Tiêu đề XI - maths - chapter 12 - PLANES (11.03.2015)-(232-251).pdf ✅

1 JEE MAINS ADVANCE 3D - PLANES Equation of a Plane:  Every first degree equation in x,y,z always represents a plane.  Plane surface is a surface in which line joining every two points P and Q on it lies entirely in the surface.  The general form of equation of plane is ax by cz d     0 , a b c , , are not all zero i.e., 2 2 2 a b c    0 Equation of Planes with Different Conditions:  i) The equation of the plane passing through the point  x y z 1 1 1 , ,  and having d.r’s of normal as (a,b,c) is a x x b y y c z z        1 1 1      0 or ax by cz ax by cz      1 1 1 ii) The equation of the plane passing through a point  x y z 1 1 1 , ,  and parallel to the plane ax by cz d     0is a x x b y y c z z        1 1 1      0 => ax by cz ax by cz      1 1 1 W.E-1 : The equation of the plane parallel to the plane 2 3 4 5 0 x y z     and passing through the point (1,1,1) is Sol : The plane is a x x b y y c z z        1 1 1      0 2 1 3 1 4 1 0  x y z                 2 3 4 9 0 x y z Equation of plane which is Parallel to lines:  i) The equation of the plane passing through the point  x y z 1 1 1 , ,  and parallel to lines whose d.r’s are a b c 1 1 1 , ,  and a b c 2 2 2 , ,  is 3D-PLANES 1 1 1 1 1 1 2 2 2 0 x x y y z z a b c a b c     ii) The equation of the plane passing through the points x y z 1 1 1 , ,  ,  x y z 2 2 2 , ,  and parallel to the line whose d.r’s are (a,b,c) is 1 1 1 2 1 2 1 2 1 0 x x y y z z x x y y z z a b c        iii) The equation of the plane passing through three non collinear points  x y z 1 1 1 , ,  , x y z 2 2 2 , ,  , x y z 3 3 3 , ,  is 1 1 1 2 1 2 1 2 1 3 1 3 1 3 1 0 x x y y z z x x y y z z x x y y z z           iv) If  x y z x y z x y z 1 1 1 2 2 2 3 3 3 , , , , , , ,     and  x y z 4 4 4 , ,  are coplanar, then 4 1 4 1 4 1 2 1 2 1 2 1 3 1 3 1 3 1 0 x x y y z z x x y y z z x x y y z z           General equation of a plane with different conditions:  i) The equation of a plane with d.r’s of normal as (a , b ,c) is ax by cz d     0 . ii) If a) a=0, b c   0, 0 Then equation by cz d    0 represents a plane which is parallel to x-axis and er  to YZ - plane. b) b a c    0, 0, 0 then equation ax cz d    0 represents a plane which is par- allel to y-axis and er  to xz -plane. c) a b c    0, 0, 0 then equation ax by d    0 represents a plane which is par- SYNOPSIS
3D - PLANES 2 JR JEE MAINS VOL - IV JEE MAINS ADVANCE Distance= 1 2 2 2 2 d d a b c    ,= 5 3 2 1 4 4 1 6     W.E-3 : The equation of the parallel plane lying midway between the parallel planes 2 3 6 7 0 x y z     and 2 3 6 7 0 x y z     is Sol : The required plane is 1 2 0 2 d d ax by cz       7 7 2 3 6 0 2 x y z       2 3 6 0 x y z    W.E-4 : The reflection of the plane in the plane x y z     3 0 is Sol : The given planes are , 2 3 4 3 0 x y z     ( 1 1 1 1 a x b y c z d     0) x y z     3 0 ax by cz d     0  Equation of the required plane be obtained using the fact reflection of 1 1 1 1 a x b y c z d     0 in the plane ax by cz d     0 is given by    1 1 1 2 aa bb cc ax by cz d         2 2 2 1 1 1 1       a b c a x b y c z d  The reflection is 4 3 2 15 0 x y z     Foot and image:  i) The foot of the perpendicular of the point P x y z  1 1 1 , ,  on the plane ax by cz d     0is Qh k l , ,  then  1 1 1  1 1 1 2 2 2 h x k y l z ax by cz d a b c a b c             ii) If Q (h, k, l) is the image of the point p x y z  1 1 1 , ,  w.r.to the plane ax by cz d     0 then  1 1 1  1 1 1 2 2 2 h x k y l z 2 ax by cz d a b c a b c             allel to z-axis and er  to XY -plane. iii) The equation of the plane passing through  x y z 1 1 1 , ,  and parallel to a) yz- plane and er  to X-axis is 1 x x  b) xy-plane and er  to Z-axis is 1 z z  c) zx-plane and er  to Y-axis is 1 y y  iv) Equation of plane parallel to the plane 1 ax by cz d     0 is of the form 2 ax by cz d     0 v) Distance between the above two parallel planes is 1 2 2 2 2 d d a b c    vi) Equation of plane parallel to 1 r n d .    is 2 r n d .    (vector form) vii) The equation of the plane, mid way between the parallel planes ax by cz d    1 0 and ax by cz d    2 0 is 1 2 0 2 d d ax by cz            viii) The equation of the plane which bisects the line joining A x y z  1 1 1 , ,  and B x y z  2 2 2 , ,  and perpendicular to AB is  x x x y y y z z z 1 2 1 2 1 2                2 2 2 2 2 2 1 1 1 2 2 2 2 x y z x y z      ix) The reflection of 1 1 1 1 a x b y c z d     0 in the plane ax by cz d     0 is given by    1 1 1 2 aa bb cc ax by cz d         2 2 2 1 1 1 1       a b c a x b y c z d W.E-2 : Distance between parallel planes 2 2 3 0 x y z     a n d 4 4 2 5 0 x y z     is Sol : The given planes are 2 2 3 0 x y z     , 5 2 2 0 2 x y z     Here 1 2 5 2, 2, 1, 3, 2 a b c d d      
