Question

 The centre of projection coincides with the origin. The projection plane passes through the point P(1,5,6) and has a normal vector (3,2,-1). Obtain the perspective projection transformation. 
d) A square has opposite vertices at (0,1) and (2,3). Shear the square 

i) by 1 unit along the x-axis with respect to y = 1.

ii) by 3 units along the y-axis wi  respect to x = 1.

06 Mar 2026
Answer :
Word Count : 502
The given problem involves two parts: obtaining the perspective projection transformation for a plane and performing shearing on a square. Let's solve step by step. Perspective Projection Transformation The centre of projection is at the origin (O(0,0,0)). The projection plane passes through (P(1,5,6)) and has a normal vector (\mathbf{n} = (3,2,-1)). The general equation of a plane is: [ n_x x + n_y y + n_z z + d = 0 ] Substitute the point (P(1,5,6)) to find (d): [ 3(1) + 2(5) + (-1)(6) + d = 0 \implies 3 + 10 - 6 + d = 0 \implies 7 + d = 0 \implies d = -7 ] So, plane equation: [ 3x + 2y - z -
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