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Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
Three dim. geometry
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Three dim. geometry

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  • 1. THREE–DIMENSIONAL GEOMETRYIntroductionThree dimensional geometry developed accordance to Einstein’sfield equations. It is useful in several branches of science like it isuseful in Electromagnetism. It is used in computer algorithms toconstruct 3D models that can be interactively experienced in virtualreality fashion. These models are used for single view metrology. 3-D Geometry as carrier of information about time by Einstein. 3-DGeometry is extensively used in quantum & black hole theory. Inthis chapter we present a vector–algebra approach to three–dimensional geometry. The aim is to present standard properties oflines and planes, with minimum use of complicated three–dimensional diagrams such as those involving similar triangles. Wesummarize the chapter: Points are defined as ordered triples of realnumbers and the distance between points P1 = (x1, y1, z1) and P2 =(x2, y2, z2) is defined by theformulaP1P2 = (x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2.Directed line segmentsPoints in space Points in space can be represented by vectors asthis is equivalent to a displacement relative to the origin of thecoordinate system.
  • 2. For example the vector shown above can represent the point(5,8,3)Section Formula:(1)Integral division: If R(x, y, z) is point dividing join of P(x1, y1, z1) & Q(x2, y2,z3) in ratio of m : n.Then, x = ,y= ,z=(2)External division: Coordinates of point R which divides join of P(x1, y1, z1) &Q(x2, y2, z2) externally in ratio m : n areIllustration: Show that plane ax + by + cz + d = 0 divides line joining(x1, y1, z1) & (x2, y2, z) in ratio ofAns: Let plane ax + by + cz + d = 0 divides line joining (x1,y1, z1) & (x2, y2, z2) in ratio K : 1 Coordinates of P
  • 3. must satisfy eq. of plane.ax + by + cz + d = 0Direction Cosines:Let is a vector , ,inclination with x, y & z-axis respectively. Then cos , cos & cosare direction cosines of. They denoted bydirection angles.& lies 0 , ,Note: (i) Direction cosines of x-axis are (1, 0, 0)Direction cosines of y-axis are (0, 1, 0)Direction cosines of z-axis are (0, 0, 1)(ii) Suppose OP be any line through origin O which hasdirection l, m, n(r cos , r cos , r cos ) where OP = rcoordinates of P are (r cos , r cos , r cos )or x = lr, y
  • 4. = mr, z = nr(iii) l2+ m2+ n2= 1 Proof: | |=| |=| |2= x2+ y2+ z2= l2| |2+ m2| |2+ n2| |2 l2 + m 2 + n 2 = 1(iv)Direction ratios: Suppose l, m, & n are direction cosines of vector& a, b, c are no.s such that a, b, c are proportional to l, m,n. These a, b, c are direction ratios. =kSuppose a, b, c are direction ratios of vectorhaving direction cosines l, m, n. Then, = l = a, m = b, n = c l2 + m 2 + n 2 = 1 ⇨ a2 2 + b2 2 + c2 2 =1 =l=± ,m=± ,n=±Note: (i) Ifhaving direction cosines l, m, n. Then, l = ,m= ,n=
  • 5. (ii) Direction ratios of line joining two given points(x1, y1, z1) & (x2, y2, z2) is (x2- x1, y2- y1, z2- z1)(iii) If direction ratios ofare a, b, c =(iv) Projection of segment joining points P(x1, y1, z1) andQ(x2, y2, z2) on a line with direction cosines l, m, n, is:(x2- x1)l + (y2- y1)m + (z2- z1)nIllustration: If a line makes angles , &with coordinate axes, prove thatsin2 +sin2+ sin2 =2Ans: Line is making , &with coordinate axes.Then, direction cosines are l = cos, m = cos & n = cos . But l2+ m2+ n2= 1cos2 + cos2 + cos2 = 1,1-sin2 +1-sin2 + 1 - sin2 = 1 sin2 + sin2 + sin2 =2Angle b/w two vectors in terms of direction cosines &
