Title XI - maths - chapter 12 - 3D Geometry-(W.E _ LEVEL-V-(11.03.2015)-(273-295).pdf ✅

THREE DIMENSIONAL GEOMETRY JEE ADVANCED - VOL - IV Narayana Junior Colleges 273 Narayana Junior Colleges Page No : 273 WORKEDOUT EXAMPLES W.E-1: The vector along the line of greatest slope in the plane x y z    3 with respect to the plane x y z    0 is Ans:      i j k 2  . Sol:     a i j k b i j k       ,     1 1 1 2 2 2  1 1 1 i j k a b i k i k              1 1 1 2 1 0 1 i j k a a b i j k          hence the vector along the line with greatest slope is      i j k 2  W.E-2. If the plane ax by   0 is rotated through an angle ' '  about it’s line of intersection with the plane z=0. Then the equation of the plane in the new position will be A) 2 2 ax by z a b     tan 0  B) 2 2 ax by z a b     tan 0  C) 2 2 bx ay z a b     tan 0  D) 2 2 bx ay z a b     tan 0  Sol: (BCD) Equation of the plane passing through the intersection of ax by   0 and z  0 is ax by z     0 --------(1) Dcs of normal to the plane are 2 2 2 2 2 2 2 2 2 , , a b a b a b a b           2 2 2 2 2 2 2 cos a b a b a b         2 2       a b tan W.E-3. A rod of length 2 units whose one end is (1,0,-1) and the other lies on the plane x y z     2 2 4 0 , then A)The rod sweeps a figure of volume  cubic units B)The area of the region which the rod traces on the plane is 2 square units C)The length of the projection of the rod on the plane is 3 units D)The centre of the region which the rod traces on the plane is 2 2 5 , , 3 3 3        Sol. (ACD) The rod sweeps a cone of height 1 and radius 3 units W.E-4. Consider a set of points R in the space which are at a distance of 2 units from the line 1 2 1 1 2 x y z      between the planes x y z     2 3 0 and x y z     2 2 0 then (IIT-JEE-2014) A)The volume of the bounded figure by the points R and the planes is 10 3 3  cubic units B)The area of the curved surface formed by the set of points R is 20 6  square units C)The volume of the bounded figure by the set of points in R and the planes is 20 6  cubic units D)The area of the curved surface formed by the set of points in R is 10 3  square units Sol: (BC) The distance between the planes is 5 6 h  Also the figure formed is a cylinder, whose radius is r units  2 2 20 6 Hencethe volume r h cubicunits     20 2 6 Areaof curved surface rh square units    W.E-5. From a point P   , , , perpendiculars PQ THREE DIMENSIONAL GEOMETRY
THREE DIMENSIONAL GEOMETRY JEE ADVANCED - VOL - IV Narayana Junior Colleges 274 Narayana Junior Colleges Page No : 274 and PR are drawn respectively on the lines y x z   , 1 and y x z     , 1. If P is such that QPR is a right angle, then the possible value(s) of  is (are) A) 2 B)1 C)-1 D) 2 sol: (C) Let Q k k   , ,1 , R l l     , , 1 1 1 1 0 x y z PQ lar to            k Q  , ,1 1 1 1 0 x y z PR lar to           l R 0 0,0, 1   Given 0   QPR 90       PQ PR   1 1 0       1, 1 if  1 then P=Q which is not possible. Hence  1 W.E-6. Supose in a tetrahedron ABC D AB C D , 1; 3;  the distance and angle between the skew lines AB and CD are 2 and 3  respectively. If the volume of the tetrahedron is V then the values of 6V is sol: b c d    1 3    shortest distence between AB and CD is 2 also  ,  3 b c d       Let equation of A B be r b  ...... 1    and equation of CD be r c c d     .... 2      n b c d              . .ˆ . ˆ ˆ c b c d c n S D projection of c on n n b c d                  [ ] [ ] sin / 3 c b c c b d b c d                0 [ ] 2 1. 3. 3 / 2  b c d     b c d 3           Volume 1 1 1 [ ] .3 6 6 2  b c c      W.E-7. Find the shortest distance of plane parallel to z-axis and containing line x y 2z 3 0 2x 3y 4z 4         from z-axis. Ans. (2) sol: Equation of line (1) is 5 2 2 0 1 x y z      Equation of line (2) is 0 0 1 x y z   Shortest distance=  . 4 2 2 a c b d units b d      W.E-8. Statement I: The plane 5 2 8 0 x z    contains the line 2 3 0 x y z     and 3 5 x y z    and is perpendicular to 2 5 3 0 x y z     Statement II: The plane 3 5 x y z    meets the line x y z      1 1 1 at the point (1,1,1). Sol: Equation of plane is 2 3 3 5 0 x y z x y z            For  1, we get 5x+2z-8=0 which is perpendicular to 2x-y-5z-3=0 as 5 2 0 1 2 5 0           W.E-9. The reflection of the plane 2 3 4 3 0 x y z     in the plane x y z     3 0 is the plane.
