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REVISION PROBLEMS FOR FIRST MID-TERM EXAM IN PROJECTION

REVISION PROBLEMS FOR FIRST MID-TERM EXAM IN PROJECTION. Next. The Main Menu. ا Previous. Example (1). Represent a point A. Given that: 1) A is at equal distances from and 2) The distance of A from the origin 0 is 8 cms

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REVISION PROBLEMS FOR FIRST MID-TERM EXAM IN PROJECTION

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  1. REVISION PROBLEMS FOR FIRST MID-TERM EXAM IN PROJECTION

  2. Next The Main Menu اPrevious Example (1) Represent a point A. Given that: 1) A is at equal distances from and 2) The distance of A from the origin 0 is 8 cms 3) The point A is above and behind , . 4) The point A lies on the left hand side of at a distance 6 cms π π y = z 1 2 A A = 8 π π 2 1 z is positive and y is negative A A x = -- 6 A

  3. 8 // Next The Main Menu اPrevious // A1= A2 = , Y is -ve , and Z is +ve A A z A1 =A2 y is -ve A3 A // // X = - 6 x A K O 12 45 . M y 45 L

  4. MISSING PROJECTIONS OF POINTS z y z z A // A = A A 1 2 A 1 y 3 A // y y y A // A A // A // A 2 x x 3 o 12 12 o o x 12 y A A = A + 2 3 // A 1 Given A and A .Find A Given A and A . Find A Given A and A . Find A 2 3 1 2 3 1 1 2 3

  5. Example (2) FIND THE MISSING PROJECTIONS z A z 1 + y x A 12 // y O + + A // A A 2 3 // B y B 3 B 2 + x + 12 O // + Given A and A B y z C 1 B 2 3 + 1 Given B and B y C 2 3 // x 12 O y C // Given C and C C 3 C 1 2 + + 2

  6. z C3 C2 m2 m3 x12 o m1 y C1 Next The Main Menu اPrevious A point C m C m 3 3 C m 2 2 C m 1 1

  7. d(A,x-axis) = 4 cms .Find the locus of A d(A,y-axis) = 4 cms. Find the locus of A 3 z 2 z Locus of A Locus of A 3 2 4 cms x x 4 cms 12 12 o o z d(A,z-axis) = 4 cms. Find the locus of A x 12 1 o 4 cms Locus of A 1

  8. EXAMPLE (3) Given the two projections of a str. Line m . Represent the following points lying on m : i) A ( 9 ,?,?) ii) B(?, -1.5,?) iii) C(?,?,2.0) iv) D if d ( D,z-axis) = 5.5 cms v) E if d( E, y-axis) = 4 cms 2 Locus of E Locus of D 1 E m 2 1 D 2 C . 2 B A 4 cms 2 1 . 1.5 cms m 2.0 cms 9 cms B 2 2 x O A 12 1 5 .5 cms C 1 D 1 E 1

  9. Next The Main Menu اPrevious 1- Traces of a straight line: Definition: The points of intersection of a straight line m with the principle planes are called TRACES The horizontal trace H = m π z = 0 1 H The vertical trace V = m y v = 0 The side trace S = m x s = 0

  10. z S2 2 m2 S = S3 O m V1 m H2 3 x 3 H V 3 H = H1 3 m1 V = V2 S1 1 y

  11. z m2 m3 S1 m1 Next The Main Menu اPrevious Example(4): S3 = S S2 V3 V2 = V H2 x12 o H3 V 1 y H = H 1

  12. EXAMPLE (5) Represent a point A lying on the given str. Line m and d(A,x-axis) = 4 cms S S 2 3 m 3 m 2 A A 2 3 LOCUS OF A 3 R= 4 H 2 H 3 H 1 A 1 m 1 S 1

  13. Example (6) : Find m and the traces of the given str. Line m 3 S S=S 2 3 m m 3 2 V H H x 2 1 3 12 45 H=H 1 S m 1 1 V=V 2 V 3

  14. Example (7) . S 2 H =H S =S B 1 3 2 B A m 3 1 2 m V = V 3 2 V 3 / -3 cms A A 3 . 2 / H O V 3 H 1 2 m 1 . Given m and m .Find i) m ii) the traces H, V and S of m iii) the point A on m if A ( ?, - 3, ?) 1 2 3 B 1 S 1 iv) The point B on m if B is the closest point to z- axis

  15. TRUE LENGTH OF A SEGMENT AB AND ITS INCLINATIONS ON π , i=1,2,3 B i z T. L B A β y x B B y A B π 2 2 y T. L B x B 3 B A B 2 y β z A B α x A O x x A A A A B 3 1 z π z B A 3 A 1 z α π A B 1 A y T.L

  16. [B] T. L z [A] [B] A3 A2 [A] T.L m3 xA m2 yA xB B3 B2 x12 yB o zA xA zB A1 xB m1 yA y yB B1 zB zA [A] T.L [B] 4- The true length of a segment of a straight line: Trapezium method β α

  17. z A3 A2 [B] [B] m2 m3 B2 B3 x12 [A] A1 m1 y Next The Main Menu اPrevious Coordinates differences method T.L T.L o T.L B1

  18. T. L T. L T. L Horizontal projection Side projection Next The Main Menu اPrevious There are three right angled triangles which are frequently used in solving problems in PROJECTION ; Vertical projection

  19. B2 A2 x12 0 1 Locus of c1 B1 A1 Example (8 ) Represent the two projections of an equilateral triangle(مثلث متساوى الاضلاع) ABC if its side AB is given and C(?, 1, ?).

  20. . Locus of c2 B2 A2 . C2 x12 0 Vertical projection. of BC B1 A1 Locus of C 2 T. L of AB AB // / 1 Locus of c1 Vertical projection. of AC C1 / // T. L of BC T. L of AC T. L of BC AB // B C 2 2 T. L of AC T.L. of AB = T.L. of AC = T.L. of BC / A C 2 2

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