# How to extract d from a plane being created by two vectors?

I want to let fall a perpendicular from a point A in space being given by $A_x, A_y$ and $A_z$ on a plane being given by two vectors $B$ and $C$.

Ultimately I want to determine the foot x0 of the perpendicular. Note: This is not the question, this is the introduction. Here the questions follow.

I found

$$x_0 = p \vec+ t_0*n$$

while

$$t_0 = \frac{(d - n*p)}{n^2}$$

what is d in my case?

Is the vector n squared the same as:

$$n^2_x = n_x*n_x$$ $$n^2_y = n_y*n_y$$ $$n^2_z = n_z*n_z$$

Is the $\mathrm{Vector}_n * \mathrm{Vector}_p$ the same as $(np = \mathrm{Vector}_n * \mathrm{Vector}_p)$?

$$np_x = n_x*p_x$$ $$np_y = n_y*p_y$$ $$np_z = n_z*p_z$$

Thanks go to the one who formatted it. As you are at it, can you put arrows over the appropriate p and n vectors? Then you can remove this phrase.

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Two points do not Uniquely determine a plane. May be, you'll have to make your question precise. – user21436 Feb 21 '12 at 15:40
"...given by two vectors B and C" The question is precise, you are unable to comprehend it, though. – Zurechtweiser Feb 21 '12 at 16:09
-1 for a "perfectly fine" attitude. Here, read this. – Rahul Feb 21 '12 at 20:13

First, you can find a vector normal to the plane in question by taking a cross product: $n = B \times C$.
Then, you want to resolve $A$ into a component parallel to $n$ and a component perpendicular to $n$. The parallel component is found by projection, which uses the dot product: $\frac{A \cdot n}{n \cdot n} n$, or if you prefer $\frac{A \cdot n}{\|n\|^2} n$. The perpendicular component is then $A$ minus the parallel component. The perpendicular component is the same as the position vector of the foot of the perpendicular.