3 Equations in 3 Unknowns

“The central problem of linear algebra is the solution of simultaneous linear equations. The most important case is when the number of unknowns equals the number of equations… we begin in three dimensions.” Gilbert Strang – Linear Algebra and its Applications

Ax + By + Cz = D is a linear equation which can be represented by a plane. The intersection of 3 planes at a single point is the solution of the 3 equations used to plot the planes.

its hard to show the intersection of 3 planes with a static image. the Region object is trimmable with the command SurfTrim.
this is the very first example set of equations from Strang’s (old version) book.

A plane is defined by 3 points. But the equation for the plane is more directly derived from a single point and a vector that is orthogonal (perpendicular) to the plane. The dot product of two vectors is

\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta

so if the angle between the vectors is 90 degrees, the scalar dot product is zero. The dot product also has the alternative computation that if vector a = (a1, a2, a3) and b = (b1, b2, b3) then

\mathbf{a} \cdot \mathbf{b} = (a1 \ast b1) + (a2 \ast b2) + (a3 \ast b3)

if N is the vector perpendicular to the plane,

\mathbf{N} =   \begin {pmatrix}  A \\  B \\  C \\  \end{pmatrix}

if P1 is the specific point in the plane, and P is a generic point in the plane,

\mathbf{P1} =   \begin {pmatrix}  x1 \\  y1 \\  z1 \\  \end{pmatrix}

\mathbf{P} =   \begin {pmatrix}  x \\  y \\  z \\  \end{pmatrix}

then a vector from P1 to P is (P – P1) and the dot product of N and this line is zero.

\mathbf{N} \cdot (\mathbf{P} - \mathbf{P1}) = 0

according to the rules, that can be written as

\mathbf{N} \cdot \mathbf{P} = \mathbf{N} \cdot \mathbf{P1}

which becomes

Ax + By + Cz = A(x1) + B(y1) + C(z1)

the numbers on the right have specific values, because we are using a specific point and a specific vector, so we can just call those D

Ax + By + Cz = D

which is the standard equation for a plane. The constants A B and C are from the normal vector.

In linear algebra, we are usually given 3 of these equations, and asked to solve it, which means find the intersection of the planes, if it exists. To plot from this equation, Ax + By + Cz = D, where A,B,C and D have actual values, to autocad or any other graphic method, we need three points. About the only way to do that i know is to substitute values. In the most general case where neither A, B or C is zero, the plane intersects all 3 axes, and the 3 intersection points can be found by solving the equation with two variables of x, y and z set to zero. Where one variable of A, B, and C is set to zero, the plane is parallel to the missing axis variable. Where two variables are zero the plane is parallel to both. Its a little tedious, but you just set up tables and solve the equation in the abstract with the missing variables visualizing the plane. Then a select case statement will be the gateway to the actual drawing of the plane.

The other method needed – besides plotting given only the standard equation – is deriving the equation from 3 given points. For that the cross product is essential. 3 non-linear points form a triangle, from which you can see 2 vectors in the plane. The cross product of 2 vectors is a third vector that is orthogonal (perpendicular) to both of them. the cross product gives you the normal vector, with ABC direction numbers. Then you use one of the 3 points, the normal vector, and derive the equation.

all this is covered in the latter third of a full calculus book, the chapter on vectors in space, or 3D vectors, or geometry of space. its also in analytic geometry books and implied or explained in linear algebra books. I am making a linear algebra homework tool.

In the most general case where neither A, B, or C is zero, the axis intercepts are calculated, a triangle can be drawn, and it can be filled with a Region object. If D is zero the plane goes thru the origin, and that is also a special case, the 3 intercepts are the same.

If you are given the std equation, you have the normal vector, but if you are given the 3 points in the plane, you need the normal vector. That is a pretty simple operation with the cross product. the cross product is a pretty fussy formula, but once it is captured to a sub its easy to use.

    Function find_norm(ptA() As Double, ptB() As Double, ptC() As Double) As Double()
        'given 3 points, not co-linear, find norm
        Dim AB() As Double
        Dim AC() As Double
        Dim norm() As Double

        AB = Minusv(ptB, ptA)
        AC = Minusv(ptC, ptA)
        norm = Cross_Product(AB, AC)

        Return norm
    End Function

    Function Cross_Product(t() As Double, v() As Double) As Double()
        Dim x, y, z As Double
        x = t(1) * v(2) - t(2) * v(1)
        y = t(2) * v(0) - t(0) * v(2)
        z = t(0) * v(1) - t(1) * v(0)
        Cross_Product = Pt(x, y, z)

        Return Cross_Product
    End Function

now having 3 points and a normal vector, its optional, but i find the centroid of the triangle as a convenient reference point. if the vertexes are F, G, H, the centroid is very simply (F + G + H) / 3. Now using a vector from the centroid to a vertex, and having the normal, you can once again use the cross product to find a vector orthogonal to both of them. For the purpose of creating a UCS (User Coordinate System) in the plane. Once you have created the UCS, its a simple matter to enlarge your plane, using a Circle center at the centroid, or a square or rectangle.

its a work in progress, here is current state of form. its not really easy to pick off the intersections graphically. the regions can be trimmed but i am going to look at other objects and methods.

How Do I Access Autocad Com Objects From Visual Studio?

I have stopped ignoring visual studio. I am having no issues accessing Com ActiveX objects from VB.Net with nearly the same code that works in VBA. I am not sure what all the naysaying is about. Every thread i have ever seen on running Com objects from Visual Studio says, Don’t do it, It’s a Bad Idea.

It seems to work fine.

as time goes by I may update this, but at this point –

Right click References in Solution Explorer, Add Reference, add the autocad type library.
To get the Autocad namespace to work across all modules, double click My Project in Solution Explorer and add Autocad to Imported Namespaces.

Get a book or study online the differences in syntax. They are pretty minor. The code editor is much improved. Arrays are a little simplified. Indexing is a little different, no static type. There is no variant type, but there is an object type that microsoft uses to read and write arrays to excel spreadsheets. (microsoft has a good one page link to connect to excel com objects HERE) Autodesk has no such single page that i can find, but here are some useful links.





All methods to be capitalized. All arguments in parentheses. No set statements.
Sometimes you have to figure out how to get the namespace shortcuts to work. The error reporting in vs really helps.
The Try, Catch, End Try error system works more cleanly.
Many more tools to use with collections and array types.
There is a stricter approach to datatypes, a setting to relax it, but i have Option Strict On. this is where a book will help. 2012 books are cheap.

So far i have pasted in VBA code, made changes per notes above, and the same objects work with the same logic – lines, polylines, dimensions, blocks, ucs, text, selection sets.

Code below is for version 2018 (“Autocad.Application.22”)

Module Module_Connect_Acad

    Public acadApp As AcadApplication
    Public acaddoc As AcadDocument
    Public acadms As AcadModelSpace

    Public Sub Connect_acad()
        acadApp = Nothing
        Dim strProgId As String = "AutoCAD.Application.22"

        Try         '' Get a running instance of AutoCAD
            'acadApp = GetObject(, strProgId)
            acadApp = CType(GetObject(, strProgId), AcadApplication)

            Try     '' Create a new instance of AutoCAD
                acadApp = CType(CreateObject(strProgId), AcadApplication)

            Catch ex As Exception
                Exit Sub
            End Try
        End Try

        acadApp.Visible = True  '' Display the application
        ' MsgBox("Now running " & acadApp.Name & " version " & acadApp.Version)

        'load whatever globals needed based on acadapp
        acaddoc = acadApp.ActiveDocument
        acadms = acadApp.ActiveDocument.ModelSpace
    End Sub

End Module