-1
$\begingroup$

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.

  • 1
    -1 for a "perfectly fine" attitude. Here, read [this](http://en.wikipedia.org/wiki/Dot_product).2012-02-21

1 Answers 1

4

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.

  • 1
    Perhaps if you'd asked Dave to explain further you'd know that that is equivalent to what he told you in his answer.2012-02-21