THREE DIMENSIONAL GEOMETRY JEE ADVANCED - VOL - IV Narayana Junior Colleges 275 Narayana Junior Colleges Page No : 275 A) 4 3 2 15 0 x y z     B) x y z     3 2 15 0 C) 4 3 2 15 0 x y z     D) x y z     3 2 5 0 Sol: (A)    ' ' ' 2 aa bb cc ax by cz d      =   2 2 2 ' ' ' ' a b c a x b y c z d       Reflection of plane 2 3 4 3 0 x y z     in the plane x y z     3 0 2 2 3 4 3 3 2 3 4 3            x y z x y z         4 3 2 15 0 x y z W.E-10. P,Q,R,S are four coplanar points on the sides AB,BC,CD,DA of a skew quadrilaterial. The product . . . AP BQ CR DS PB QC RD SA equals A)-2 B)-1 C)2 D)1 sol: (D) Let the vertices A,B,C,D of a quadrilaterial be  x y z x y z x y z 1 1 1 2 2 2 3 3 3 , , , , , , , ,     , x y z 4 4 4 , , . The equation of plane PQRS be u ax by cz d      0 Let r 1 1 1 u a x b y c z d     where r 1,2,3,4 Then . . . AP BQ CR DS PB QC RD SA 1 2 4 3 2 3 4 1 1 u u u u u u u u                               W.E-11. Perpendiculars are drawn from points on the line 2 1 2 1 3 x y z      to the plane x y z    3. The feet of perpendiculars lie on the line . (IIT JEE-2013) A) 1 2 5 8 13 x y z      B) 1 2 2 3 5 x y z      C) 1 2 4 3 7 x y z      D) 1 2 2 7 5 x y z      Sol: (D) Any point B on line is 2 2, 1,3        Point B lies on the plane for some ‘ ’        2 2 1 3 3       3 5 9 4 6 1, , 2 2 2   B              The foot of the perpendicular from point (-2,-1,0) on the plane is the point A(0,1,2)   7 5 . ' 1 , 2, 7,5 2 2 D r s of AB           Hence the required line is 1 2 2 7 5 x y z      W.E-12. Consider the lines 1 1 3 : 2 1 1 x y z L      , 2 4 3 3 : 1 1 2 x y z L      and the planes 1P x y z : 7 2 3    , 2 P x y z :3 5 6 4    . Let ax by cz d    be the equation of the plane passing through the point of intersection of lines L1 and L2 and perpendicular to planes P1 & P2 match ListI with ListII and select the correct answer using the code given below the lists. (IIT JEE 2013) List-I List-II P) a= 1) 13 Q) b= 2) -3 R) c= 3) 1 S) d= 4) -2 codes: P Q R S
THREE DIMENSIONAL GEOMETRY JEE ADVANCED - VOL - IV Narayana Junior Colleges 276 Narayana Junior Colleges Page No : 276 A 3 2 4 1 B 1 3 4 2 C 3 2 1 4 D 2 4 1 3 Sol: (A) Plane perpendicular to P1 and P2 has D.r’s of normal 7 1 2 16 48 32 .....(1) 3 5 6 i j k     i j k  For point of intersection of lines 2 1, , 3 4, 3,2 3       1 1 2 2 2                 2 1 4 2 3     1 2 1 2 or           1 2 1 2 3 3 or 1 2      2, 1  Point is (5,-2,-1) ..............(2) From (1) and (2) required plane is        1 5 3 2 2 1 0  x y z      or x y z    3 2 13        a b c d 1, 3, 2, 13 W.E-13. Let A 4,0,3 ,B 14,2, 5         then which of the following points lie on the bisector of the angle between OA and OB( Where ‘O’ is the orgin of reference) is(are) A) (2, 1, -1) B) (2, 11, 5) C) (10, 2, -2) D) (1, 1, 2) Sol : (D) OA = 4 i 3k     OB 14 i 2 j 5k       4 i 3k 14 i 2 j 5k OA ,OB , 5 15              Any vector in the direction of angle bisectors of the from λ a + b         λ = 2i +2 j 4k 15            Answer is D with 15 2   Passage: Two lines whose equations are 3 2 1 2 3 x y z       and 2 3 2 3 2 3 x y z      lie in the same plane then W.E-14.The value of   1 sin sin   is equal to A)3 B) 3 C)4 D)  4 W.E-15. Point of intersection of the lines lie on A)3 20 x y z    B)3 25 x y z    C)3 2 24 x y z    D)3 2 10 x y z    Sol. 14 (d) Both lines are coplanar 2 3 3 2 3 0 1 1 1               2 2 3 3 3 3 3 2 0           4     1 1 sin sin 4 sin sin 4 4         15 (b) Let 1 3 2 1 2 3 4 x y z r       1 1 1        x r y r z r 3 2 , 3 2, 4 1 it will lie on 1 2 3 2 1 3 2 3 x y z r        So, point of intersection is (5,5,5)