Question

Pr. #1) In computer graphics and perspective drawing, we need to represent objects seen by the eye in space as images on a two-dimensional plano. Suppose that the west 0.01) as shown below and that we want to represent a point ( V ) as a point on the plane. We do this by projecting Ponto the plane with a ray from E. The point will be portrayed as the point P(0.y.). The problem for us as graphic designers is to find y and a given E and P. (a) Write a vector equation that holds between EP and EP. Use the equations to express y and of To 1. 9. and 2 in terms (b) Test the formula obtained for y and in part (a) by investigating their behavior atr and by seeing what happens as to 0. What do you find? 0 and 1' 1 P(0...) . . . . ( ...) Er,0,0)

Answer 1

Solution - The position vectors for the eye E and point p are P = X i + b + ok P = xi ? Youj+z,k_ @ The vector equation for E and P is given by = Ětt (Ě -P,] the line passing through Where raxityj+Zk teo _xity i+Zk = xoi+t(xo - X,)i-4-zik) for the coordinates of point p which is in the y-z plane x 20.92 4. ZEZ Thus the vector equation reduces to yg +Zk z Xoit t[(xo2X1 - Yoj -2,k] - Koi + yj + zk = t[CX0-2) i – Yj-z, By comparing the coefficient of ifik -no y = 2 20-, -y,_-Z, L 220 g, - Xo-x, za Xo 21 7 - X,

16 We observe that at x=0 - y = 4 Z-21 This means that the point P, is itself located in the y z plane. | We observe that at a=ko. - 9 = X, 9, Za Boz, Thus values are indeterminate, since the condition implies that the point P is located such that the line PP does I not intersect y-z plane further as nos ,we observe that y0 i tij | 2 *0 1 This implies the point E is infinitely I placed the inclination of the line I PP, will cease and will seem to I become coincident with the x- axis