  • 6. Cos = cos = l1l2 + m1m2 + n1n2(ii) IF a1, b1, c1 and a2, b2, c2 are d.r.s of & . Then&cos =cos =Note: (i) If two lines are then, cos = 0 orl1l2+ m1m2+ c1c2=0or a1a2+ b1b2+ c1c2=0(ii)If two lines are || thencos =1 oror
  • 7. Illustration: Find angle b/w lines whose direction cosines are & .Ans: Let be anglecos =l1l2+m1m2+n1n2cos =- =1200Straight lineVectorial eq. of line:Let a line passing through a point of P.V. & || to given line EF( ).Then, eq. of lineProof: Let P be any point on line AP & its P.V. isThen, = = (By law)
  • 8. Note: Eq. of line through origin & || to is =Cartesion eq. of straight line:Passing through point & parallel to given vector .Let line is passing through A(x1, y1, z1) & || to EF whose d.r.s are a,b, c.d.r.s of AP = (x - x1, y - y1, z - z1)d.r.s of EF = (a, b, c) Since EF || AP, then, Eq. of line in parametric form.Illustration: Find eq. of line || to & passing through pt.(1, 2, 3) ?Ans: A(1, 2, 3) = eq. of line passing through & || to =( )+ ( )Vector eq. of line passing through two given points :Vector eq. of line passing through two points whose P.V. S & is: = + ( - )Proof: is collinear with
  • 9. =Cartesian form :Eq. of line passing through (x1, y1, z1) & (x2, y2, z2)D.R.s of AB = (x2- x1, y2- y1, z2- z1)D.R.s of AP = (x - x1, y - y1, z - z1)Since AB || APIllustration: Find cartesian eq. of line are 6x + 2 = 3y - 1 = 2z + 2.Find its direction ratios.Ans: is cartesion eq. of line6x + 2 = 3y - 1 = 2z + 26(x + ) = 3(y - ) = 2(z + 1)
  • 10. on comparing a = ,b= ,c=Angle b/w Two lines: =Angle b/w two lines cos =If =Condition for perpendicularity: . = 0 a1a2+ b1b2+ c1c2 =0Condition for parallelism :Dumb Question: Why angle between pair of linesAns: &
  • 11. Perpendicular distance of point from a line :(a)Castesian form :Suppose L is foot of line of . Since L lies on line AB so,coordinate of L is =i.e. L(x1+ a, y1+ b, z1+ c) direction ratios of PL are: (x1+ a , y1+ b- , z1 + c- )also direction ratios of AB are (a, b, c) Since PL ABa(x1+ a- ) + b( b- ) + c(z1+ c- )=0Vector Form:
  • 12. Since L lies on line AB , P.V. of L = P.V. of line AB = +drs ofSince P.V. of L is +Reflection of point in straight line:Castesian Form:
  • 13. From above, we get coordinate of L(foot of )But L is mid point of PQ = x1+ a , = y1 + b , = z1 + c = 2(x1 + a )- , = 2(y1 + b )- , = 2(z1 + c )-Vector Form:From above, we get , P.V. of L, +Let P.V. of Q isSince L is mid point of PQ
  • 14. Illustration: Find reflection of point P(2, 3, 1) in lineAns:Since L lies on line AB coordinate of L (3 + 2, 2 + 1, 4 - 3)DRs of PL are = (3 + 2 - 2, 2 + 1 - 3, 4 - 3 - 1)= (3 , 2 - 2, 4 - 4) , DRs of AB are (3, 2, 4)Since PL AB3(3 ) + 2(2 - 2) + 4(4 - 4) = 09 +4 - 4 + 16 - 16 = 0 29 = 20 =Since L is mid point of PQSo, =3 + 2, =2 + 1, =4 -3
  • 15. = (6 + 4 - 2), = (4 + 2 - 3), =8 -6-1 =6 + 2, =4 - 1, =8 -7 where =Skew lines: Those lines which do not lies in same plane.Shortest distence b/w two skew lines:The line which is to both line l1 & l2 are c/d line of shortestdistance.Vector Form:Let l1 & l2 are: & respectively.Since is to both l1 & l2 which are parallel to & is || to xLet be unit vector along , then =±
