-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.

  • 0
    Two points do not Uniquely determine a plane. May be, you'll have to make your question precise.2012-02-21
  • 2
    "...given by two vectors B and C" The question is precise, you are unable to comprehend it, though.2012-02-21
  • 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.

  • 5
    @Zurechtweiser Your attitude isn't going to encourage people to answer your question.2012-02-21
  • 0
    @Zurechtweiser Indeed, you should be a bit more appreciative of people who are spending their time trying to help you, for free.2012-02-21
  • 0
    @Zev Chonoles Oh please don't give the "help for free"-rap. Instead: Where is YOUR contribution, my friend?2012-02-21
  • 0
    I flagged these comments for a moderator only to be declined. So, It makes me think people appreciate this attitude here. Let's only see how far this gets.2012-02-21
  • 0
    @Kannappan: The only two options available to moderators for flagged comments are to dismiss or to delete the comment. In general I would like to avoid deleting comments that are not directly abusive or spam; if Zurechtweiser wants to show everyone his bad attitude that is his right. I have now deleted one comment above that I considered abusive. The next comment here that is not about the mathematics of the question at hand will result in a suspension.2012-02-21
  • 0
    I got the answer elsewhere. It was simply "ax + by + cz + d = 0". abc are for the normalized cross product of the two vectors creating the plane, xyz is one arbitrary point of B or C. So you figure out d. And yes, I am an ungrateful fellow who always insults people without thinking. I deserve to never receive help from you again ..... I am serious, never help me again - cause I can live better without what you call help.2012-02-21
  • 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