Vector multiplications using A/Vlisp

Hi everyone,

Are there built in functions to perform vector transformation ( perhaps using dot and cross products)?. If so, how can I access them.

Thanking in advance,

Aloy

What are you trying to do Aloy ? a bit more info is required .

May want to check the mathematical functions kindly offered by LeeMac.

;; Vector Dot Product  -  Lee Mac
;; Args: u,v - vectors in R^n

(defun vxv ( u v )
   (apply '+ (mapcar '* u v))
)

;; Vector Cross Product  -  Lee Mac
;; Args: u,v - vectors in R^3

(defun v^v ( u v )
   (list
       (- (* (cadr u) (caddr v)) (* (cadr v) (caddr u)))
       (- (* (car  v) (caddr u)) (* (car  u) (caddr v)))
       (- (* (car  u) (cadr  v)) (* (car  v) (cadr  u)))
   )
)

Hi, Tharwat, What I am trying to do is explained in the following link:

http://www.surveydrawing.net/find-point-3d-surface.html

Since I posted the above I found the following function in the net that might help; have not tested yet to solve equations:

(defun C:getNormal ( Point1 Point2 Point3 )

(setq Edge1 (vectorDiff Point2 Point1)

Edge2 (vectorDiff Point3 Point1))

(setq crossed (cross-product Edge1 Edge2))

(setq normVal (vectorNorm crossed))

(setq unitVector (list (/ (car crossed) normVal)

(/ (cadr crossed) normVal)

(/ (caddr crossed) normVal)))

(eval 'unitVector)

)

; vector norm (length)

(defun vectorNorm (Point1)

(eval '(sqrt (+ (expt (car Point1) 2.0)

(expt (cadr Point1) 2.0)

(expt (caddr Point1) 2.0))))

)

; vector cross-product

; A x B = (Ay*Bz - Az*By) (Az*Bx - Ax*Bz) (Ax*By - Ay*Bx)

(defun cross-product (Point1 Point2)

(eval '(list (- (* (cadr Point1) (caddr Point2)) (* (caddr Point1) (cadr Point2)))

(- (* (caddr Point1) (car Point2)) (* (car Point1) (caddr Point2)))

(- (* (car Point1) (cadr Point2)) (* (cadr Point1) (car Point2)))))

)

; subtract vector

; A - B = (Ax - Bx) (Ay - By) (Az - Bz)

(defun vectorDiff (Point1 Point2)

(eval '(list (- (car Point1) (car Point2))

(- (cadr Point1) (cadr Point2))

(- (caddr Point1) (caddr Point2)) ))

)

Hi, Mircea,

Thanks for directing to Lee Mac's solutions. As explained to Tharwat, I already got another solution but what you have given appears much more elegant.

Regards,

Aloy