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1 fo/u fopkjr Hkh# tu] ugha vkjehks dke] foifr ns[k NksM+s rqjar e/;e eu dj ';kea iq#"k flag ladyi dj] lgrs foifr vusd] ^cuk^ u NksM+s /;s; dks] j?kqcj jk[ks VsdAA jfpr% ekuo /kez i.ksrk ln~q# Jh j.knksm+nklth egkjkt STUDY PACKAGE Subject : Mathematics Topic : VECTORS & DIMENSIONAL GEOMETRY Available Online : R Inde. Theor. Short Revision. Eercise (E ). Assertion & Reason 5. Que. from Compt. Eams 6. 9 Yrs. Que. from IIT-JEE(Advanced) 7. 5 Yrs. Que. from AIEEE (JEE Main) Student s Name : Class Roll No. : : Address : Plot No. 7, III- Floor, Near Patidar Studio, Above Bond Classes, Zone-, M.P. NAGAR, Bhopal : , , WhatsApp

2 THREE DIMENSIONAL GEOMETRY Coordinate of a point in space There are infinite number of points in space. We want to identif each and ever point of space with the help of three mutuall perpendicular coordinate aes OX, OY and OZ. Three mutuall perpendicular lines OX, OY, OZ are considered as the three aes. The plane formed with the help of and aes is called - plane, similarl & aes form - plane and and aes form - plane. Consider an point P in the space, Drop a perpendicular from that point to - plane, then the algebraic length of this perpendicular is considered as -coordinate and from foot of the perpendicular drop perpendiculars to and aes. These algebraic lengths of perpendiculars are considered as and coordinates respectivel.. Vector representation of a point in space If coordinate of a point P in space is (,, ) then the position vector of the point P with respect to the same origin is î + ĵ + kˆ.. Distance formula Distance between an two points (,, ) and (,, ) is given as ) + ( ) + ( ) ( Vector method We know that if position vector of two points A and B are given as OA and OB then AB OB OA AB ( i + j + k) ( i + j + k) AB ( ) + ( ) + (. Distance of a point P from coordinate aes ) Let PA, PB and PC are distances of the point P(,, ) from the coordinate aes OX, OY and OZ respectivel then PA +, PB +, PC + Eample : Show that the points (0, 7, 0), (, 6, 6) and (, 9, 6) form a right angled isosceles triangle. Solution: Let A (0, 7, 0), B (, 6, 6), C (, 9, 6) AB (0 + ) + (7 6) + (0 6) 8 AB Similarl BC & AC 6, Clearl AB +BC AC ABC 90 Also AB BC Hence ABC is right angled isosceles. Eample : Show b using distance formula that the points (, 5, 5), (0,, ) and (,, ) are collinear. Solution Let A (, 5, 5), B (0,, ), C (,, ). AB ( 0) + (5 + ) + ( 5 ) BC (0 ) + ( + ) + ( + ) 8 AC ( ) + (5 + ) + ( 5 + ) 8 BC + AC AB Hence points A, B, C are collinear and C lies between A and B. Eample : Find the locus of a point which moves such that the sum of its distances from points A(0, 0, α) and B(0, 0, α) is constant. Solution. Let the variable point whose locus is required be P(,, ) Given PA + PB constant a (sa) ( 0) + ( 0) + ( + α) + ( 0) + ( 0) + ( α) a a + + ( α) α + α a α α a + + ( α) α a a + + ( α) α a + a α α α Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page of 77

3 or, + + α + a a α or, + This is the required locus. a α a Self practice problems :. One of the vertices of a cuboid is (,, ) and the edges from this verte are along the +ve -ais, +ve -ais and + ais respectivel and are of length,, respectivel find out the vertices. Ans. (,, 5), (,, 5), (,, ), (,, ), (,, 5), (,, 5), (,, ).. Show that the points (0,, ), (,, ), (, 5, 0) and (, 6, ) are the vertices of a square.. Find the locus of point P if AP BP 8, where A (,, ) and B (,, ) Ans Section Formula If point P divides the distance between the points A (,, ) and B (,, ) in the ratio of m : n, then coordinates of P are given as m + n m + n m + n,, m + n m + n m + n Note :- Mid point + + +,, 5. Centroid of a triangle G,, 6. Incentre of triangle ABC: Where AB c, BC a, CA b a + b + c a + b + c a + b + c,, a + b + c a + b + c a + b + c 7. Centroid of a tetrahedron A (,, ) B (,, ) C (,, ) and D (,, ) are the vertices of a tetrahedron then coordinate of its centroid (G) is given as i i i,, Eample : Show that the points A(,, ), B(,, ) and C(,, 0) are collinear. Also find the ratio in which C divides AB. Solution: Given A (,, ), B (,, ), C (,, 0). A (,, ) B (,, ) Let C div ide AB internall in the ratio k :, then k + k + k + C,, k + k + k + k + k 6 k k + k + k + For this value of k,, and 0 k + k + Since k < 0, therefore k divides AB eternall in the ratio : and points A(5, A, B,, C 6) are collinear. Eample : The vertices of a triangle are A(5,, 6), B(,, ) and C(,, ). The internal bisector of BAC meets BC in D. Find AD. Solution AB AC + + Since AD is the internal bisector of BAC BD AB 5 DC AC D divides BC internall in the ratio 5 : ( ) 5 + D,, AD B D C (,, ) (,, ) 9 or, D,, Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page of 77

4 50 unit 8 Eample : If the points P, Q, R, S are (, 7, 8), (,, ), (,, ) and (,, 5) respectivel, show that PQ and RS intersect. Also find the point of intersection. Solution Let the lines PQ and RS intersect at point A. Let A divide PQ in the ratio λ :, then P(, 7, 8) S(,, 5) λ + λ + 7 λ + 8 A,,.... () λ l λ + λ + λ + Let A divide RS in the ratio k :, then A k + k + 5k + K A,,... () k + k + k + R(,, ) Q(,, ) From () and (), we have, λ + k + λ + k +... () λ + 7 k + λ + k +... () λ + 8 5k + λ + k + From (), λk λ + k + λk + λ + k +... (5) or λk + λ k 0 From (), λk λ + 7k + 7 λk + λ + k +... (6) or λk + 5λ 5k 0... (7) Multipling equation (6) b, and subtracting from equation (7), we get λ + k 0 or, λ k Putting λ k in equation (6), we get λ + λ λ 0 or, λ ±. But λ, as the co-ordinates of P would then be underfined and in this case PQ RS, which is not true. λ k. Clearl λ k satisfies eqn. (5). Hence our assumption is correct A,, 5 9 or, A,,. Self practice problems:. Find the ratio in which plane divides the line joining the points A (,, ) and B (,, 6). Ans. :. Find the co-ordinates of the foot of perpendicular drawn from the point A(,, ) to the line joining the point B(,, 6) and C(5,, ). Ans. (,, 5) 8. Two vertices of a triangle are (, 6, ) and (,, ) and its centroid is,,. Find the third verte. Ans. (, 5, ). If centroid of the tetrahedron OABC, where co-ordinates of A, B, C are (a,, ), (, b, ) and (,, c) respectivel be (,, ), then find the distance of point (a, b, c) from the origin. Ans Show that,, 0 is the circumcentre of the triangle whose vertices are A (,, 0), B (,, ) and C (,, ) and hence find its orthocentre. Ans. (,, 0) 8. Direction Cosines And Direction Ratios (i) Direction cosines: Let α, β, γ be the angles which a directed line makes with the positive directions of the aes of, and respectivel, then cos α, cosβ, cos γ are called the direction cosines of the line. The direction cosines are usuall denoted b (l, m, n). Thus l cos α, m cos β, n cos γ. (ii) If l, m, n be the direction cosines of a line, then l + m + n (iii) Direction ratios: Let a, b, c be proportional to the direction cosines l, m, n then a, b, c are called the direction ratios. If a, b, c, are the direction ratios of an line L then a î + bĵ + ckˆ will be a vector parallel to the line L. Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page of 77

5 (iv) (v) (vi) If l, m, n are direction cosines of line L then l î + m ĵ + nkˆ is a unit vector parallel to the line L. If l, m, n be the direction cosines and a, b, c be the direction ratios of a vector, then a b c l,m,n a + b + c a + b + c a + b + c a b c or l, m, n a + b + c a + b + c a + b + c If OP r, when O is the origin and the direction cosines of OP are l, m, n then the coordinates of P are (lr, mr, nr). I f d i r e c t i o n c o s i n e s o f t h e l i n e A B a r e l, m,n, AB r, and the coordinates of A is (,, ) then the coordinates of B is given as ( + rl, + rm, + rn) If the coordinates P and Q are (,, ) and (,, ) then the direction ratios of line PQ are, a, b & c and the direction cosines of line PQ are l, PQ m PQ and n PQ (vii) Direction cosines of aes: Since the positive ais makes angles 0º, 90º, 90º with aes of, and respectivel. Therefore Direction cosines of ais are (, 0, 0) Direction cosines of ais are (0,, 0) Direction cosines of ais are (0, 0, ) Eample : If a line makes angles α, β, γ with the co-ordinate aes, prove that sin α + sin β + sinγ. Solution Since a line makes angles α, β, γ with the co-ordinate aes, hence cosα, cosβ, cosγ are its direction cosines cos α + cos β + cos γ ( sin α) + ( sin β) + ( sin γ) sin α + sin β + sin γ. Eample : Find the direction cosines l, m, n of a line which are connected b the relations l + m + n 0, mn + ml nl 0 Solution Given, l + m + n 0... () mn + ml nl 0 From (), n (l + m).... () Putting n (l + m) in equation (), we get, m(l + m) + ml + (l + m) l 0 or, ml m + ml + l + ml 0 or, l + ml m 0 or, or l m l + 0 [dividing b m m ] l ± + 8 m ±, Case I. when m l : In this case m l From (), l + n 0 n l l : m : n : : Direction ratios of the line are,, Direction cosines are ±, ± + + ( ) + + ( ), ± or,,, or 6, 6, 6 Case II. When m l : In this case l m From (), m + m + n 0 n m l : m : n m : m : m : : Direction ratios of the line are,,. Direction cosines are,, ( ) + + ( ) + + ( ) + + ( ) + + or,,, Self practice problems:. Find the direction cosine of a line ling in the plane and making angle 0 with -ais. Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page 5 of 77

6 Ans. m ±, l, n 0. A line makes an angle of 60 with each of and aes, find the angle which this line makes with -ais. Ans. 5. A plane intersects the co-ordinates aes at point A(a, 0, 0), B(0, b, 0), C(0, 0, c) O is origin. Find the direction ratio of the line joining the verte B to the centroid of face AOC. a c Ans., b,. A line makes angles α, β, γ, δ with the four diagonals of a cube, prove that cos α + cos β + cos γ + cos δ. 9. Angle Between Two Line Segments: If two lines have direction ratios a, b, c and a, b, c respectivel then we can consider two vectors parallel to the lines as a i + b j + c k and a i + b j + c k and angle between them can be given as. aa + bb + cc cos θ. a + b + c a + b + c a b c (i)the line will be perpendicular if a a + b b + c c 0 (ii)the lines will be parallel if a b c (iii) Two parallel lines have same direction cosines i.e. l l, m m, n n Eample : What is the angle between the lines whose direction cosines are,, ;,,? Solution Let θ be the required angle, then cosθ l l + m m + n n θ 0, 6 6 Eample : Find the angle between an two diagonals of a cube. Solution The cube has four diagonals Y cosθ (0, a, a) D E C(0, 0, a) G(0, a, 0) F(a, a, 0) O (0, 0, 0) (a, a, a) B(a, 0, a) Z OE, AD, CF and GB The direction ratios of OE are a, a, a or,,, its direction cosines are,,. Direction ratios of AD are a, a, a. or,,,. its direction cosines are,,. Similarl, direction cosines of CF and GB respectivel are,, and,,. We take an two diagonals, sa OE and AD Let θ be the acute angle between them, then A (a, 0, 0) X or, θ cos. Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page 6 of 77

7 Eample : If two pairs of opposite edges of a tetrahedron are mutuall perpendicular, show that the third pair will also be mutuall perpendicular. Solution: Let OABC be the tetrahedron where O is the origin and co-ordinates of A, B, C be (,, ), (,, ), (,, ) respectivel. A (,, ) O (0, 0, 0) B C (,, ) (,, ) Let OA BC and OB CA. We have to prove that OC BA. Now, direction ratios of OA are 0, 0, 0 direction ratios of BC are ( ), ( ), ( ). or,,, OA BC. ( ) + ( ) + ( ) 0 Similarl,... () OB CA ( ) + ( ) + ( ) 0 Adding equations () and (), we get... () ( ) + ( ) + ( ) 0 OC BA [ direction ratios of OC are,, and that of BA are ( ), ( ), ( )] Self practice problems:. Find the angle between the lines whose direction cosines are given b l + m + n 0 and l + m n 0 Ans. 60. P (6,, ) Q (5,, ) R (,, 5). are vertices of a find Q. Ans. 90 Show that the direction cosines of a line which is perpendicular to the lines having directions cosines l m n and l m n respectivel are proportional to m n m n, n l n l, l m l m 0. Projection of a line segment on a line (i) If the coordinates P and Q are (,, ) and (,, ) then the projection of the line segments PQ on a line having direction cosines l, m, n is l( ) + m( ) + n( ) a.b (ii) Vector form: projection of a vector a on another vector b is a. bˆ In the above case we can consider PQ as ( ) î + ( ) ĵ + ( ) kˆ in place of a and l î + m ĵ + n kˆ in place of b. (iii) b l r, m r & n r are the projection of r in OX, OY & OZ aes. (iv) r r (l î + m ĵ + n kˆ ) Solved Eample : Find the projection of the line joining (,, ) and (,, ) on the line having direction ratios,, 6. Solution Let A (,, ), B (,, ) B A P L M Q Direction ratios of the given line PQ are,, ( 6) 7 direction cosines of PQ are 6,, Projection of AB on PQ l ( ) + m( ) + n( ) ( ) + ( ) ( ) Self practice problems:. A (6,, ), B (5,,,), C(,, ) D (0,, 5) Find the projection of line segment AB on CD line. Ans. 5/7 Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page 7 of 77

8 . The projections of a directed line segment on co-ordinate aes are,, 6. Find its length and direction cosines. Ans. ;,,. Find the projection of the line segment joining (,, ) and (,, 5) on a line which makes equal 7 acute angles with co-ordinate aes. Ans. A PLANE If line joining an two points on a surface lies completel on it then the surface is a plane. OR If line joining an two points on a surface is perpendicular to some fied straight line. Then this surface is called a plane. This fied line is called the normal to the plane... Equation Of A Plane (i) (ii) (iii) (iv) Normal form of the equation of a plane is l + m + n p, where, l,m n are the direction cosines of the normal to the plane and p is the distance of the plane from the origin. General form: a + b + c + d 0 is the equation of a plane, where a, b, c are the direction ratios of the normal to the plane. The equation of a plane passing through the point (,, ) is given b a ( ) + b( ) + c ( ) 0 where a, b, c are the direction ratios of the normal to the plane. Plane through three points: The equation of the plane through three non collinear points (,, ), (,, ), (,, ) is (v) Intercept Form: The equation of a plane cutting intercept a, b, c on the aes is + + (vi) Vector form: The equation of a plane passing through a point having position vector a & normal to vector n is ( r a ). n 0 or r. n a. n Note: (a) Vector equation of a plane normal to unitvector nˆ and at a distance d from the origin is r. n d (b) Coordinate planes (i) Equation of plane is 0 (ii) Equation of plane is 0 (iii) Equation of plane is 0 (c) Planes parallel to the aes: If a 0, the plane is parallel to ais i.e. equation of the plane parallel to the ais is b + c + d 0. Similarl, equation of planes parallel to ais and parallel to ais are a + c +d 0 and a + b + d 0 respectivel. (d) Plane through origin: Equation of plane passing through origin is a + b + c 0. (e) Transformation of the equation of a plane to the normal form: To reduce an equation a + b + c d 0 to the normal form, first write the constant term on the right hand side and make it positive, then divide each term b a + b + c, where a, b, c are coefficients of, and respectivel e.g. a b c d + + ± a + b + c ± a + b + c ± a + b + c ± a + b + c Where (+) sign is to be taken if d > 0 and ( ) sign is to be taken if d < 0. (f) An plane parallel to the given plane a + b + c + d 0 is a + b + c + λ 0. Distance between two parallel planes a + b + c + d 0 and a + b + c + d 0 is d d given as a + b + c (g) Equation of a plane passing through a given point & parallel to the given vectors: The equation of a plane passing through a point having position vector a and parallel to (h) b & c is r a + λ b + µ c (parametric form) where λ & µ are scalars. or r. (b c) a. (b c) (non parametric form) A plane a + b + c + d 0 divides the line segment joining (,, ) and (,, ). in the ratio a + b a + b + c + c + d + d 0 Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page 8 of 77

9 (i) (j) The plane divides the line segment joining the points (,, ) and (,, ) in the ratio. Similarl plane in and plane in Coplanarit of four points The points A( ), B( ) C( ) and D( ) are coplaner then 0 ver similar in vector method the points A ( r ), B( r ), C( r ) and D( r ) are coplanar if [ r r, r r, r r ] 0 Eample : Find the equation of the plane upon which the length of normal from origin is 0 and direction ratios of this normal are,, 6. Solution If p be the length of perpendicular from origin to the plane and l, m, n be the direction cosines of this normal, then its equation is l + m + n p... () Here p 0; Direction ratios of normal to the plane are,, Direction cosines of normal to the required plane are + l 7, m 7, n 7 6 Putting the values of l, m, n, p in (), equation of required plane is or, Eample : Show that the points (0,, 0), (,, ), (,, ), (,, 0) are coplanar. Solution Let A (0,, 0), B (,, ), C (,, ) and D (,, 0) Equation of a plane through A (0,, 0) is a ( 0) + b ( + ) + c ( 0) 0 or, a + b + c + b 0 If plane () passes through B (,, ) and C (,, )... () Then a + b c 0... () and a + b + c 0 From () and (), we have... () + or, k (sa) Putting the value of a, b, c, in (), equation of required plane is k k( + ) + k 0 or, + 0 Clearl point D (,, 0) lies on plane ()... () Thus points D lies on the plane passing through A, B, C and hence points A, B, C and D are coplanar. Eample : If P be an point on the plane l + m + n p and Q be a point on the line OP such that OP. OQ p, show that the locus of the point Q is p(l + m + n) + +. Solution Let P (α, β, γ), Q (,, ) Direction ratios of OP are α, β, γ and direction ratios of OQ are,,. Since O, Q, P are collinear, we have α β γ k (sa) As P (α, β, γ) lies on the plane l + m + n p,... () lα + mβ + nγ p or k( + m + n ) p... () Given, OP. OQ p + β + γ + + α p or, k ( + + ) + + p or, k ( + + ) p... () On dividing () b (), we get, l + m + n + + p or, p (l + m + n ) + + Hence the locus of point Q is p (l + m + n) + +. Eample : A point P moves on a plane + +. A plane through P and perpendicular to OP meets the Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page 9 of 77

10 co-ordinate aes in A, B and C. If the planes through A, B and C parallel to the planes 0, 0, 0 intersect in Q, find the locus of Q. Solution Given plane is () Let P (h, k, l) h k l Then () OP Direction cosines of OP are h, k, l Equation of the plane through P and normal to OP is h + k + l or, (h + k + l ) A h C h + k + 0, 0, l Let Q (α, β, γ), then Now α h From (), h Similarl + k + l h, 0, 0, B l, β h + k + l k + + α β γ ( ) h h a k b α h + k + l aα + k + l bβ and l h c 0, k (h, γ + k + k + l cγ, 0, k h + + l + l h + + aα bβ cγ a or, aα bβ cγ α β γ Required locus of Q (α, β, γ) is ) + l k l + b c... ()... () [from ()] [from ()] Self practice problems :. Check wether this point are coplanar if es find the equation of plane containing them A (,, ) B (0,, 0) C (,, ) D (,, 0) Ans. es, +. Find the plane passing through point (,, ) and perpendicular to the line joining the points (, 6, ) and (,, 0). Ans Find the equation of plane parallel to and cutting intercepts on the aes whose rum 000 is 50. Ans Find the equation of plane passing through (,, ) and (9,, 6) and perpendicular to the plane Ans Find the equation of the plane to î + ĵ + kˆ and î ĵ and passing through (,, ). Ans Find the equation of the plane passing through the point (,, ) and perpendicular to the planes and + 0. Ans Sides of a plane: A plane divides the three dimensional space in two equal parts. Two points A ( ) and B ( ) are on the same side of the plane a + b + c + d 0 if a + b + c + d and Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page 0 of 77

11 a + b + c + d are both positive or both negative and are opposite side of plane if both of these values are in opposite sign. Eample : Show that the points (,, ) and (,, ) lie on opposite sides of the plane Solution Since the numbers and + are of opposite sign., the points are on opposite sides of the plane.. A Plane & A Point a' + b' + c' + d (i) Distance of the point (,, ) from the plane a + b + c+ d 0 is given b. a + b + c (ii) The length of the perpendicular from a point having position vector a to plane r. n d is a.n d given b p. n (iii) The coordinates of the foot of perpendicular from the point (,, ) to the plane ' ' ' (a a + b + c + d 0 aregain b + b + c + d) a + b + c (iv) To find image of a point w.r.t. a plane. (a) Let P (,, ) is a given point and a + b + c + d 0 is given plane Let (,, ) is the image point. then λa, λb, λc λa +, λb +, λc + (b) a + b + c 0 from (i) put the values of,, in (ii) and get the values of λ and resubtitute in (i) to get ( ). The coordinate of the image of point (,, ) w.r.t the plane a + b + c + d 0 are given ' ' ' (a b + b + c + d) a + b + c (v) The distance between two parallel planes a + b + c + d 0 and a + b + c + d 0 is d d' given b a + b + c Eample : Find the image of the point P (, 5, 7) in the plane Solution Given plane is () P (, 5, 7) Direction ratios of normal to plane () are,, Let Q be the image of point P in plane (). Let PQ meet plane () in R then PQ plane () Let R (r +, r + 5, r + 7) Since R lies on plane () (r + ) + r r R (,, ) or, 6r r Let Q (α, β, γ) Since R is the middle point of PQ α + α 9 β + 5 β γ + 7 γ Q ( 9,, ). Eample : Find the distance between the planes + and Solution Given planes are () and () a b c We find that Hence planes () and () are parallel. Plane () ma be written as () Required distance between the planes ( ) +. 9 Eample : A plane passes through a fied point (a, b, c). Show that the locus of the foot of perpendicular to it from the origin is the sphere + + a b c 0 Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page of 77

12 Solution Let the equation of the variable plane be O(0, 0, 0) FREE Download Stud Package from website: & P( α, β, γ) l + m + n + d 0 Plane passes through the fied point (a, b, c)... () la + mb + nc + d 0... () Let P (α, β, γ) be the foot of perpendicular from origin to plane (). Direction ratios of OP are α 0, β 0, γ 0 i.e. α, β, γ From equation (), it is clear that the direction ratios of normal to the plane i.e. OP are l, m, n ; α, β, γ and l, m, n are the direction ratios of the same line OP α β γ (sa) l m n k l kα, m kβ, n kγ... () Putting the values of l, m, n in equation (), we get kaα + kbβ + kcγ + d 0... () Since α, β, γ lies in plane () lα + mβ + nγ + d 0... (5) Putting the values of l, m, n from () in (5), we get or kα + kβ + kγ + d 0 kα + kβ + kγ kaα kbβ kcγ 0... (6) [putting the value of d from () in (6)] or α + β + γ aα bβ cγ 0 Therefore, locus of foot of perpendicular P (α, β, γ) is + + a b c 0... (7) Self practice problems:. Find the intercepts of the plane on the aes. Also find the length of perpendicular from origin to this line and direction cosines of this normal. Ans. a 8, b, c, p 7 ;,, Find : (i) (ii) perpendicular distance foot of perpendicular (iii) image of (, 0, ) in the plane Ans. (i) 6 (ii),, 6 6. Angle Between Two Planes: (i) 5 (iii), Consider two planes a + b + c + d 0 and a + b + c + d 0. Angle between these planes is the angle between their normals. Since direction ratios of their normals are (a, b, c) and (a, b, c ) respectivel, hence θ, the angle between them, is given b cos θ ' ' ' a + b aa' + bb' + cc' + c a + b + c, 7 Planes are perpendicular if aa + bb + cc 0 and planes are parallel if (ii) The angle θ between the planes r. n d and r. n d is given b, cos θ Planes are perpendicular if 5. Angle Bisectors (i) (ii) (iii) n. n 0 & planes are parallel if n λ n. The equations of the planes bisecting the angle between two given planes a + b + c + d 0 and a + b + c + d 0 are a + b + c + d a + b + c + d a a' b c b' c' n.n n. n ± a + b + c a + b + c Equation of bisector of the angle containing origin: First make both the constant terms positive. a Then the positive sign in + b + c + d a ± + b + c + d gives the bisector of a + b + c a + b + c the angle which contains the origin. Bisector of acute/obtuse angle: First make both the constant terms positive. Then a a + b b + c c > 0 origin lies on obtuse angle Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page of 77

