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Analytical Solid Geometry

Analytical Solid Geometry. Concept of Octant. Z. P(x 1 , y 1 , z 1 ). . M. X. . . Y. O. Z. P(x 1 , y 1 , z 1 ). . Y. M. X. ELEMENTARY RESULTS. r. O. COS  ?. Let direction cosines of OP be l, m, n. Then OM = y 1. COS  = m = OM/OP  cos  = y 1 / r

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Analytical Solid Geometry

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  1. Analytical Solid Geometry

  2. Concept of Octant

  3. Z P(x1, y1, z1)  M X   Y O

  4. Z P(x1, y1, z1)  Y M X ELEMENTARY RESULTS r O COS  ? Let direction cosines of OP be l, m, n. Then OM = y1 . COS  = m = OM/OP  cos  = y1 / r y1 = r cos  = rm

  5. Similarly x1 = rl ; z1 = rn Hence coordinates of P (lr, mr, nr) Note : When r = 1 , coordinates of P becomes (l, m, n). We have , r2 = (x1)2 +(y1)2+(z1)2 =(lr)2 + (mr)2 + (nr)2 = (l2 + m2 + n2)r2 l2 + m2 + n2 = 1

  6. Definition: If there exists a, b, c such that

  7. If a, b, c are the direction ratios a, b, c are the direction ratios of a straight line and l, m, n are the corresponding direction cosines then, l2/a2 = m2/b2 = n2/c2 = (l2 +m2 + n2) / (a2 + b2 + c2) = 1 / (a2 + b2 + c2) l2 = a2/(a2 + b2 + c2)  l =  a/((a2+b2+c2)) m2 = b2/(a2 + b2 + c2)  m =  b/((a2+b2+c2)) n2 = c2/(a2 + b2 + c2)  n =  c/((a2+b2+c2))

  8. To find direction ratios of the line joining P(x1, y1, z1) and Q(x2, y2, z2)

  9. Z Q(x2, y2, z2)  P(x1, y1, z1) R O Y X COS  ? Let direction cosines of the line PQ are l, m, n and length PQ = r. QPR =  and PR = y2 – y1 We have from  PQR , cos = ( y2– y1) / r y2 – y1 = rcos 

  10. Hence direction ratios of the line joining P(x1, y1, z1) and Q(x2, y2, z2) are x 2 - x1 , y2 – y1 , z2 – z1 .

  11. Angle between two lines

  12. L 1 l1, m1 , n1 Z L 2 P l2 , m2 , n2 Q  Y O Cosine rule ? X Draw OP = OQ = 1thatare parallel to L1 and L2 respectively as shown in the fig. Then coordinates P are (l1, m1, n1) and that of Q are (l2, m2, n2).

  13. By cosine rule applying to the  OPQ we get , PQ2 = OP2 + OQ2 – 2(OP.OQ) cos  = 1+1-2cos  = 2 - 2cos  () Also PQ2 = (l2 – l1)2 + (m2 - m1)2 + (n2 – n1)2 = (l12 + m12 + n12) + (l22 + m22 + n22) – 2(l1l2 + m1m2+ n1n2) = 1 + 1 - 2(l1l2 + m1m2 + n1n2) (**) By () and () we have Cos  = l1l 2 + m1m2 + n1n2

  14. Note : i. If (a1, b1, c1) and ( a2, b2, c2) are the direction ratios of the lines L1 & L2 respectively then ii. If two lines with these direction cosines l1, m1 ,n1 and l2, m2, n2 are perpendicular to each other then, cos  = cos /2 = l1l2 + m1m2 + n1n2 i.e., l1l2 + m1m2+ n1n2 = 0

  15. Exercises: 1. If the lines are parallel to each other then prove l1 = l2 , m1 = m2 and n1 = n2 2. If the lines have the direction ratios a1, b1, c1 and a2, b2, c2 then they are parallel when a1/a2 = b1/b2 = c1/c2 and they are perpendicular when a1a2 + b1b2 + c1c2 = 0

  16. Q  R P M N L Definition: If PQ is the line segment and  is the angle made by PQ with L then, projection of PQ on L = MN = PR = PQ. cos  = rcos   In general projection = rcos 

  17. To find the projection of a line segment joining P (x1, y1, z1) and Q(x2, y2, z2) on a line with direction cosines l, m, n.

