# Thread: Distance between vector help

1. ## Distance 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.

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