13 a a + b b + c c < 0 origin lies in acute angle 6. Famil of Planes (i) An plane passing through the line of intersection of non parallel planes or equation of the plane through the given line in serval form. a + b + c + d 0 & a + b + c + d 0 is a + b + c + d + λ (a + b + c + d ) 0 (ii) The equation of plane passing through the intersection of the planes r. n d & r. n d is r. (n + λ n ) d + λd where λ is arbitrar scalar Eample : The plane is rotated through 90 about its line of intersection with the plane + +. Find its equation in the new position. Solution Given planes are... () and () Since the required plane passes through the line of intersection of planes () and () its equation ma be taken as k ( ) 0 or ( + k) + ( k) + ( k) k 0... () Since planes () and () are mutuall perpendicular, ( + k) ( k) ( k) 0 or, + k + k + k 0 or, k Putting k in equation (), we get, This is the equation of the required plane. Eample : Find the equation of the plane through the point (,, ) which passes through the line of intersection of the planes and Solution Given planes are () and () Given point is P (,, ). Equation of an plane through the line of intersection of planes () and () is k ( ) 0... () If plane () passes through point P, then k ( ) 0 or, k From () required plane is Eample : Find the planes bisecting the angles between planes and Which of these bisector planes bisects the acute angle between the given planes. Does origin lie in the acute angle or obtuse angle between the given planes? Solution Given planes are and ()... () Equations of bisecting planes are + 9 ± + + ( ) + ( ) + ( ) + ( ) + () or, [ + 9] ± ( + + ) or, ,... () [Taking +ve sign] and () [Taking ve sign] Now a a + b b + c c ( ) () + ( ) ( ) + ( ) () < 0 Bisector of acute angle is given b a a + b b + c c < 0, origin lies in the acute angle between the planes. Eample : If the planes c b 0, c + a 0 and b + a 0 pass through a straight line, then find the value of a + b + c + abc. Solution Given planes are c b 0... () c + a 0 b + a 0... ()... () Equation of an plane passing through the line of intersection of planes () and () ma be taken as c b + λ (c + a) 0 or, ( + λc) (c + λ) + ( b + aλ) 0... () If planes () and () are the same, then equations () and () will be identical. + cλ (c + λ) b + aλ b a (i) (ii) (iii) From (i) and (ii), a + acλ bc bλ Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page of 77

14 (a + bc) or, λ... (5) (ac + b) From (ii) and (iii), (ab + c) c + λ ab + a λ or λ... (6) a From (5) and (6), we have, (a + bc) (ab + c) ac + b. ( a ) or, a a + bc a bc a bc + ac + ab + bc or, a bc + ac + ab + a + a bc a 0 or, a + b + c + abc. Self practice problems:. A tetrahedron has vertices at O(0, 0, 0), A(,, ), B(,, ) and C(,, ). Prove that the angle 9 between the faces OAB and ABC will be cos. 5. Find the equation of plane passing through the line of intersection of the planes 5 and and the point (,, ). Ans , λ. Find the equations of the planes bisecting the angles between the planes + + 0, and sepecif the plane which bisects the acute angle between them. Ans , ; Show that the origin lies in the acute angle between the planes and Prove that the planes , and have a common line of intersection. 7. Area of a triangle: Let A (,, ), B (,, ), C (,, ) be the vertices of a triangle, then ( + + ) where, and Vector Method From two vector AB and AC. Then area is given b i j k AB AC Eample : Through a point P (h, k, l) a plane is drawn at right angles to OP to meet the co-ordinate aes in A, B and C. If OP p, show that the area of ABC is p 5. hkl Solution OP p Direction cosines of OP are h, k, l Since OP is normal to the plane, therefore, equation of the plane will be, h + k + l or, h + k + l p... () A p, 0, 0 h, B p 0,, 0 k, C p 0, 0, l Now area of ABC, A + A + A Now A area of projection of ABC on -plane area of AOB Mod of p h p k 0 p hk Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page of 77

15 Similarl, A p kl 8 and A p lh p p p + + h k k l l h p 8 h k l 8 p (l + k + h ) h k l 8. Volume Of A Tetrahedron: 8 0 or, p 5. hkl Volume of a tetrahedron with vertices A (,, ), B(,, ), C (,, ) and D (,, ) is given b V 6 9. Equation Of A Line (i) (ii) (iii) (iv) (v) A LINE A straight line in space is characterised b the intersection of two planes which are not parallel and therefore, the equation of a straight line is a solution of the sstem constituted b the equations of the two planes, a + b + c + d 0 and a + b + +c + d 0. This form is also known as non smmetrical form. The equation of a line passing through the point (,, ) and having direction ratios a, b, c is a r. This form is called smmetric form. A general point on the line is given b ( b c + ar, + br, + cr). Vector equation: Vector equation of a straight line passing through a fied point with position vector a and parallel to a given vector b is r a + λ b where λ is a scalar. The equation of the line passing through the points (,, ) and (,, ) is Vector equation of a straight line passing through two points with position vectors a & b is r a + λ ( b a ). (vi) Reduction of cartesion form of equation of a line to vector form & vice versa Note: Straight lines parallel to co-ordinate aes: r ( î + ĵ + kˆ ) + λ (a î + b ĵ + c kˆ ). Straight lines Equation Straight lines Equation (i) Through origin m, n (v) Parallel to ais p, q (ii) ais 0, 0 (vi) Parallel to ais h, q (iii) ais 0, 0 (vii) Parallel to ais h, p (iv) ais 0, 0 Eample : Find the equation of the line through the points (,, 7) and (,, 6) in vector form as well as in cartesian form. Solution Let A (,, 7), B (,, 6) Now a OA b OB i + j 7 k, i j + 6 k Equation of the line through A( a ) and B( b ) is r a + t ( b a ) or r i + j 7 k + t ( i 5 j + k )... () Equation in cartesian form : Equation of AB is or, Eample : Find the co-ordinates of those points on the line which is at a distance of 6 units from point (,, ). Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page 5 of 77

16 + Solution Given line is... () 6 Let P (,, ) Direction ratios of line () are,, 6 6 Direction cosines of line () are,, Equation of line () ma be written as +... () Co-ordinates of an point on line () ma be taken as 6 r +, r, r Let Q r, r, r Distance of Q from P r According to question r r ± Putting the value of r, we have 5 9 Q,, or Q,, Eample : Find the equation of the line drawn through point (, 0, ) to meet at right angles the line + + Solution Given line is () Let P (, 0, ) Co-ordinates of an point on line () ma be taken as Q (r, r +, r ) Direction ratios of PQ are r, r +, r Direction ratios of line AB are,, Since PQ AB (r ) ( r + ) ( r ) 0 9r 6 + r + r + 0 r 7 r Therefore, direction ratios of PQ are,, 7 or,,, 7 Equation of line PQ is 0 7 or, 7 Eample : Show that the two lines and intersect. Find also the 5 point of intersection of these lines. Solution Given lines are... () 0 and... () 5 An point on line () is P (r +, r +, r +) and an point on line () is Q (5λ +, λ +, λ) Lines () and () will intersect if P and Q coincide for some value of λ and r. r + 5λ + r 5λ... () r + + λ + r λ... () r + λ r λ... () Solving () and (), we get r, λ Clearl these values of r and λ satisf eqn. () Now P (,, ) Hence lines () and () intersect at (,, ). Self practice problems:. Find the equation of the line passing through point (, 0, ) having direction ratio,, 5. Prove that this line passes through (,, 7). Ans Find the equation of the line parallel to line and passing through the point (, 0, 5). 9 Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page 6 of 77

17 5 Ans.. Find the coordinates of the point when the line through (,, ) and (5,, 6) crosses the plane. Ans.,, Reduction Of Non-Smmetrical Form To Smmetrical Form: Let equation of the line in non smmetrical form be a + b + c + d 0, a + b + c + d 0. To find the equation of the line in smmetrical form, we must know (i) its direction ratios (ii) coordinate of an point on it. (i) Direction ratios: Let l, m, n be the direction ratios of the line. Since the line lies in both the planes, it must be perpendicular to normals of both planes. So a l + b m + c n 0, a l + b m + c n 0. From these equations, proportional values of l, m, n can be found b cross multiplication as i l bc bc j k c m a ca a n b ab Alternative method The vector a b c i (b c b c ) + j (c a c a ) + k (a b a b ) will be parallel a b c to the line of intersection of the two given planes. hence l : m: n (b c b c ): (c a c a ): (a b a b ) (ii) Point on the line Note that as l, m, n cannot be ero simultaneousl, so at least one must be non ero. Let a b a b 0, then the line cannot be parallel to plane, so it intersect it. Let it intersect plane in (,, 0). Then a + b + d 0 and a + b + d 0.Solving these,we get a point on the line. Then its equation becomes. bd bd da da 0 ab ab ab ab 0 or bc bc ca ca ab ab bc bc ca ca ab ab Note: If l 0, take a point on plane as (0,, ) and if m 0, take a point on plane as (, 0, ). Alternative method a b If a b Put 0 in both the equations and solve the equations a + b + d 0, a + b + d 0 otherwise Put 0 and solve the equations a + c + d 0 and a + c + d 0 Eample : Find the equation of the line of intersection of planes + 5, 8 + in the smmetric form. Solution Given planes are () and () Let l, m, n be the direction ratios of the line of intersection : then l + m 5n 0... () and 8l + m n 0 l m n l m n l m n or, or, Hence direction ratios of line of intersection are,,. Here 0, therefore line of intersection is not parallel to -plane. Let the line of intersection meet the -plane at P (α, β, 0). Then P lies on planes () and () α + β + 0 or, α + β 0... (5) and or, 8α + β 0 α + β (6) Solving (5) and (6), we get α β α β or, α, β 0 Hence equation of line of intersection in smmetrical form is. Eample : Find the angle between the lines 0, and + 0, Solution Given lines are () + 0 and () Let l, m, n and l, m, n be the direction cosines of lines () and () respectivel Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page 7 of 77

18 line () is perpendicular to the normals of each of the planes 0 and l m + 0.n 0... () and 0l + m n 0... () Solving equations () and (), we get, l m n l m n or, k (let). 0 0 ( ) 0 Since line () is perpendicular to the normals of each of the planes + 0 and + 6 0, l + m 0... (5) and m n 0... (6) l l m or, m n and n m or, m. l m n t (let). If θ be the angle between lines () and (), then cosθ l l + m m + n n (k) ( t) + (k) (t) + (k) (t) 9kt + kt + 8kt 0 θ 90. Self practice problems:. Find the equation of the line of intersection of the plane Ans.. Show that the angle between the two lines defined b the equations and is cos. Prove that the three planes + 7 0, + 5 0, intersect at a point and find its co-ordinates. Ans. (, 7, ). Foot, Length And Equation Of Perpendicular From A Point To A Line: a b c (i)cartesian form: Let equation of the line be r (sa)...(i) l m n and A (α, β, γ) be the point. An point on line (i) is P (lr + a, mr + b, nr + c)... (ii) If it is the foot of the perpendicular from A on the line, then AP is perpendicular to the line. So l (lr + a α) + m (mr + b β) + n (nr + c γ) 0 i.e. r (α a) l + (β b) m + (γ c)n since l + m + n. Putting this value of r in (ii), we get the foot of perpendicular from point A on the given line. Since foot of perpendicular P is known, then the length of perpendicular is given b AP ( γ l r + a α) + (mr + b β) + (nr + c ) the α β γ equation of perpendicular is given b (ii) Vector Form: Equation of a line lr + a α mr + b β nr + c γ passing through a point having position vector α and perpendicular to the lines r a + λ b and r a + λ b is parallel to b b. So the vector equation of such a line is r α + λ ( b b ). Position vectorβ of the image of a point α in a straight line r a + λ b is given b β a (a α). b b α. Position vector of b the foot of the perpendicular on line is f a (a α). b b. The equation of the perpendicular is r α + µ b (a α).b (a α) b. b. To find image of a point w. r. t a line Let L is a given line Let (,, ) is the image of the point P (,, ) with respect to the line L. Then (i) a ( ) + b ( ) + c ( ) 0 Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page 8 of 77

19 + + + (ii) λ a b c from (ii) get the value of,, in terms of λ as aλ +, bα, cλ + now put the values of,, in (i) get λ and resubtitute the value of λ to to get ( ). + + Eample : Find the length of the perpendicular from P (,, ) to the line. + + Solution Given line is... () P (,, ) Co-ordinates of an point on line () ma be taken as Q (r, r +, r ) Direction ratios of PQ are r, r + 6, r Direction ratios of AB are,, Since PQ AB (r ) + (r + 6) ( r ) 0 5 or, r r Q,, PQ 7 Second method : Given line is + + P (,, ) Direction ratios of line () are,, RQ length of projection of RP on AB ( + ) + PR ( ) 5 5 (+ ) units. P (,, ) A R Q B (,, ) PQ PR RQ Self practice problems: Find the length and foot of perpendicular drawn from point (,, 5) to the line. 0 Also find the image of the point in the line. Ans., N (,, ), Ι (0, 5, ). Find the image of the point (, 6, ) in the line. Ans. (, 0, 7) Find the foot and hence the length of perpendicular from (5, 7, ) to the line Find also the equation of the perpendicular. Ans. (9,, 5) ; ; 6. Angle Between A Plane And A Line: (i) If θ is the angle between line l m n and the plane a + b + c + d 0, then sin θ al + bm + cn. ( a + b + c ) l + m + n (ii) Vector form: If θ is the angle between a line r ( a + λ b ) and r. n.n d then sin θ b. b n l m n (iii) Condition for perpendicularit b n 0 (iv) Condition for parallel al + bm + cn 0 b. n 0 Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page 9 of 77

20 . Condition For A Line To Lie In A Plane (i) Cartesian form: Line l m n (ii) would lie in a plane a + b + c + d 0, if a + b + c + d 0 & al + bm + cn 0. Vector form: Line r a + λ b would lie in the plane r. n d if b. n 0 & a. n d 5. Coplanar Lines: (i) (ii) If the given lines are α l for intersection/coplanarit is two lines Alternative method: is β γ n α α' β β' m α l l l' β m m m' and γ ' ' α l γ γ ' n n' ' ' β γ ' ' m n, then condition 0 & plane containing the above l' m' n' Condition of coplanarit if both the lines are in general form Let the lines be a + b + c + d 0 a + b + c + d & α + β + γ + δ 0 α + β + γ + δ d The are coplanar if a' α α' b' β β' c' γ γ' d' δ δ' 0 n 0 get vector along the line of shortest distance as Now get unit vector along this vector û li + mj + nk i l j m k n l m n Let v (α α ) î + (B B ) ĵ + ( ) S. D. u. v Eample : Find the distance of the point (, 0, ) from the plane 9 measured parallel to the + 6 line. 6 Solution Given plane is 9... () + 6 Given line AB is... () 6 Equation of a line passing through the point Q(, 0, ) and parallel to line () is + r.... () 6 Co-ordinates of point on line () ma be taken as P (r +, r, 6r ) If P is the point of intersection of line () and plane (), then P lies on plane (), (r + ) (r) ( 6r ) 9 r or, P (,, 9) Distance between points Q (, 0, ) and P (,, 9) PQ ( ) + ( 0) + ( 9 ()) A P B Q (, 0, ) Eample: Find the equation of the plane passing through (,, 0) which contains the line +. Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page 0 of 77

21 Solution Equation of an plane passing through (,, 0) ma be taken as a ( ) + b ( ) + c ( 0) 0... () where a, b, c are the direction ratios of the normal to the plane. Given line is +... () If plane () contains the given line, then a + b c 0 Also point (,, ) on line () lies in plane ()... () a ( ) + b ( ) + c ( 0) 0 or, a b + c 0 Solving equations () and (), we get,... () or, k (sa).... (5) 6 Substituting the values of a, b and c in equation (), we get, 6 ( ) + ( ) + ( 0) 0. or, This is the required equation. + Eample : Find the equation of the projection of the line on the plane B Solution A Let the given line AB be + Given plane is... () Let DC be the projection of AB on plane ()... () D C Clearl plane ABCD is perpendicular to plane (). Equation of an plane through AB ma be taken as (this plane passes through the point (,, ) on line AB) a ( ) + b ( + ) + c ( ) 0... () where a b + c 0... () [ normal to plane () is perpendicular to line ()] Since plane () is perpendicular to plane (), a + b + c 0 Solving equations () & (5), we get,... (5). 9 5 Substituting these values of a, b and c in equation (), we get 9 ( ) ( + ) 5 ( ) 0 or, (6) Since projection DC of AB on plane () is the line of intersection of plane ABCD and plane (), therefore equation of DC will be (i) and... (7) (ii) Let l, m, n be the direction ratios of the line of intersection of planes (i) and (ii) 9l m 5n 0... (8) and l + m + n 0... (9) l m n Eample : Show that the lines and are coplanar. Also find the equation of the plane containing them. Solution Given lines are + + r (sa)... () and R (sa) If possible, let lines () and () intersect at P.... () An point on line () ma be taken as (r +, r, r ) P (let). An point on line () ma be taken as ( R + 7, R, R 7) P (let). r + R + 7 or, r + R... () Also or, r R r R... () Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page of 77

