Sorry if I don't make sense... it's a little hard to explain. English is my second language.

Thanks in advance for all the help.

 

[stapel]

Hi, I stumbled across this formula but I'm a bit confused.
http://mathworld.wolfram.com/Point-LineDistance3-Dimensional.html

From (9)-(11), do the X1's and X2's and X0's (in bold text) in the formula represent the x-coordinates of 3 points X1, X2, X0 respectively? Or do I have to do the calculation for the y-coordinates and z-coordinates to solve for d?

The page defines its notation near the beginning:

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. . . . .Let a line in three dimensions be specified by two points x₁ = (x₁, y₁, z₁) and x₂ = (x₂, y₂, z₂) lying on it, so a vector along the line is given by:

. . . . . . . . . .v = \(\displaystyle \left[\begin{array}{c}x_1\, +\, (x_2\, -\, x_1)\,t\\y_1\, +\, (y_2 \, -\, y_1)\,t\\z_1\, +\, (z_2\, -\, z_1)\, t \end{array}\right]\)

. . . . .The squared distance between a point on the line with parameter t and a point x₀ = (x₀, y₀, z₀) is therefore:

. . . . . . . . . .\(\displaystyle d^2\, =\, \left[(x_1\, -\, x_0)\, +\, (x_2\, -\, x_1)\, t\right]^2\, +\, \left[(y_1\, -\, xy_0)\, +\, (y_2\, -\, xy_1)\, t\right]^2\, +\, \left[(z_1\, -\, z_0)\, +\, (z_2\, -\, z_1)\, t\right]^2\)

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So the bolded x's are points.