Find distance of point (2, 0, 1) from plane

$\displaystyle

\frac{x-2}{3} + \frac{y+1}{4} + \frac{z-1}{2} = 0

$

I know how to solve for this if I'm given the parameters but I'm stuck on how to solve for the distance given this formula.

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- Oct 20th 2008, 03:44 PMkhuezyDistance between vector help
Find distance of point (2, 0, 1) from plane

$\displaystyle

\frac{x-2}{3} + \frac{y+1}{4} + \frac{z-1}{2} = 0

$

I know how to solve for this if I'm given the parameters but I'm stuck on how to solve for the distance given this formula. - Oct 21st 2008, 07:18 AMearboth
I assume that you use:

1. Equation of the plane $\displaystyle p:Ax+By+Cz+D=0$

2. Distance of the point $\displaystyle P\left(x_0, y_0, z_0\right)$ to the plane is:

$\displaystyle d = \dfrac{Ax_0+By_0+Cz_0+D}{\sqrt{A^2+B^2+C^2}}$

If so:

Multiply the equation by the common denominator (here it is 12) to get rid of the fractions, expand the brackets and collect the constants. You have now an equation in the well known form:

3. $\displaystyle \frac{x-2}{3} + \frac{y+1}{4} + \frac{z-1}{2} = 0 ~\implies~4x-8+3y+3+6z-6 = 0~\implies~4x+3y+6z-11=0$

4. Go ahead!