Find point on plane

[markraz]

Hi, If I have a plane..for instance P = (2x+1y-2z = 4) and I have a point q<0,0,0>

and I already know the distance of q perpendicular to the plane P is 4/3

What is the easiest method to calculate the X,Y,Z point of q onto P ?? I guess it could be considered a projection of q onto p ?? What is that point of intersection?

is this easy to do?

thanks

 

[pka]

Hi, If I have a plane..for instance P = (2x+1y-2z = 4) and I have a point q<0,0,0>
What is the easiest method to calculate the X,Y,Z point of q onto P ?? I guess it could be considered a projection of q onto p ?? What is that point of intersection?

Find where the line \(\displaystyle \ell :<2t,t,-2t>\) intersects the plane \(\displaystyle \bf{P}\).

 

[markraz]

Find where the line \(\displaystyle \ell :<2t,t,-2t>\) intersects the plane \(\displaystyle \bf{P}\).

thanks

what is the simplest method to do that with the data I already have?? is it possible to calculate this intersection with q and the 4/3?

thanks

 

[pka]

what is the simplest method to do that with the data I already have?? is it possible to calculate this intersection with q and the 4/3?

The simplest way is to quite being lazy and learn the fundamentals. It would also help if you had correct answers. The distance Q is from that plane is 12.
Suppose that \(\displaystyle \Pi : (<x,y,z>-Q>\cdot N=0\) is a plane and \(\displaystyle P\) is a point. Then the distance \(\displaystyle \delta(P;\Pi)=\dfrac{|\vec{PQ}\cdot\vec{N}|}{\| \vec{N}\|}\)

Now if we want the point where the 'foot' of the of the perpendicular from \(\displaystyle P \) to \(\displaystyle \Pi \) we must find the intersection \(\displaystyle \ell: P + t\vec{N}\cap\Pi \).

 

[markraz]

The simplest way is to quite being lazy and learn the fundamentals. It would also help if you had correct answers. The distance Q is from that plane is 12.
Suppose that \(\displaystyle \Pi : (<x,y,z>-Q>\cdot N=0\) is a plane and \(\displaystyle P\) is a point. Then the distance \(\displaystyle \delta(P;\Pi)=\dfrac{|\vec{PQ}\cdot\vec{N}|}{\| \vec{N}\|}\)

Now if we want the point where the 'foot' of the of the perpendicular from \(\displaystyle P \) to \(\displaystyle \Pi \) we must find the intersection \(\displaystyle \ell: P + t\vec{N}\cap\Pi \).

thank you sir

so this video is wrong? https://www.youtube.com/watch?v=SgJo7_4mp6w

bummer

maybe that's why I'm having so much trouble while trying not to be lazy ........... :lol: