3D-COORDINATE GEOMETRY

Transcription

1 J-Mathematics. INTRODUCTION : P OI N T In earlier classes we have learnt about points, lines, circles and conic section in two dimensional geometry. In two dimensions a point represented by an ordered pair (x, y) (where x & y are both real numbers) In space, each body has length, breadth and height i.e. each body exist in three dimensional space. Therefore three independent quantities are essential to represent any point in space. Three axes are required to represent these three quantities.. RCTANGUL AR CO- ORDINAT SYSTM : In cartesian system of the three lines are mutually perpendicular, such a system is called rectangular cartesian co-ordinate system. Co-ordinate axes and co-ordinate planes : When three mutually perpendicular planes intersect at a point, then mutually perpendicular lines are obtained and these lines also pass through that point. If we assume the point of intersection as origin, then the three planes are known as co-ordinate planes and the three lines are known as co-ordinate axes. Octants : very plane bisects the space. Hence three co-ordinate plane divide the space in eight parts. These parts are known as octants.. COORDINATS OF A POINT IN SPAC : Let O be a fixed point, known as origin and let OX, OY and OZ be three mutually perpendicular lines, taken as x-axis, y-axis and z-axis respectively, in such a way that they form a right handed system. The plane s XOY, YOZ and ZOX are know n as xy-plane, yz-plane and zx-plane respectively. Let P be a point in space and distances of P from yz, zx and xy planes be x, y, z respectively (with proper signs) then we say that coordinates of P are (x, y, z). Also OA = x, OB = y, OC = z 4. DISTANC FORMUL A : The distance between two points A (x, y, z ) and B (x, y, z ) is given by AB [(x x ) (y y ) (z z ) ] ( a ) Dista nce from Origi n : Let O be the origin and P (x, y, z) be any point, then OP (x y z ) ( b ) Dista nce of a poi nt from coordi nate a xe s : Illustration : D-COORDINAT GOMTRY 54 X y' y' F x A i Z C k O z' x' j z O z' P(x, y, z) Let P(x, y, z) be any point in the space. Let PA, PB and PC be the perpendiculars drawn from P to the axes OX, OY and OZ respectively. Then PA (y z ) ; PB (z x ) ; PC (x y ) Prove by using distance formula that the points P (,, ), Q (,, ) and R (, 5, 7) are collinear. D B x' y Y NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65

2 J-Mathematics Solution : We have PQ = ( ) ( ) ( ) = Q R = ( ) (5 ) (7 ) = and PR = ( ) (5 ) (7 ) Since QR = PQ + PR. Therefore the given points are collinear. A n s. Illustration : Find the locus of a point the sum of whose distances from (, 0, 0) and (, 0, 0) is equal to 0. Solution : Let the points A(,0,0), B (,0,0) and P(x,y,z) Given : PA + PB =0 (x ) (y 0) (z 0) + (x ) (y 0) (z 0) = 0 (x ) y z = 0 (x ) y z Squaring both sides, we get ; (x ) + y + z = 00 + (x + ) + y + z 0 (x ) y z 4x 00 = 0 (x ) y z x + 5 = 5 (x ) y z Again squaring both sides we get x + 50x + 65 = 5 {(x +x +) + y +z } 4x + 5y + 5z 600 = 0 i.e. required equation of locus A n s. 5. SCTION FORMULA : Let P(x, y, z ) and Q(x, y, z ) be two points and let R (x, y, z) divide PQ in the ratio m : m. Then co-ordinates of R(x, y, z) = mx m x my m y mz m z,, m m m m m m If (m /m ) is positive, R divides PQ internally and if (m /m ) is negative, then externally. Mid-Point : Mid point of PQ is given by x x y y z z,, NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65 Illustration : Find the ratio in which the plane x y + z = 7 divides the line joining the points (, 4, 7) and (, 5, 8). Solution : Let the required ratio be k : The co-ordinates of the point which divides the join of (, 4, 7) and (, 5, 8) in the ratio k : are k 5k 4 8k 7,, k k k Since this point lies on the plane x y + z 7 = 0 k 5k 4 8k k k k (k ) ( 5k + 4) + (8k + 7) = 7 k + 7 k + 0k + 4k 7k = k 7k = 6 0k = 6 ; k = 0 0 Hence the required ratio = k : = 0 : = : 0 A n s. 55

3 J-Mathematics Do yourself : ( i ) Find the distance between the points P(, 4, 5) and Q(,, ). ( i i ) Show that the points A(0, 7, 0), B(, 6, 6) and C( 4, 9, 6) are vertices of an isosceles right angled triangle. (iii) Find the locus of a point such that the difference of the square of its distance from the points A(, 4, 5) and B(,, 7) is equal to k. ( i v ) Find the co-ordinates of points which trisects the line joining the points A(,, 4) and B(0, 4, 7) ( v ) Find the ratio in which the planes (a) xy (b) yz divide the line joining the points P(, 4, 7) and Q(, 5, 8). 6. CNTROID OF A TRIANGL : Let A(x, y, z ), B(x, y, z ), C(x, y, z ) be the vertices of a triangle ABC. Then its centroid G is given by x x x y y y z z z G,, Illustration Solution : 4 : If the centroid of a tetrahedron OABC where A, B, C, are given by (a,, ), (, b, ) and (,, c) respectively be (,, ), then distance of P (a, b, c) from origin is - (A) 07 (B) 4 (C) 07 4 (D) none of these Centroid is F I HG K J a x, 4 y, 4 z (,, ) OP a b c b 0 c 0 a =, b = 5, c = Ans. (A) 7. DIRCTION COSINS OF LIN : If be the angles made by a line with x-axis, y-axis & z-axis respectively then cos, cos & cos are called direction cosines of a line, denoted by, m & n respectively. Note : (i) If line makes angles with x, y & z axis respectively then is another set of angle that line makes with principle axes. Hence if, m & n are direction cosines of line then, m & n are also direction cosines of the (ii) same line. Since parallel lines have same direction. So, in case of lines, which do not pass through the origin. We can draw a parallel line passing through the origin and direction cosines of that line can be found. Important points : ( i ) Direction cosines of a line : Take a vector A ai ˆ bj ˆ ckˆ parallel to a line whose D.C s are to be found out. A.i ˆ a A cos a a cos similarly, A b cos ; A cos cos cos m n c cos A 56 x ai + bj + ck z y NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65

4 ( i i) Direction cosine of axes : Since the positive x-axes makes angle 0, 90, 90 with axes of x, y and z respectively, D.C. s of x axes are, 0, DIRCTION R ATIOS : D.C. s of y-axis are 0,, 0 D.C. s of z-axis are 0, 0, J-Mathematics Any three numbers a, b, c proportional to direction cosines, m, n are called direction ratios of the line. i.e. m n a b c There can be infinitely many sets of direction ratios for a given line. Direction ratios and Direction cosines of the line joining two points : Let A(x, y, z ) and B(x, y, z ) be two points, then d.r. s of AB are x x, y y, z z and the d.c. s of AB are r (x x ), r (y y ), r (z z ) where r [ (x x ) ] 9. RL ATION BT WN D.C S & D.R S : m n a b c m n m n a b c a b c a ; m = a b c b a b c ; n = c a b c Important point : Direction cosines of a line are unique but Dr's of a line in no way unique but can be infinite. NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65 7. PROJCTIONS : ( a ) Project ion of li ne segment OP on co- ordi nate a xe s : Let line segment make angle with x-axis Thus, the projections of line segment OP on axes are the absolute values of the co-ordinates of P. i.e. Projection of OP on x-axis = x Projection of OP on y-axis = y Projection of OP on z-axis = z Now, in OAP, angle A is a right angle and OA = x OP = x y z cos x x y z x OP if OP = r, then x = OP cos = r Similarly y = OP cos = mr, z = nr, where, m, n are DC s of line ( b ) Project ion of a li ne segment A B on coordi nate a xe s : x A z O P(x,y,z) Projection of the point A(x, y, z ) on x-axis is (x, 0, 0). Projection of point B(x, y, z ) on x-axis is F(x, 0, 0). Hence projection of AB on x-axis is F = x x. y 57

5 J-Mathematics Similarly, projection of AB on y and z-axis are y y, z z respectively. Note : Projection is only a length therefore it is always taken as positive. ( c ) Projection of line segment AB on a line having direction cosines, m, n : Let A(x, y, z ) and B(x, y, z ). z Now projection of AB on F = CD = AB cos A(x,y,z ) B(x,y,z ) = (x x ) (y y ) (z z ) (x x ) (y y )m (z z )n (x x ) (y y ) (z z ) x C D F y (x x ) (y y )m (z z )n = Illustration 5 : A line OP makes with the x-axis an angle of measure 0 and with y-axis an angle of measure 60. Find the angle made by the line with the z-axis. Solution : = 0 and = 60 cos = cos 0 = and cos = cos 60 = but cos + cos + cos = + cos = cos = cos = ± 4 4 = 45 or 5 A n s. Illustration 6 : Find the projection of the line segment joining the points (, 0, ) and (, 5, ) on the line whose direction ratios are 6,,. Solution : Do yourself - : The direction cosines, m, n of the line are given by 6, m,n The required projection is given by = (x x ) + m(y y ) + n(z z ) = 6 [ ( )] (5 0) ( ) = 6 5 = = m n m n A n s. 7 7 ( i ) Find the projections of the line segment joining the origin O to the point P(,, 5) on the axes. ( i i ) Find the projections of the line joining the points P(,, 5) and Q(0,, 8) on the axes. (iii) Find the direction ratios & direction cosines of the line joining the points O(0, 0, 0) and P(,, 4).. ANGL BTWN TWO LINS : Let be the angle between the lines with d.c. s, m, n and, m, n then cos = + m m + n n. If a, b, c and a, b, c be D.R. s of two lines then angle between them is given by cos (a a b b c c ) (a b c ) (a b c ) 49 NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65

6 Illustration 7 : If a line makes angles,, with four diagonals of a cube, then cos + cos + cos + cos equals - (A) (B) 4 (C) 4/ (D) /4 J-Mathematics Solution : Let OA, OB, OC be coterminous edges of a cube and OA = OB = OC = a, then co-ordinates of its vertices are O(0, 0, 0), A(a, 0, 0), B(0, a, 0), C(0, 0, a), L(0, a, a), M(a, 0, a), N(a, a, 0) and P (a, a, a) Direction ratio of diagonal AL, BM, CN and OP are Z C L,,,,,,,,,,, M P Let, m, n be the direction cosines of the given line, then O Y B A N cos = m n m n X Similarly cos = m n m n m n, cos and cos cos + cos + cos + cos = 4 Ans. (C) Illustration 8 : (a) Find the acute angle between two lines whose direction ratios are,, 6 and,, respectively. (b) Find the measure of the angle between the lines whose direction ratios are,, 7 and,,. Solution : (a) a =, b =, c = 6; a =, b =, c =. If be the angle between two lines whose d.r s are given, then cos a a b b c c a b c a b c 6 = 6 = NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65 0 cos (b) ( ) 7 54 ( ) ( ) 4 The actual direction cosines of the lines are 7,, and,, If is the angle between the lines, then 7 cos = = 0 90 A n s. 59

7 J-Mathematics. PRPNDICUL AR AND PAR ALLL LINS : Let the two lines have their d.c. s given by, m, n and, m, n respectively then they are perpendicular if = 90 i.e. cos = 0, i.e. + m m + n n = 0. Also the two lines are parallel if = 0 i.e. sin = 0, i.e. m m n n No te : If instead of d.c. s, d.r. s a, b, c and a, b, c are given, then the lines are perpendicular if a a + b b + c c = 0 and parallel if a b c a b c. Illustration 9 : If the lines whose direction cosines are given by a + bm + cn = 0 and fmn + gn + hm = 0 are perpendicular, then f g h equals - a b c (A) 0 (B) (C) (D) none of these a bm Solution : liminating n between the given relations, we find that (fm + g) + hm = 0 c or ag (af bg ch) bf 0 m m...(i) Let and m m, are roots of (i), then bf m m ag f / a m m g / b...(ii) Similarly m m g / b n n h / c...(iii) From (ii) and (iii), we get Do yourself - : mm nn f / a g / b h / c =.f/a ; m m =.g/b ; n n =.h/c + m m + n n = f g h a b c f g h a b c = 0 { + m m + n n = 0} Ans. (A) ( i ) Find the angle between the lines whose direction ratios are,, and 4,,. ( i i ) If a line makes and angle with axes, then prove that sin + sin + sin =. (iii) Find the direction cosines of the line which is perpendicular to the lines with direction cosines proportional to (,, ) & (0,, ). 60 NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65

8 J-Mathematics PLAN. DFINITION : A geometrical locus is a plane, such that if P and Q are any two points on the locus, then every point on the line PQ is also a point on the locus. 4. QUATIONS OF A PL AN : The equation of every plane is of the first degree i.e. of the form ax + by + cz + d = 0, in which a, b, c are constants, not all zero simultaneously. ( a ) quat ion of pla ne passi ng t hrough a fixed poi nt : Vector form : If a be the position vector of a point on the plane and n be a vector normal to the plane then it s vectorial equation is given by r a.n 0 r. n = d, where d a.n constant. Car te sia n form : If a(x, y, z ) and n ai ˆ bj ˆ ckˆ, then car tesian equat ion of plane w i ll be a(x x ) + b(y y ) + c(z z ) = 0 ( b ) Plane Parallel to the Coordinate Planes : (i) quation of yz plane is x = 0. (ii) quation of zx plane is y = 0. (iii) quation of xy plane is z = 0. (iv) quation of the plane parallel to xy plane at a distance c is z = c or z = c. (v) quation of the plane parallel to yz plane at a distance c is x = c or x = c (vi) quation of the plane parallel to zx plane at a distance c is y = c or y = c. ( c ) quations of Planes Parallel to the Axes : If a = 0, the plane is parallel to x-axis i.e. equation of the plane parallel to x-axis is by + cz + d = 0. Similarly, equations of planes parallel to y-axis and parallel to z-axis are ax + cz + d = 0 and ax + by + d = 0, respectively. NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65 ( d ) quat ion of a Pla ne i n Intercept Form : quation of the plane which cuts off intercepts a, b, c from the axes x, y, z respectively is x + y + z =. a b c ( e ) quat ion of a Pla ne i n Normal Form : O Vector form : If ˆn is a unit vector normal to the plane from the origin and d be the perpendicular distance of plane from origin then its vector equation is r. n ˆ = d. Cartesian form : If the length of the perpendicular distance of the plane from the origin is p and direction cosines of this perpendicular are (, m, n), then the equation of the plane is x + my + nz = p. ( f ) quation of a Plane through three points : Vector form : If A, B, C are three points having P.V.'s a, b, c respectively, then vector equation of the plane is [r a b ] [r b c] [r c a ] [a b c]. r n 6

9 J-Mathematics Cartesian form :The equation of the plane through three non-collinear points (x, y, z ), (x, y, z ) and (x, y, z ) is x x y y z z x x y y z z = 0 x x y y z z Illustration 0 : Find the equation of the plane through the points A(,, ), B(, 4, ) and C(7, 0, 6). Solution : The general equation of a plane passing through (,, ) is a (x ) + b (y ) + c (z + ) = 0...(i) It will pass through B (, 4, ) and C (7, 0, 6) if a ( ) + b (4 ) + c ( + ) = 0 or a + b + c = 0...(ii) and a (7 ) + b (0 ) + c (6 + ) = 0 or 5a b + 7c = 0...(iii) Solving (ii) and (iii) by cross-multiplication, we have a b c a b c or (say) Illustration a = 5, b = and c = Substituting the values of a, b and c in (i), we get 5 (x ) + (y ) (z + ) = 0 or 5(x ) + (y ) (z + ) = 0 5x + y z = 7, which is the required equation of the plane A n s. : A plane meets the co-ordinates axis in A,B,C such that the centroid of the ABC is the point (p,q,r) show that the equation of the plane is Solution : Let the required equation of plane be : x y z =...(i) a b c 6 x y z = p q r Then, the co-ordinates of A, B and C are A(a, 0, 0), B(0, b, 0), C(0, 0, c) respectively So the centroid of the triangle ABC is a b c,, But the co-ordinate of the centroid are (p,q,r) a = p, b = q, c = r Putting the values of a, b and c in (i), we get the required plane as Do yourself - 4 : x y z = p q r x y z = A n s. p q r ( i ) quation of a plane is x + 4y + 5z = 7. (a) (b) (c) (d) Find the direction cosines of its normal Find the points where it intersects the axes. Find its intercept form. Find its equation in normal form (in cartesian as well as in vector form) ( i i ) Find the equation of the plane passing through the points (,, ), (, 0, ) and (,, ). NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65

10 J-Mathematics 5. ANGL BTWN TWO PLANS : Vector form : If r.n d and r.n d be two planes, then angle between these planes is the angle between their normals n. n cos n n Planes are perpendicular if n.n 0 and they are parallel if n n Cartesian form : Consider two planes ax + by + cz + d = 0 and a' x + b' y + c' z + d' = 0. Angle between these planes is the angle between their normals. cos aa ' b b ' cc ' a b c a ' b ' c ' Planes are perpendicular if aa' + bb' + cc' = 0 and they are parallel if a b c. a ' b ' c ' Planes parallel to a given Plane : quation of a plane parallel to the plane ax + by + cz + d = 0 is ax + by + cz + d' = 0. d' is to be found by other given condition. Illustration : Find the angle between the planes x + y + z = 9 and x y + z = 5 Solution : We know that the angle between the planes a x + b y + c z + d = 0 and a x + b y + c z + d = 0 is given by cos = a a b b c c a b c a b c Therefore, angle between x + y + z = 9 and x y + z = 5 is given by cos = ()() ()( ) ()() ( ) = A n s. Illustration : Find the equation of the plane through the point (, 4, ) and parallel to the plane x + y z = 7. NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65 Solution : Let the equation of a plane parallel to the plane x + y z = 7 be x + y z + k = 0 This passes through (, 4, ), therefore ( ) () + 4 ( ) + k = k = 0 k = 8 Putting k = 8 in (i), we obtain x + y z 8 = 0 or x + y z = 8 A n s. This is the equation of the required plane. Do yourself - 5 : (i) Prove that the planes x y + z + 7 = 0 and 4x + y 6z 5 = 0 are perpendicular. (ii) Find the angle between the planes x + 4y + z + 7 = 0 and x + y z = 5 6. A PLAN THROUGH TH LIN OF INTRSCTION OF TWO GIVN PLANS : Consider two planes u ax + by + cz + d = 0 and v a' x + b' y + c' z + d' = 0. The equation u + v = 0, a real parameter, represents the plane passing through the line of intersection of given planes and if planes are parallel, this represents a plane parallel to them. 6

11 J-Mathematics Illustration 4 : Find the equation of plane containing the line of intersection of the plane x + y + z 6 = 0 and x + y + 4z + 5 = 0 and passing through (,,). Solution : The equation of the plane through the line of intersection of the given planes is, (x + y + z 6) + (x + y + 4z + 5) = 0...(i) If it is passes through (,,) ( + + 6) + ( ) =0 4 Putting =/4 in (i); we get (x + y + z 6) + (x + y + 4z + 5) = 0 4 0x + y + 6z 69 = 0 A n s. 7. PRPNDICUL AR DISTANC OF A POINT FROM TH PL AN : Vector form : If r.n d be the plane, then perpendicular distance p, of the point A( a ) a.n d p n Distance between two parallel planes r.n d r.n d & d d is. n Cartesian form : Perpendicular distance p, of the point A(x, y, z ) from the plane ax + by + cz + d = 0 is given by ax by cz d p (a b c ) Distance between two parallel planes ax + by + cz + d = 0 & ax + by + cz + d = 0 is d d a b c Illustration 5 : Find the perpendicular distance of the point (,, 0) from the plane x + y +z + 5 = 0 Solution : We know that the perpendicular distance of the point (x, y, z ) from the plane ax + by + cz + d = 0 is so required distance = ax by cz d a b c Illustration 6 : Find the distance between the parallel planes x y + z + = 0 and 4x y + 4z + 5 = 0. Solution : Let P(x, y, z ) be any point on x y + z + = 0, then x y + z + = 0 Do yourself - 6 : The length of the perpendicular from P(x, y, z ) to 4x y + 4z + 5 = 0 is 4 x y 4z 5 (x y z ) 5 ( ) 5 4 ( ) Therefore, the distance between the two given parallel planes is 6 64 [using (i)] ( i ) Find the perpendicular distance of the point P(,, ) from the plane x + y + z + = 0. ( i i ) Find the equation of the plane passing through the line of intersection of the planes x + y + z = 5 and x + y + z + 5 = 0 and passing through the point (0, 0, 0). A n s. A n s. NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65

12 8. BISCTORS OF ANGLS BTWN TWO PLANS : Let the equations of the two planes be ax + by + cz + d = 0 and a x + b y + c z + d = 0. Then equations of bisectors of angles between them are given by ax by cz d a x b y c z d (a b c ) (a b c ) J-Mathematics ( a ) quat ion of bisector of the angle contai ni ng origi n : First make both constant terms positive. Then positive sign give the bisector of the angle which contains the origin. ( b ) Bisector of acute/obtuse angle : First making both constant terms positive, aa + bb + cc > 0 origin lies in obtuse angle aa + bb + cc < 0 origin lies in acute angle Illustration 7 : Find the equation of the bisector planes of the angles between the planes x y + z + = 0 and x y + 6z + 8 = 0 and specify the plane which bisects the acute angle and the plane which bisects the obtuse angle. Solution : The two given planes are x y + z + = 0 and x y + 6z + 8 = 0 where d, d > 0 and a a + b b + c c = > 0 a x b y c z d a b c a x b y c z d = a b c (acute angle bisector) NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65 and ax by cz d a x by cz d = a b c a b c (obtuse angle bisector) i.e., x y z x y 6z 8 = ± (4x 7y + 4z + ) = ± (9x 6y + 8z + 4) Taking positive sign on the right hand side, we get 5x y 4z = 0 (obtuse angle bisector) and taking negative sign on the right hand side, we get x y + z + 45 = 0 (acute angle bisector) A n s. 9. POSITION OF TWO POINTS W.R.T. A PL AN : Two points P(x, y, z ) & Q(x, y, z ) are on the same or opposite sides of a plane ax + by + cz + d = 0 according to ax + by + cz + d & ax + by + cz + d are of same or opposite signs. The plane divides the line joining the points P & Q externally or internally according to P and Q lying on same or opposite sides of the plane. Do yourself - 7 : ( i ) Find the position of the poi nt P(,, ), Q(, 0, ) and R(,, 8) w.r.t. the plane x y + 4z 7 = 0. ( i i ) Two given planes are x + y z + 5 = 0 and 6x y + z 7 = 0. Find (a) equation of plane bisecting the angle between the planes. (b) equation of a plane parallel to the plane bisecting the angle between both the two planes and passing through the point (,, 0). (c) specify which plane is acute angle bisector and which one is obtuse angle bisector. 65

13 J-Mathematics STRAIGHT LIN 0. DFINITION : A straight line in space is characterised by the intersection of two planes which are not parallel and, therefore, the equation of a straight line is present as a solution of the system constituted by the equations of the two planes : a x + b y + c z + d = 0; a x + b y + c z + d = 0 This form is also known as unsymmetrical form. Some particular straight lines : Straight lines q u a t i o n (i) Through the origin y = mx, z = nx (ii) x-axis y = 0, z = 0 or (iii) y-axis x = 0, z = 0 or (iv) z-axis x = 0, y = 0 or (v) parallel to x-axis y = p, z = q (vi) parallel to y-axis x = h, z = q (vii) parallel to z-axis x = h, y = p x y z 0 0 x y z 0 0 x y z 0 0. QUATION OF A STR AIGHT LIN IN SYMMTRICAL FORM : ( a ) One point form : Let A(x, y, z ) be a given point on the straight line and, m, n be the d.c s of the line, then its equation is x x y y z z r m n (say) It should be noted that P(x + r, y + mr, z + nr) is a general point on this line at a distance r from the point A(x, y, z ) i.e. AP = r. One should note that for AP = r;, m, n must be d.c. s not d.r. s. If a, b, c are direction ratios of the line, then equation of the line is x x y y z z r but here AP r a b c ( b ) quation of the line through two points A(x, y, z ) and B(x, y, z ) is x x y y z z x x y y z z Illustration 8 : Find the co-ordinates of those points on the line x y z which is at a distance of 6 units from point (,, ). Solution : Here, x y z...(i) 6 is the given straight line Let, P = (,,) on the straight line Here direction ratios of line (i) are (,,6) Direction cosines of line (i) are : 6,, Units P(,,) Q(r+,r,6r+) NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65

14 J-Mathematics quations of line(i) any may be written as x y z...(ii) / 7 / 7 6 / 7 Co-ordinates of any point on the line (ii) may be taken as 6 r, r, r Let, Q r, r, r Given r =, r = ± 5 9 Putting the value of r, we have Q,, or Q =,, A n s.. ANGL BTWN A LIN AND A PLAN : Let equations of the line and plane be x x y y z z m n the angle which line makes with the plane. Then ( ) is the angle between the line and the normal to the plane. and ax + by + cz + d = 0 respectively and be n line So, sin a bm cn (a b c ) ( m n ) Line is parallel to plane if = 0 i.e. if a + bm + cn = 0. Line is perpendicular to the plane if line is parallel to the normal of the plane i.e. if a b c. m n NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65 Illustration 9 : Find the angle between the line Solution : The given line is x y z x y z and the given plane is x + 4y + z + 5 = 0... (i)... (ii) and the plane x + 4y + z + 5 = 0. If the line (i) makes angle with the plane (ii), then the line (i) will make angle (90 ) with the normal to the plane (i). Now direction-ratios of line (i) are <,, > and direction-ratios of normal to plane (ii) are <, 4, > Hence Do yourself - 8 : ()() ( )(4) ( )() cos(90 ) sin sin ( i ) Find the equation of the line passing through the point (4,, ) and having direction ratios,, ( i i ) Find the symmetrical form of the line x y + z = 5, x + y + z = 6. (iii) Find the angle between the plane x + 4y + 5 = 0 and the line x y z. 0 ( i v ) Prove that the line x y 4 z 5 is parallel to the plane 4x + 4y 5z + = 0. 4 A n s. 67

15 J-Mathematics. CONDITION IN ORDR THAT TH LIN MAY LI ON TH GIVN PLAN : x x y y z z The line m n will lie on the plane Ax + By + Cz + D = 0 if ( a ) A + Bm + Cn = 0 and ( b ) Ax + By + Cz + D = 0 4. IMAG OF A POINT IN TH PLAN : P In order to find the image of a point P(x, y, z ) in a plane ax + by + cz + d = 0, assume it as a mirror. Let Q(x, y, z ) be the image of the point P(x, y, z ) in the plane, then x x y y z z ( a ) Line PQ is perpendicular to the plane. Hence equation of PQ is a b c x x y y z z ( b ) Hence, Q satisfies the equation of line then r. The plane passes through a b c the middle point of line PQ and the middle point satisfies the equation of the plane i.e. x x y y z z a b c d 0. The co-ordinates of Q can be obtained by solving these equations. 5. FOOT, LNGTH AND QUATION OF PRPNDICUL AR FROM A POINT TO A LIN : Let equation of the line be x x y y z z m n r (say) and A ( ) be the point. Any point on the line (i) is P(r + x, mr + y, nr + z ) If it is the foot of the perpendicular, from A on the line, then AP is to the line, so (r + x ) + m (mr + y ) + n (nr + z ) = 0 i.e. r = ( x ) + ( y ) m + ( z ) n since + m + n = Putting this value of r in (ii), we get the foot of perpendicular from point A to the line. Length : Since foot of perpendicular P is known, length of perpendicular, AP [( r x ) (mr y ) (nr z ) ] quation of perpendicular is given by x y z r x mr y nr z (i)... (ii) r R P Q A( ) Illustration 0 : Find the co-ordinates of the foot of the perpendicular from (,, ) on the line joining (5, 4, 4) and (, 4, 6). Solution : Let A (,, ), B (5, 4, 4) and C (, 4, 6) be the given points. Let M be the foot of the perpendicular from A on BC. If M divides BC in the ratio :, then A(,,) co-ordinates of M are,, Direction ratios of BC are 5, 4 4, 6 4 i.e. 4, 0, 90 D.R.'s of AM are ,, B M C (5,4,4) (,4,6) 4 5,, 4, +, 5 + Since AM BC (4) + 0( + ) (5 + ) = = 0 = Hence the co-ordinates of M are (, 4, 5) A n s. L NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65

16 J-Mathematics Illustration : Find the length of perpendicular from P(,,) to the line x y z Solution : Given line is and P(,,) x y z...(i) Co-ordinates of any point on (i) may be taken as (r,r+, r ) Let Q = (r, r +, r ) Direction ratio's of PQ are : (r, r +6, r ) Direction ratio's of AB are : (,, ) Since, PQ AB (r ) + (r + 6) ( r ) = 0 r = 5 4 Q =,, PQ = A Q P(,,) 90 B PQ = 5 4 units A n s. Do yourself - 9 : ( i ) Find the image of point P(,, ) in the plane x y + z + = 0 as well as the foot of the perpendicular drawn from the point (,, ). ( i i ) Find the distance of the point (,, ) from the plane x y + z = 5 measured parallel to the line x y z 6 NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65 (iii) Prove that x y z 5 lies in the plane x + y z = QUATION OF PL AN CONTAINING TWO INTRSCTING LINS : Let the two lines be and x y z m n x y z m n These lines will coplanar if the plane containing the two lines is m n 0 m n x y z... (i)... (ii) m n 0 m n (It is condition for intersection of two lines) 69

17 J-Mathematics Illustration : Find the equation of the plane containing the line x 4 y z. 5 x y 6 z 4 and parallel to the line Solution : Any plane containing the line 4 is a(x ) + b(y + 6) + c(z + ) = 0... (i) where, a + 4b + c = 0... (ii) Also, it is parallel to the second line and hence, its normal is perpendicular to this line a b + 5c = 0... (iii) a b c Solving (ii) & (iii) by cross multiplication, we get k 6 7 a = 6k, b = k & c = 7k Putting these values in (i), we get 6k(x ) k(y + 6) 7k(z + ) = 0 6x y 7z = 09, which is the required equation of the plane. 7. LIN OF GRATST SLOP : Consider two planes G-plane and H-plane. H-plane is treated as a horizontal plane or reference plane. G-plane is a given plane. Let AB be the line of intersection of G-plane & H-plane. Line of greatest slope is a line which is contained by G-plane & perpendicular to line of intersection of G-plane & H-plane. Obviously, infinitely many such lines of greatest slopes are contained by G-plane. Generally an additional information is given in problem so that a unique line of greatest slope can be found out. A Q B G-plane P H-plane Illustration : Assuming the plane 4x y +7z =0 to be horizontal, find the equation of the line of greatest slope through the point (,,) in the plane x +y 5z = 0. Solution : The required line passing through the point P (,,) in the plane x + y 5z = 0 and is having greatest slope, so it must be perpendicular to the line of intersection of the planes x + y 5z = 0...(i) and 4x y + 7z = 0...(ii) Let the d.r'.s of the line of intersection of (i) and (ii) be a, b, c a + b 5c = 0 and 4a b + 7c = 0 {as dr'.s of straight line (a, b,c) is perpendicular to d.r'.s of normal to both the planes} a b c Now let the direction ratio of required line be proportional to, m, n then its equation be x y z m n where + m 5n = 0 and 4 +7m + 5n = 0 m n so, Thus the required line is x y z 70 A n s. NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65

18 8. ARA OF TRIANGL : To find the area of a triangle in terms of its projections on the co-ordinates planes. J-Mathematics Let x, y, z be the projections of the plane area of the triangle on the planes yoz, zox, xoy respectively. Let, m, n be the direction cosines of the normal to the plane of the triangle. Then the angle between the plane of the triangle and yoz plane is the angle between the normal to the plane of the triangle and the x-axis. x = Similarly y = m ; z = n x y z If A(x, y, z ), B(x, y, z ), C(x, y, z ) be the three vertices of the triangle then y z y z y z x, x z x z x z y, x y x y x y z Do yourself - 0 : (i) Prove that the lines x y z 5 and x y 4 z 6 are coplanar. Find their point of intersection. ( i i ) Find the area of the triangle whose vertices are the points (,, ), (,, 4), (, 4, ). Miscellaneous Illustrations : Illustration 4 : If a variable plane cuts the coordinate axes in A, B and C and is at a constant distance p from the origin, find the locus of the centroid of the tetrahedron OABC. Solution : Let A (a, 0, 0), B (0, b, 0) and C (0, 0, c) NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65 x y quation of plane ABC is z a b c = Now p = length of perpendicular from O to plane (i) = a b c or p = a b c Let G() be the centroid of the tetrahedron OABC, then = a 4, = b 4, = c 4 or, a = 4, b = 4, c = 4 a a 4 4 Putting these values of a, b, c in equation (ii), we get 6 p = 6 p or x z A(a, 0, 0) C(0, 0, c) O B(0, b, 0) y locus of () is x + y + z = 6 p A n s. 7

19 J-Mathematics Illustration 5 : Through a point P(h, k, ) a plane is drawn at right angles to OP to meet the coordinate axes in A, B and C. If OP = p, show that the area of ABC is 5 p hk. Solution : OP = h k = p Direction cosines of OP are h h k, h k k, h k Since OP is normal to the plane, therefore, equation of the plane will be, h h k x + h k k y + h k z = h k or, hx + ky + z = h + k + = p A p, 0, 0, B p h 0,, 0 k, C p 0, 0, Now area of ABC, = A xy + A yz + A zx Now A xy = area of projection of ABC on xy-plane = area of AOB = Mod of p h 0 0 p k 0 0 = 4 p hk Similarly, A yz = 4 p k 4 p and A zx = h = or = p p p p 4 h k 4 k 4 h 4h k 5 p hk Illustration 6 : Find the locus of a point, the sum of squares of whose distances from the planes : x z = 0, x y + z = 0 and x + y + z = 0 is 6 Solution : Given planes are x z = 0, x y + z = 0 and, x + y + z =0 Let the point whose locus is required be P(,,). According to question = 6 6 or ( + ) ( ) =6 6 or = 6 6 or + + = 6 7 A ns. Hence, the required equation of locus is x + y + z = 6 A n s. NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65

20 J-Mathematics Illustration 7 : Direction ratios of normal to the plane which passes through the point (, 0, 0) and (0,, 0) which makes angle /4 with x + y = are - Solution : (A),, (B),, (C),, (D),, The plane by intercept form is x y z c d.r. s of normal are,, c and of given plane are,, 0. cos c 0 c c c 4 c d.r. s are,, Ans. (D) NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65 : (i) (iii) 8x + y + 4z ± k + 9 = 0 (iv) ( v ) ( a ) 7 : 8, externally ( b ) : internally : ( i ),, 5 ( i i ), 4, (iii),, 4 & : ( i ) 4 : (i) (a) (d) (iii) 4,, 5 5 ( i i ) x + y + z = 7 5 : ( i i ) cos 56 6 : ( i ) 8 6,, (b) x 4y z 7 & (ii) x + 4y + z = 0 7 : ( i ) P, Q same side & R opposite side 8 : ( i ) 9 : ( i ) 0 : ( i ) 7, 0, 0, 7 0,, 0 4 & 7 0, 0, 5 8,, 5 4,, ˆ 4 ˆ ˆ 7 r. i j k ( i i ) ( a ) 4x + y 5z + 4 = 0 & x y + z 56 = 0 ( b ) 4x + y 5z 4 = 0 & x y + z 70 = 0 ( c ) & 0,, 6 x y z 7 / 7 / 4 7 / 5 ( c ) 4x + y 5z + 4 = 0 (acute angle bisector) & x y + z 56 = 0 (obtuse angle bisector) x 4 y z ( i i ) 5,, & 4,,,, ANSWRS FOR DO YOURSLF x / 4 y 9 / 4 z ( i i ) ( i i ) 8 6 (iii) sin 5 5 7

21 J-Mathematics XRCIS - 0 CHCK YOUR GRASP SLCT TH CORRCT ALTRNATIV (ONLY ON CORRCT ANSWR). The plane XOZ divides the join of (,, 5) and (,, 4) in the ratio :, then is - (A) (B) / (C) (D) /. Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the area of triangles ABC, ACD and ADB be, 4 and 5 sq. units respectively. Then the area of the triangle BCD, is - (A) 5 (B) 5 (C). Which one of the following statement is INCORRCT? (A) If n. a = 0, n. b = 0 and n. c = 0 for some non zero vector n, then [a b c] (B) there exist a vector having direction angles = 0 and = 45 (C) locus of point in space for which x = and y = 4 is a line parallel to the z-axis whose distance from the z-axis is 5 (D) In a regular tetrahedron OABC where 'O' is the origin, the vector OA + OB + OC is perpendicular to the plane ABC. 4. Consider the following 5 statements (I) 74 5 (D) 5 There exists a plane containing the points (,, ) and (,, 4) and perpendicular to the vector V = î + ĵ ˆk (II) There exist no plane containing the point (, 0, 0); (0,, 0); (0, 0, ) and (,, ) (III) If a plane with normal vector N is perpendicular to a vector V then N V = 0 (IV) (v) If two planes are perpendicular then every line in one plane is perpendicular to every line on the other plane Let P and P are two perpendicular planes. If a third plane P is perpendicular to P then it must be either parallel or perpendicular or at an angle of 45 to P. Choose the correct alternative. (A) exactly one is false (B) exactly are false (C) exactly are false (D) exactly four are false 5. Let L be the line r = î + ĵ ˆk + ( î + ˆk ) and let L be the line r = î + ĵ + µ( î + ĵ ˆk ). Let be the plane which contains the line L and is parallel to L. The distance of the plane from the origin is - (A) / 7 (B) /7 (C) 6 (D) none of these 6. The intercept made by the plane r. n = q on the x-axis is - q ˆi. n (A) (B) (C) ˆi. n q ( î.n ) q q (D) n 7. If from the point P(f, g, h) perpendiculars PL, PM be drawn to yz and zx planes then the equation to the plane OLM is - (A) x f (C) x f + y g + z h = 0 (B) x f y g + z h = 0 (D) x f + y g z h = 0 + y g + z h = 0 = 0 NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65

22 J-Mathematics 8. The line which contains all points (x, y, z) which are of the form (x, y, z) = (,, 5) + (,, ) intersects the plane x y + 4z = 6 at P and intersects the YZ plane at Q. If the distance PQ is a a, b N and a > then (a + b) equals - (A) (B) 95 (C) 7 (D) none of these b, where 9. A plane passes through the point P(4, 0, 0) and Q(0, 0, 4) and is parallel to the y-axis. The distance of the plane from the origin is - (A) (B) 4 (C) (D) 0. The distance between the parallel planes given by the equations, r. ( î ĵ + ˆk ) + = 0 and r. (4 î 4 ĵ + ˆk ) + 5 = 0 is - (A) / (B) / (C) /4 (D) /6. If the plane x y + 6z = 0 makes an angle sin (k) with x-axis, then k is equal to - (A) (B) 7 (C) (D). A variable plane forms a tetrahedron of constant volume 64K with the coordinate planes and the origin, then locus of the centroid of the tetrahedron is - (A) x + y + z = 6K (B) xyz = 6k (C) x + y + z = 4K (D) x + y + z = 4k. The expression in the vector form for the point r of intersection of the plane r n = d and the perpendicular line r = r 0 + t n where t is a parameter given by - (A) r = r d r 0 n 0 + n r 0 n (B) r n = r 0 n (C) r = r r n d 0 0 n (D) r = 0 n r r n 0 + n n n NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65 x y z 4. The equation of the plane containing the line and the point (0, 7, 7) is - (A) x + y + z = (B) x + y + z = (C) x + y + z = 0 (D) none of these SLCT TH CORRCT ALTRNATIVS (ON OR MOR THAN ON CORRCT ANSWRS) 5. Consider the plane r.n d and r.n d, then which of the follwoing are true - (A) they are perpendicular if n.n 0 n.n (B) angle between them is cos n n d d (C) normal form of the equation of plane are r.n & r.n n n (D) none of these 6. The equation of the plane which contains the lines r ˆi ˆj k ˆ (i ˆ ˆj k) ˆ and r ˆi ˆj k ˆ µ(i ˆ ˆj k) ˆ must be - (A) r.(7i ˆ 4ˆj k) ˆ 0 (B) 7(x ) 4(y ) (z + ) = 0 (C) r.(i ˆ ˆj k) ˆ 0 (D) r.(i ˆ ˆj k) ˆ 0 75

23 J-Mathematics 7. The plane containing the lines r a ta ' and r a ' sa - (A) must be parallel to a a ' (B) must be the perpendicular to a a ' (C) must be [r, a, a '] = 0 (D) (r a).(a a ') 0 8. The points A(5,, ), B(7, 4, 7), C(, 6, 0) and D(,, 4) are the vertices of a - (A) parallelogram (B) rectangle (C) rhombus (D) square 9. If P, P, P denotes the perpendicular distances of the plane x y + 4z + = 0 from the parallel planes x y + 4z + 6 = 0, 4x 6y + 8z + = 0 and x y + 4z 6 = 0 respectively, then - (A) P + 8P P = 0 (B) P = 6P (C) 8P = P (D) P + P + P = 9 CHCK YOUR GRASP ANSWR KY X R CI S - Que A n s. D A B D A A B A D D Q u e A n s. B B A C A, B A, B B, C, D A, C A,B,C,D 76 NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65

24 J-Mathematics XRCIS - 0 BRAIN TASRS SLCT TH CORRCT ALTRNATIVS (ON OR MOR THAN ON CORRCT ANSWRS). If the line r = î ĵ + ˆk + ( î + ĵ + ˆk ) makes angles,, with xy, yz and zx planes respectively then which one of the following are not possible? (A) sin + sin + sin = and cos + cos + cos = (B) tan + tan + tan = 7 and cot + cot + cot = 5/ (C) sin + sin + sin = and cos + cos + cos = (D) sec + sec + sec = 0 and cosec + cosec + cosec = 4/. A plane meets the coordinate axes in A, B, C such that the centroid of the triangle ABC is the point (, r, r ). The plane passes through the point (4, 8, 5) if r is equal to - (A) (B) (C) 5 (D) 5. Indicate the correct order statements - (A) The lines x 4 = y 6 = z 6 and x = y = z are orthogonal (B) The planes x y 4z = and the plane x y z = are orthogonal. (C) The function f(x) = n(e + e x ) is monotonic increasing x R. (D) If g is the inverse of the function, f(x) = n(e + e x ) then g(x) = n(e x e ) 4. The coordinates of a point on the line x = y = z at a distance 4 4 from the point (,, 0) are- (A) (9,, 4) (B) ( 8 4 +, 4, 4 4 ) (C) ( 7,, 4) (D) ( 8 4 +, 4, 4 4 ) NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65 5. Let 6x + 4y 5z = 4, x 5y + z = and x 9 (A) the angle between them must be (C) the plane containing them must be x + y z = 0 6. The lines x = y = z and x = y = z = y 4 = z 5 be two lines then- (B) the angle between them must be cos 5 6 (D) they are non-coplanar are - (A) coplanar for all (B) coplanar for = 9/ (C) if coplanar then intersect at 4,, (D) intersect at 7. If two pairs of opposite edges of a tetrahedron are perpendicular then -,, (A) the third is also perpendicular (B) the third pair is inclined at 60 (C) the third pair is inclined at 45 (D) (B), (C) are false 8. The equation of a plane bisecting the angle between the plane x y + z + = 0 and x y + 6z + 8 = 0 is - (A) 5x y 4z 45 = 0 (B) 5x y 4z = 0 (C) x y + z + 45 = 0 (D) x y + z + 5 = 0 77

25 J-Mathematics 9. A non-zero vector a is parallel to the line of intersection of the plane determined by the vectors î, î + ĵ and the plane determined by the vectors î ĵ, î ˆk,. The possible angle between a and î ĵ + ˆk is - (A) / (B) /4 (C) /6 (D) /4 0. If, m, n and, m, n are DCs of the two lines inclined to each other at an angle, then the DCs of the bisector of the angle between these lines are- (A) m m n n,, sin / sin / sin / (B) m m n n,, cos / cos / cos / (C) m m n n,, sin / sin / sin / (D) m m n n,, cos / cos / cos /. Points that lie on the lines bisecting the angle between the lines x y z 6 and x y z are - (A) (7,, 4) (B) (0,, 4) (C) (, 0, 0) (D) (, 6, ) BRAIN TASRS ANSWR KY XRCIS - Que A n s. A,B, D B, C C, D A, C A, C B, C A, D B, C B, D B, C Q u e. A n s. A,B,C,D 78 NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65

26 J-Mathematics XRCIS - 0 MISCLLANOUS TYP QUSTIONS TRU / FALS. If the plane xbc + yac + zab = abc cuts x, y & z-axis in A, B & C respectively then area of ABC is a b b c c a.. The angle between the line r = a + b and plane r n = d is cos b n b n. The perpendicular distance of the plane r ˆn = d, from the origin is d where d > 0 4. If A(,, ), B(, 6, ) and C(,, 4) are collinear then value of is The projection of line segment on the axes of reference are, 4 and respectively. The length of such a line segment is MATCH TH COLUMN Following questions contains statements given in two columns, which have to be matched. The statements in Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given statement in Column-I can have correct matching with ON statement in Column-II.. Match the following pair of planes with their lines of intersections : Column-I Column-II (A) x + y = 0 = y + z (p) x 0 = y 007 = z 004 (B) x =, y = (q) x 0 (C) x =, y + z = (r) x = y = z = y = z NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65 (D) x =, x + y + z = (s). Consider three planes P x + y + z = P x y + z = P x y + z = 5 x 0 = y 0 The three planes intersects each other at point P on XOY plane and at point Q on YOZ plane. O is the origin. Column-I Column-II (A) The value of is (p) (B) The length of projection of PQ on x-axis is (C) If the co-ordinates of point R situated at a minimum (q) (D) distance from point 'O' on the line PQ are (a, b, c), then value of 7a + 4b + 4c is (r) 4 If the area of POQ is a, then value of a b is b (s) = z 79

27 J-Mathematics Following question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given statement in Column-I can have correct matching with ON OR MOR statement(s) in Column-II.. Consider the following four pairs of lines in column I and match them with one or more entries in column II Column-I Column-II (A) L : x= + t, y=t, z= 5t (p) non coplanar lines L : r = (,, ) + (,, 0) (B) L : x = y z = 80 (q) lines lie in a unique plane x y 6 z L : (C) L : x = 6t, y= + 9t, z= t (r) infinite planes containing both the lines L : x= +s, y=4 s, z=s (D) L : x = y L : x 4 = y = z = z ASSRTION & RASON These questions contains, Statement I (assertion) and Statement II (reason). (s) lines are not intersecting (A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I. (B) Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-i. (C) Statement-I is true, Statement-II is false. (D) Statement-I is false, Statement-II is true.. Statement - I : If a plane contains point A( a ) and is parallel to vectors b and c, then its vector equation is r = a + b + µ c, where & µ are parameters and b c. B e c a u s e Statement - II : If three vectors are co-planar, then any one can be expressed as the linear combination of other two. (A) A (B) B (C) C (D) D. Statement - I : If ax + by + cz = a b c be a plane and (x, y, z ) and (x, y, z ) be two points on this plane then a(x x ) + b(y y ) + c(z z ) = 0. B e c a u s e Statement - II : If two vectors p î + p ĵ + p ˆk and q î + q ĵ + p ˆk are orthogonal then p q + p q + p q = 0. (A) A (B) B (C) C (D) D. Statement - I : If the lines x y z a b c a b c B e c a u s e = x y z a b c a b c x x a = y y b = z z c and x x a = y y b Statement - II : If the two lines are coplanar then shortest distance between them is zero. (A) A (B) B (C) C (D) D = z z c are coplanar then NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65

28 J-Mathematics 4. Statement - I : ABCDA B C D is a cube of edge unit. P and Q are the mid points of the edges B A, and B C respectively. Then the distance of the vertex D from the plane PBQ is 8. B e c a u s e Statement - II : Perpendicular distance of point (x, y, z ) from the plane ax + by + cz + d = 0 is given by ax by cz d a b c. (A) A (B) B (C) C (D) D 5. Statement - I : If a + b + 6c = 4, where a, b & c R, then the minimum value of a + b + c is 4. B e c a u s e Statement - II : The perpendicular distance of the plane px + qy + rz = from origin is p q r. (A) A (B) B (C) C (D) D 6. Consider following two planes P [r p a b] 0 P [r p c d] 0 such that (a b) (c d) 0 & let x be any vector in space. Statement-I : x.{(a b) (c d)} 0 B e c a u s e Statement-II : x.{t a tb} 0 x.{t a t b} 0, t, t R t, t R x.{(a b) (c d)} 0. (A) A (B) B (C) C (D) D 7. Consider planes P : (r ˆi).{(i ˆ ˆj k) ˆ (i ˆ k)} ˆ 0 and P : (r (i ˆ ˆj k)).{(i ˆ ˆ k) ˆ (i ˆ ˆj k)} ˆ 0 NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65 and line L : r 5i ˆ (i ˆ ˆj k) ˆ Statement-I : P & P are parallel planes. B e c a u s e Statement-II : L is parallel to both P & P. (A) A (B) B (C) C (D) D COMPRHNSION BASD QUSTIONS Comprehension # : Given four points A(,, 0); B(, 0, ); C(, 0, ) and D(0, 0, ). The point D lies on a line L orthogonal to the plane determined by the point A, B, and C On the basis of above information, answer the following questions :. quation of the plane ABC is - (A) x + y + z = 0 (B) y + z = 0 (C) x + z = 0 (D) y + z = 0. quation of the line L is - (A) r = ˆk + ( î + ˆk ) (B) r = ˆk + ( ĵ + ˆk ) (C) r = ˆk + ( ĵ + ˆk ) (D) none of these 8

29 J-Mathematics. Perpendicular distance of D from the plane ABC, is - (A) (B) Comprehension # : x x y y If a line passes through P (x, y, z ) and having Dr s a, b, c, then the equation of line is = z z = a b c and equation of plane perpendicular to it and passing through P is a(x x ) + b(y y ) + c(z z ) = 0. Further equation of plane through the intersection of the two planes a x + b y + c z + d = 0 and a x + b y + c z + d = 0 is (a x + b y + c z + d ) + k(a x + b y + c z + d ) = 0 On t he basis of above i nformat ion, a nswer t he fol low i ng que st ions :. The distance of the point (,, ) from the plane x y + z = 5 measured parallel to the line x = y = z 4 (A) 5 is - (B) (C) (C) 5. The equation of the plane through (0,, 4) and containing the line x = y = z 4 is - (A) x y + 4z = 0 (B) 5x + y + 9z 8 = 0 (C) 0x y 9z + 60 = 0 (D) 7x + 5y z + = 0. The plane x y z = is rotated through 90 about its line of intersection with the plane x + y + z =. Then equation of this plane in new position is - (A) 5x + 4y + z 0 = 0 (B) 4x + 5y + z = 0 (C) x + y + z = 9 (D) x + 4y 5z = 9 Comprehension # : Consider a triangular pyramid ABCD the position vectors of whose angular point are A(, 0, ); B(, 4, ); C(5,, ) and D(0, 5, 4). Let G be the point of intersection of the medians of the triangle BCD. On t he basis of above i nformat ion, a nswer t he fol low i ng que st ions :. The length of the vector AG is- 5 (A) 7 (B). Area of the triangle ABC in sq. units is- (A) 4 (B) 8 6 (C) 4 6 (D) none of these. The length of the perpendicular from the vertex D on the opposite face is - (A) 4 (B) (C) (D) none of these quation of the plane ABC is - (A) x + y + z = 5 (B) x y z = (C) x + y z = 4 (D) x + y z = MISCLLANOUS TYP QUSTION Tr ue / False. F. T. T 4. T 5. F Match the Column. (A) (r); (B) (s); (C) (p); (D) (q). (A) (r); (B) (p); (C) (q); (D) (s). (A) (r); (B) (q); (C) (q,s); (D) (p,s) Assertion & Reason (C) 5 9 ANSWR KY. C. A. C 4. D 5. A 6. D 7. B Comprehension Based Questions Comprehension # :. B. C. D Comprehension # :. B. C. A Comprehension # :. B. C. A 4. D (D) (D) (D) X RCIS-0 NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65

30 J-Mathematics XRCIS - 04 [A] CONCPTUAL SUBJCTIV XRCIS. Find the angle between the two straight lines whose direction cosines, m, n are given by + m n = 0 and mn + n + m = 0.. A variable plane is at a constant distance p from the origin and meets the coordinate axes in points A, B and C respectively. Through these points, planes are drawn parallel to the coordinates planes. Find the locus of their point of intersection.. P is any point on the plane x + my + nz = p. A point Q taken on the line OP (where O is the origin) such that OP.OQ = p. Show that the locus of Q is p(x + my + nz) = x + y + z. 4. The plane x + my = 0 is rotated about its line of intersection with the plane z = 0 through an angle. Prove that the equation to the plane in new position is x + my ±z m tan = 0 5. Find the equations of the two lines through the origin which intersect the line x angle of = y = z at an 6. Find the equation of the line which is reflection of the line x y + 0z = 6 x 9 = y = z in the plane 7. Find the point where the line of intersection of the planes x y + z = and x + y z = 5, intersects the plane x + y + z + 6 = 0 8. Find the foot and hence the length of the perpendicular from the point (5, 7, ) to the line x 5 line lie. = y 9 8 = 5 z 5. Also find the equation of the plane in which the perpendicular and the given straight 9. Find the equation of the plane containing the straight line x plane x y + z + = 0 0. Find the equation of the plane containing the line x Find also the S.D. between two lines. = y = z = y = z 5 x and parallel to the line are perpendicular to the = y 5 = z. 4 NOD6\\Data\04\Kota\J-Advanced\SMP\Maths\Unit#0\NG\0-D-COORDINAT GO.p65 CONCPTUAL SUBJCTIV XRCIS ANSWR KY X RCIS-04 ( A ). = 90. x 6. x 4 9 = y = z 7 + y + z = p 5. x = y = z or x = y = z 7. (,, 4) 8. (9,, 5) ; 4; 9x 4y z = 4 9. x + y +z + 4 = 0 0. x y + z = 0; units 8