Question

a line PQ 100 mm long has its end p in the first quadrant 25mm from both HP and VP and the other end Q in the second quadrant 50 mm from both HP and VP.Draw it's projection and determine the actual inclination​

Answer 1

Answer:

Projection of lines

1. ORTHOGRAPHIC PROJECTIONS OF POINTS, LINES, PLANES, AND SOLIDS. TO DRAW PROJECTIONS OF ANY OBJECT, ONE MUST HAVE FOLLOWING INFORMATION A) OBJECT { WITH IT’S DESCRIPTION, WELL DEFINED.} B) OBSERVER { ALWAYS OBSERVING PERPENDICULAR TO RESP. REF.PLANE}. C) LOCATION OF OBJECT, { MEANS IT’S POSITION WITH REFFERENCE TO H.P. & V.P.} TERMS ‘ABOVE’ & ‘BELOW’ WITH RESPECTIVE TO H.P. AND TERMS ‘INFRONT’ & ‘BEHIND’ WITH RESPECTIVE TO V.P FORM 4 QUADRANTS. OBJECTS CAN BE PLACED IN ANY ONE OF THESE 4 QUADRANTS. IT IS INTERESTING TO LEARN THE EFFECT ON THE POSITIONS OF VIEWS ( FV, TV ) OF THE OBJECT WITH RESP. TO X-Y LINE, WHEN PLACED IN DIFFERENT QUADRANTS. STUDY ILLUSTRATIONS GIVEN ON HEXT PAGES AND NOTE THE RESULTS.TO MAKE IT EASY HERE A POINT A IS TAKEN AS AN OBJECT. BECAUSE IT’S ALL VIEWS ARE JUST POINTS.

Answer 2

Answer:

Projection of lines

1. ORTHOGRAPHIC PROJECTIONS OF POINTS, LINES, PLANES, AND SOLIDS. TO DRAW PROJECTIONS OF ANY OBJECT, ONE MUST HAVE FOLLOWING INFORMATION A) OBJECT { WITH IT’S DESCRIPTION, WELL DEFINED.} B) OBSERVER { ALWAYS OBSERVING PERPENDICULAR TO RESP. REF.PLANE}. C) LOCATION OF OBJECT, { MEANS IT’S POSITION WITH REFFERENCE TO H.P. & V.P.} TERMS ‘ABOVE’ & ‘BELOW’ WITH RESPECTIVE TO H.P. AND TERMS ‘INFRONT’ & ‘BEHIND’ WITH RESPECTIVE TO V.P FORM 4 QUADRANTS. OBJECTS CAN BE PLACED IN ANY ONE OF THESE 4 QUADRANTS. IT IS INTERESTING TO LEARN THE EFFECT ON THE POSITIONS OF VIEWS ( FV, TV ) OF THE OBJECT WITH RESP. TO X-Y LINE, WHEN PLACED IN DIFFERENT QUADRANTS. STUDY ILLUSTRATIONS GIVEN ON HEXT PAGES AND NOTE THE RESULTS.TO MAKE IT EASY HERE A POINT A IS TAKEN AS AN OBJECT. BECAUSE IT’S ALL VIEWS ARE JUST POINTS

It’s given 10<n<15, which means n=11,12,13,14

When n=11,

r =4(no.of students)

Each student should get at least one toffee, therefore 11-4=7(new value of n)

To find how to distribute those 7 toffees we use the form (n+r-1)C(r-1)

(7+4-1)C(4-1) = 10C3 = (10*9*8)/3*2 = 120

Similarly for n=12,

12-4=8

(8+4-1)C(4-1) = 11C3 = (11*10*9)/(3*2) = 165

For n=13,

13-4=9

(9+4-1)C(4-1)= 12C3 = (12*11*10)/(3*2) = 220

For n=14,

14-4=10

(10+4-1)C(4-1) = 13C3 = (13*12*11)/(3*2)= 286

Therefore,

120+165+220+286=791

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