3 JEE MAINS ADVANCE 3D - PLANES P x y z  1 1 1 , ,  and Q x y z  2 2 2 , ,  lie on opposite sides of the plane ax by cz d     0 W.E-7 : If the plane 2 3 5 2 0 x y z     divides the line segment joining (1, 2, 3) and (2, 1, k) in the ratio 9 : 11 then k = Sol : 1 1 1 2 2 2 ( ) 9 11 ax by cz d ax by cz d             2 1 3 2 5 3 2 9 2 2 3 1 5 2 11 k               ,   k 2 Normal form of a plane:  i) If l,m,n are the direction cosines of normal to plane  and p is the er  distance from origin to the plane then the equation of plane is lx my nz p    ii) The normal form of the plane representing by the equation ax by cz d     0 is a) If d < 0 2 2 2 2 2 2 a b x y a b c a b c       2 2 2 2 2 2 c d z a b c a b c       b) If d > 0 2 2 2 2 2 2 a b x y a b c a b c         2 2 2 2 2 2 c d z a b c a b c       W.E-8 : If the equation of the plane 2 3 6 7 x y z    in the normal form is lx my nz p    then l p   Sol : Equation of the plane in the normal form is 2 3 6 7 4 9 36 4 9 36 4 9 36 4 9 36 x y z             2 3 6 1 7 7 7 x y z    ,lx my nz p      2 9 1 7 7 l p     iii) If ‘d ’ is the distance from the origin and l m n , ,  are the dc’s of the normal to the plane through the origin, then the foot of the perpenducular is ld md nd , ,  W.E-5 : The foot of the perpendicular from the point P(1,3,4) to the plane 2 3 0 x y z     is Sol : The given plane is 2 3 0 x y z     Here a b c d      2, 1, 1, 3,  x y z 1 1 1 , , 1,3, 4      1 1 1 h x k y l z a b c        1 1 1 2 2 2 ax by cz d a b c         h k l , , 1, 4,3     W.E-6 : If the image of the point (-1,3,4) in the plane x y   2 0 is  x y z 1 1 1 , ,  then 1z  Sol : Given that h k l x y z , , , ,    1 1 1   x y z 1 1 1 , , 1,3,4      1 1 1 1 3 4 1 2 0 x y z              2 2 2 1 2 3 1 2 0                     z1 4 0,  z1 4 Ratio formula:  i) The ratio in which the plane ax + by + cz + d = 0 divides the line segment joining  x y z 1 1 1 , ,  and  x y z 2 2 2 , ,  is     ax by cz d 1 1 1 :ax by cz d 2 2 2     ii) Position of the points w.r.to the plane a) If 1 1 1 2 2 2 0 ax by cz d ax by cz d        then the points P x y z  1 1 1 , ,  and Q x y z  2 2 2 , ,  lie on same side of the plane ax by cz d     0 b) If 1 1 1 2 2 2 0 ax by cz d ax by cz d        then the points
3D - PLANES 4 JR JEE MAINS VOL - IV JEE MAINS ADVANCE a) X – axis , Y –axis is 1 ab 2 Sq. units b) Y– axis, Z– axis is 1 bc 2 Sq. units c) Z– axis, X– axis is 1 ca 2 Sq. units ii) If the plane 1 x y z a b c    meets the co- ordinate axes in the points A,B,C. then the area of the triangle ABC is       1 2 2 2 2 ab bc ca   . W.E-10 : The area of the triangle formed by the plane 2 3 6 9 0 x y z     with Y - axis, Z- axis is (in Sq.units) Sol : The plane is 1 9 3 3 2 2 x y z       Here 9 3 , 3, 2 2 a b c       The area of the triangle = 1 2 bc =   1 3 3 2 2         = 9 4 Sq.units W.E-11 : The plane 1 2 3 4 x y z    cuts the axes in A, B,C then the area of the  ABC is (squ) Sol : a b c    2, 3, 4  Area of the ABC=       1 2 2 2 2 ab bc ca   ,  61 . Angle between Two Planes:  i) The angle between two planes is equal to the angle between the perpendiculars from the origin to the planes. ii) If ' ' is the angle between the planes Perpendicular distance from point to the plane:  i) The perpendicular distance from  x y z 1 1 1 , ,  to the plane ax + by + cz + d = 0 is 1 1 1 2 2 2 | | ax by cz d a b c      ii) The perpendicular distance of the plane ax+by+cz+d=0 from the origin is 2 2 2 d a b c   . W.E-9 : If the perpendicular distance from (1, 2, 4) to the plane 2 2 0 x y z k     is 3 then k - 4 = Sol : 2 4 4 3 4 4 1    k      K 7 , k     4 7 4 3 Intercept form of a plane:  i) If a plane cuts X-axis at A a ,0,0 , Y-axis at B b 0, ,0 and Z-axis at C c 0,0,  then a,b,c are called X-intercept,Y-intercept, Z-intercept of the plane. ii) The equation of the plane in intercept form is 1 x y z a b c    iii) If ax by cz d     0 is a plane if a b c    0, 0, 0 then X-intercept d a   Y-intercept d b   , Z-intercept d c   iv) The equation of the plane whose intercepts are K times the intercepts made by the plane ax by cz d     0 on corresponding axes is ax by cz kd     0 . Areas:  i) Area of the triangle formed by the plane 1 x y z a b c    with