Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
2  
"...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  
-1 for a "perfectly fine" attitude. Here, read this. –  Rahul Feb 21 '12 at 20:13

1 Answer 1

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.

share|improve this answer
5  
@Zurechtweiser Your attitude isn't going to encourage people to answer your question. –  Chris Taylor Feb 21 '12 at 18:16
    
@Zurechtweiser Indeed, you should be a bit more appreciative of people who are spending their time trying to help you, for free. –  Zev Chonoles Feb 21 '12 at 19:59
    
@Zev Chonoles Oh please don't give the "help for free"-rap. Instead: Where is YOUR contribution, my friend? –  Zurechtweiser Feb 21 '12 at 20:04
    
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. –  user21436 Feb 21 '12 at 20:11
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. –  Zev Chonoles Feb 21 '12 at 21:04

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.