Find intersection of point perpedicular to plane

[markraz]

Hi, how do I go about finding the intersection of a point (red dot) projected perpendicular to plane (yellow)??
Is this linear algebra? is this basic algebra? any thoughts on where to begin?

Thanks

 

[pka]

Hi, how do I go about finding the intersection of a point (red dot) projected perpendicular to plane (yellow)?

Suppose the plane is \(\displaystyle \Pi: ~ax+by+cz+d=0\) and the point is \(\displaystyle P:~ (m,n,j)\).
Now consider the line \(\displaystyle \ell:~<at+m,~bt+n,~ct+j>\). Solve \(\displaystyle a(at+m)+b(bt+n)+c(ct+j)+d=0\) for \(\displaystyle t\). That value gives the point of intersection you need.

 

[markraz]

Suppose the plane is \(\displaystyle \Pi: ~ax+by+cz+d=0\) and the point is \(\displaystyle P:~ (m,n,j)\).
Now consider the line \(\displaystyle \ell:~<at+m,~by+n,~cz+j>\). Solve \(\displaystyle a(at+m)+b(by+n)+c(cz+j)+d=0\) for \(\displaystyle t\). That value gives the point of intersection you need.

thanks, is this \(\displaystyle \ell:~ \) a parametric equation?

 

[HallsofIvy]

Yes, those are parametric equations for a line with "t" as the parameter.

What pka is doing is constructing the line perpendicular to the given plane that goes through the given point. Where that line crosses the plane is the perpendicular projection of the point onto the plane.
-)))
Basically repeating what pka said, if the plane is given by Ax+ By+ Cz= D (equivalent to pka's "ax+ by+ cz= d" with A= a, B= b, C= c, D= -d) then < A, B, C> is a vector perpendicular to the plane. Any line perpendicular to the plane can be written, in parametric equations, as x= At+ x_0, y= Bt+ y_0, z= Ct+ z_0, where (x_0, y_0, z_0) is any point on the line. Since we want the point (m, n, j) to be on the line, x= At+ m, y= Bt+ n, z= Ct+ j.

To find the point where that line crosses the given plane, replace x, y, z in the equation of the plane with those parametric equations and solve for t.

For example- find the projection of (1, 3, 0) on the plane 2x+ y- z= 4.
A vector perpendicular to that plane is <2, 1, -1>. A line through (1, 3, 0) perpendicular to the plane is x= 2t+ 1, y= t+ 3, z= -t. Putting those into the equation of the plane, 2(2t+ 1)+ (t+ 3)- (-t)= 4t+ 2+ t+ 3+ t= 6t+ 5= 4. 6t= -1, t= -1/6. Putting that into the equation of the line, x= 2(-1/6)+ 1= 2/3, y= (-1/6)+ 3= 17/6, z= -(-1/6)= (1/6).

The perpendicular projection of the point (1, 3, 0) onto the plane 2x+ y- z= 4 is
(2/3, 17/6, 1/6).

 

[bforest]

It only looks like you have to find a line that is perpendicular to the Line which contains z as a function of y. If that's the case, then you can treat x as a constant in z=f(x,y)
if z= x+ y, then treating x as a constant, z= c+y. From there it's as simple as taking the negative reciprocal of the slope: a line parallel to z= c+y would be z=c-y, or even more simply z=-y. The constant in this case does not affect the slope and thus has no bearing on the orthogonality of the lines. However, if the line need pass through a point, say, (1,0, 1) then c should be chosen to be 1, so as to correspond wit the point. More over, in order to insure that the point lies on the line, the intercept b may have to be added to the equation. So it becomes, z=-y+c+b. Z=1, c=x=1, and y= 0: 1=0+1+b. In this case, b= 0. In general. To find the perpendicular line traveling through point (X,Y,Z). You have to first treat x as a constant in the function z=f(x,y). Say this time. z=xy. Then z = cy. The line parallel to that becomes z=-(-1/c)y. And because this line passes through point (X,Y,Z), then we can say that Z=(-1/X)Y+b. Then solving algebraically b=-XZ/Y . So the equation of the line becomes z= -(1/x)y + -XZ/Y. But then again I'm not really sure... This has all been really more or less a guess of how this works. Actually kinda think about it that's an equation of a plane not a line. It's possibke the line is just z=-(-1/X)y- XZ/Y. Where X is the constant from the point. I just did some sketches and that's exactly it. Ok so here's exactly how it works, say you have the point (a,b,c). Then in order to find the equation of a line perpendicular to the functions plane and passing through the point, first rewrite z=f(x,y) as z=(a,y). Because a is a constant, this becomes a simple linear function. Secondly, find the slope in z=(a,y) and take the negative reciprocal: -1/m. Thirdly, solve for d in the equation c=f(a,-y/m)+b . b=c/f(a,-y/m). And there it is. z=f(a,-y/m)+c/f(a,-y/m). Assuming f(a,y) is linear... It'll work

 

[HallsofIvy]

I believe I responded to this in another forum, but:
Any plane can be written in the form Ax+ By+ Cz= D, for some numbers A, B, and C. The vector <A, B, C> is perpendicular to the plane. Let the given point be (p, q, r). The projection of the point on the plane is the point where a line through that point, perpendicular to the plane, intersects the plane.

Since <A, B, C> is perpendicular to the plane it is in the direction of the desired line. A line through (p, q, r) in the direction of <A, B, C> is given by x= At+ p, y= Bt+ q, z= Ct+ r with parameter t. That line passes through the plane where the x, y, z, values satisfy the equation of the plane:
A(At+ p)+ B(Bt+ q)+ C(Ct+ r)= D. Solve that linear equation for t and then solve for x, y, and z.

 

[pka]

I believe I responded to this in another forum,

You did indeed answer, a good answer as well. This raises the question: Why was this same question posted a second time? Could it be that the user is simply seeking a completely worked, 'camera-ready, solution?

 

[HallsofIvy]

I don't see how he could expect a "camera ready" solution without giving a specific problem.

 

[markraz]

thank you so much...
Is this type of problem covered in Calculus III?

 

[HallsofIvy]

Yes.