  18. R M N L direction cosines ? Q(x2, y2, z2) P(x1, y1, z1) l, m, n direction cosines ?

  19. Direction ratios of the line segment PQ are x 2 - x1 , y 2 - y1 , and z 2 - z1. The corresponding direction cosines are (x 2 - x1)/ r, (y 2 - y1)/r , (z2 - z1)/r where

  20. The angle between PQ and the line L is given by The projection is given by = rcos  = l (x2- x1) + m (y2 – y1 ) + n (z2- z1 )

  21. direction ratios ? direction ratios ?

  22. Z B’(0,0,a) A’(0,a,a) C’(a,0,a) P (a, a, a) C (0,0,a) O(0,0,0) Y A(a,0,0) B (a,a,0) X

  23. Planes

  24. Equation of a plane (normal form) : Z P(x, y, z) Q Y O X Consider a plane with OQ as normal to the plane from the origin O. Let l, m, n be the direction cosines of the normal OQ. Let length OQ = p. Let P (x, y, z) be a general point on the plane. Observe that projection OP on OQ is OQ. i.e., p = l (x- 0) + m (y - 0) + n( z – 0) i.e. lx + my + nz = p

  25. Let a, b, c be the direction ratios of a normal to the plane then

  26. Then equation of the plane is Or Thus a general linear equation in x, y, z namely ax + by + cz + d = 0 represents a plane with a, b, c as direction ratios of a normal to the plane. Thus for the plane ax + by + cz + d = 0 the perpendicular distance from the origin is given by

  27. Z (0,0,c) c b Y a (0,b,0) (a,0,0) X 2. Equation of a plane in the intercept form : Consider a plane Ax +By + Cz + d = 0 which makes intercept a with x-axis , b with y-axis and c with z-axis.

  28. As (a, 0, 0) is a point on the plane, Aa + d= 0 A = - d/a As (0, b, 0) is a point on the plane, B b + d = 0 B = - d/b As (0, 0, c) is a point on the plane, C c + d = 0 C = - d/c Substituting we get, x/a + y/b + z/c = 1

  29. Equation of a plane passing through P (x1 , y1 , z1)

  30. Consider a plane ax + by + cz + d = 0. If it passes through the point P (x1, y1, z1). Then ax1 + by1 + cz1 + d = 0 or d = -ax1 –by1 -cz1 Equation of the plane becomes ax + by + cz +d – (ax1 + by1 + cz1 +d) = 0 i.e., a(x-x1)+ b(y-y1)+ c(z-z1) = 0 Thus equation of a plane which passes through the point P (x1 , y1 , z1) is a (x-x1) + b(y-y1) + c(z - z1) = 0

  31. Consider a plane ax + by + cz + d = 0. if it passes through the point (x1, y1, z1), (x2 ,y2, z2) & (x3, y3, z3) If the plane passes through P1 then its equation is given by a(x-x1) + b(y-y1) + c(z-z1) = 0 (1) If P2 is on the plane, then, a(x2-x1) + b(y2-y1) + c(z2-z1) = 0 (2) If P3 is a point on the plane then a(x3-x1) + b(y3-y1) + c(z3-z1) = 0 (3)

  32. Eliminating a, b, c from (1), (2) and (3) we get,

  33. Points ( x1, y1, z1) , (x2, y2, z2),(x3, y3, z3) and (x4, y4, z4) are coplanar if

  34. P(x1, y1, z1) Q(f, g, h) R ax+by+cz=d Example : Find the perpendicular distance from the point (x1 , y1, z1) to the plane ax + by + cz + d = 0 From the figure PR = projection of PQ on PR The direction cosines of PQ are

  35. Hence ( because (f, g, h) is a point on ax + by + cz = 0, af + bg + ch = -d)

  36. Eg: A variable plane is at a constant distance p from the origin and cuts the axis at A, B and C. Prove that the locus of the centroid of the triangle ABC is 1/x2 + 1/y2 + 1/z2 = q/p2 Solution : Let (a, 0, 0), (0, b, 0) and (0, 0, c) be the coordinates of A, B, C respectively. The equation of the plane is given by x/a + y/b + z/c = 1 The perpendicular distance p from the origin is given by • 1/a2 +1/b2 +1/c2 = 1/p2 (1)

  37. Substituting in (1) we get, 1/(9 2) + 1/(9 2) +1/(9 2 ) = 1/P2 1/2 + 1/ 2 +1/2 = 9/P2Hence locus of (, , ) is given by 1/x2 + 1/ y2 +1/z2 = 9/P2 Let (, , ) be the centroid of the triangle ABC. Then  = (a + 0 + 0)/3 = a/3,  = b/3 and  = c/3i.e., a = 3  , b = 3  and c = 3 

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