22 and r R 7 or, r R (5) Solving equations () and (), we get, r, R Clearl r, R satisfies equation (5). Hence lines () and () intersect. lines () and () are coplanar. Equation of the plane containing lines () and () is or, ( ) ( 6 ) ( + ) ( + ) + ( + ) ( 9) 0 or, 7 ( ) 7 ( + ) 7 ( + ) 0 or, or, Self practice problems: + 6. Find the values of a and b for which the line is perpendicular to the plane a + b Ans. a, b. Prove that the lines and are coplanar. Also find the equation 5 of the plane in which the lie. Ans Find the plane containing the line Ans and parallel to the line + +. Show that the line & are intersecting each other. Find thire 5 intersection and the plane containing the line. Ans. (,, ) & Show that the lines r ( î ĵ 5 kˆ ) + λ ( î 5 ĵ 7 kˆ ) & r ( î + ĵ + 6 kˆ ) + µ ( î + ĵ + 7 kˆ ) are coplanar and find the plane containing the line. Ans. r. ( î ĵ + kˆ ) 6. Skew Lines: (i) (ii) 0 The straight lines which are not parallel and non coplanar i.e. non intersecting are called α' α β' β γ' γ skew lines. If l m n 0, then lines are skew. l' m' n' Shortest distance: Suppose the equation of the lines are α β γ α' β' γ' and l m n l' m' n' ( α α' ) (mn' m' n) + ( β β' ) (nl n' l) + ( γ γ' ) ( lm' l' m) S.D. (mn' m' n) α' α β' β γ' γ l m n (mn m n) l' m' n' (iii) Vector Form: For lines a + λ b & a + λ b to be skew ( b b ). ( a a ) 0 or [ b b ( a a )] 0. (iv) Shortest distance between the two parallel lines r a + λ b & r a + µ b ( a a) b is d. b Eample : Find the shortest distance and the vector equation of the line of shortest distance between the lines given b Solution r r + 8 j + k + λ r Given lines are j + k and r r + 8 j + k + λ i j + k... () r r 7 j + 6 k + µ i + j + k Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page of 77

23 and r r 7 j + 6 k + µ i + j + k... () Equation of lines () and () in cartesian form is 8 AB : λ and CD : µ Let L (λ +, λ + 8, λ + ) and M ( µ, µ 7, µ + 6) Direction ratios of LM are λ + µ + 6, λ µ + 5, λ µ. Since LM AB (λ + µ + 6) ( λ µ + 5) + (λ µ ) 0 or, λ + 7µ 0... (5) Again LM CD (λ + µ + 6) + ( λ µ + 5) + (λ µ ) 0 or, 7λ 9µ 0... (6) Solving (5) and (6), we get λ 0, µ 0 L (, 8, ), M (, 7, 6) Hence shortest distance LM ( + ) + (8 + 7) + ( 6) 70 0 units Vector equation of LM is A C 90 L 90 M r i + 8 j + k + t 6 i + 5 j k 8 Note : Cartesian equation of LM is. 6 5 Eample : Prove that the shortest distance between an two opposite edges of a tetrahedron formed b the planes + 0, + 0, + 0, + + a is a. Solution Given planes are (i) (ii) (iii) + + a... (iv) Clearl planes (i), (ii) and (iii) meet at O(0, 0, 0) Let the tetrahedron be OABC Let the equation to one of the pair of opposite edges OA and BC be + 0, () O (0, 0, 0) + 0, + + a... () equation () and () can be epressed in smmetrical form as () C 0 0 a and, 0 d. r. of OA and BC are (, ) and (,, 0).... () Let PQ be the shortest distance between OA and BC having direction cosine (l, m, n) PQ is perpendicular to both OA and BC. l + m n 0 and l m 0 Solving (5) and (6), we get, l m n k (sa) also, l + m + n O A k + k + k k 6 l, m, n B C Shortest distance between OA and BC i.e. PQ The length of projection of OC on PQ O P ( ) l + ( ) m + ( ) n a. a Self practice problems: C Q (0, 0, a) A B B D P Q A D Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page of 77

24 . Find the shortest distance between the lines and 5. Find also 5 FREE Download Stud Package from website: & its equation. Ans., Prove that the shortest distance between the diagonals of a rectangular parallelopiped whose sides are bc ca ab a, b, c and the edges not meeting it are,, b + c c + a a + b 7. Sphere: General equation of a sphere is given b u + v + w + d 0 ( u, v, w) is the centre and u + v + w d is the radius of the sphere. Eample : Find the equation of the sphere having centre at (,, ) and touching the plane Solution: Given plane is () Let H be the centre of the required sphere. Given H (,, ) H Radius of the sphere, HP length of perpendicular from H to plane () P + + Equation of the required sphere is ( ) + ( ) + ( ) or Eample : Find the equation of the sphere if it touches the plane r.( i j k ) 0 and the position vector of its centre is i + 6 j k Solution Given plane is r.( i j k ) 0... () Let H be the centre of the sphere, then OH i + 6 j k c (sa) Radius of the sphere length of perpendicular from H ot plane () c.( i j k ) ( i j k i j k i + 6 j k ).( i j k ) 6 + a (sa) Equation of the required sphere is or r c a i + j + k ( i + 6 j k ) or ( ) + ( 6) + ( + ) 9 or ( ) i + ( 6) j + ( + ) k 9 or 9 ( ) or Eample : Find the equation of the sphere passing through the points (, 0, 0), (0,, 0), (0, 0, ) and whose centre lies on the plane + + Solution Let the equation of the sphere be u + v + w + d 0 Let A (, 0, 0), B (0,, 0), C (0, 0, )... () Since sphere () passes through A, B and C, 9 + 6u + d 0 v + d 0... ()... () w + d 0... () Since centre ( u, v, w) of the sphere lies on plane + + u v w... (5) () () 6u + v 8 () () v + w... (6)... (7) v 8 From (6), u... (8) 6 From (7), w + v... (9) Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page of 77

25 Putting the values of u, v and w in (5), we get v + 8 v v v + 8 v 6 v v From (8), u 6 From (9), w w From (), d v 0 From (), equation of required sphere is or Eample : Find the equation of the sphere with the points (,, ) and (,, ) as the etremities of a diameter. Find the co-ordinates of its centre. Solution Let A (,, ), B (,, ) Equation of the sphere having (,, ) and (,, ) as the etremities of a diameter is ( ) ( ) + ( ) ( ) + ( ) ( ) 0 Here,,,,, required equation of the sphere is ( ) ( ) + ( ) ( ) + ( ) ( ) 0 or Centre of the sphere is middle point of AB 5 Centre is,, Self practice problems:. Find the value of k for which the plane + + k touches the sphere Ans. ±. Find the equation to the sphere passing through (,, ), (, 5, ) and (,, 0) which has its centre in the plane Ans Find the equation of the sphere having centre on the line 0, and passing through the points (0,, ) and (,, ). Ans Find the centre and radius of the circle in which the plane intersects the sphere Ans. units 5. A plane passes through a fied point (a, b, c) and cuts the aes in A, B, C. Show that the locus of the centre of the sphere OABC is + +. Teko Classes, Maths : Suhag R. Karia (S. R. K. Sir), Bhopal Phone : , page 5 of 77