  • 16. PQ = Projection of on = .Dumb Question: How PQ = Projection on ?Ans: Form fig., it is clear that PQ is projection of on . PQ =Cartesian form: Two skew lines &Shortest distance :Condition for lines to intersect:Two lines intersects if shortest distance = 0
  • 17. orShortest distance b/w parallel line:Let l1 & l2 are respectively& BM is shortest distance b/w l1 & l2sin = BM = AB sin =| | sin | x |=| | | | sin( - ) =| || | sin = (| | sin )| |=Dumb Question: Why we have taken sin( - )?Ans:
  • 18. Since direction of vector is opposite to || lines. So, we havetaken ( - ) instead of BM =Illustration: Find shortest distance b/w lines:Ans:On comparing
  • 19. Plane:(i) Eq. of plane passing through a given point (x1, y1, z1) is: a(x - x1) + b(y - y1) + c(z - z1) = 0 where a, b, c constants.Proof: General eq. of plane is ax + by + cz + d = 0....................(i)It is passes through (x1, y1, z1) ax1 + by1 + cz1 + d = 0 ............................ (ii)By (i) - (ii), we get a(x - x1) + b(y - y1) + c(z - z1) = 0Concurrence and CoplanarityFour planes a x+b y+c z+d =0, a x+b y+c z+d =0, a x+b y+c z+d =0and a x+b y+c z+d =0 are concurrent if and only ifFour points (x ,y ,z ), (x ,y ,z ), (x ,y ,z ) and (x ,y ,z ) are coplanarif and only if(Both of these assertions remain true in oblique coordinates.)Intercept form of a plane:eq. of plane of intercepting lengths a, b & c with x, y & z-axisrespectively is,
  • 20. Illustration: A variable plane moves in such a way that sum ofreciprocals of its intercepts on 3 coordinate axes is constant. Showthat plane passes through fixed point.Ans: Let eq. of plane is . Then, intercepts ofplane with axes are: A(a, 0, 0), B(0, b, 0), c(0, 0, c) = constant (k) (given) = 1 & comparing with fixed point x= ,y= ,z=This shows plane passes through fixed point ( , , )Vector eq. of plane passing through a given point & normal to givenvector:Vector eq. of plane passing through u point of P.V. & normal tovector is ( - ). =0Dumb Question: What is normal to vector ?Ans: Plane normal to vector means the every line in plane is tothat given vector.Proof: Let plane passes through A( ) & normal to vector &be P.V. of every point P on plane.
  • 21. Since lies in plane & is normal to plane. . =0 ( - ). =0 ( = - )eq. of plane ( - ). =0Eq. of plane in normal form: Vector eq. of plane normal to unitvector & at O distance d from origin is . = d.Proof:ON is to plane such that & =Since . =0 ( - ). =0 ( -d ).d =0 .d - d2 . =0⇨ d( . ) - d2 = 0 . =d
  • 22. eq. of plane is . =dCartesian Form: Let l, m, n be d.r.s of normal to given plane & P islength of from origin to plane, then eq. of plane is lx + my + nz =P.Eq. of plane passing through 3 given points:Eq. of plane passing through three points A, B, C having P.V.s , , respectivelt. Let be P.V. of any point P is the plane.So, vectors, = - , = - , = - arecoplanarHence, .( x )=0 ( - ).{( - )x( - )} = 0 ( - ).( x - x - x - x )=0 ( - ).( x + x + x )=0 .( x+ x + x )= .( x )+ .( x )+ .( x ) [ ]+[ ]+[ ]=[ ]Note: If P is length of from origin on this plane,
  • 23. then P = n=| x + x + x |Eq. of plane that passes through a point A with position vector &is || to given vectors & :Derivation:Let be P.V. of any point P in planeThen, = - = -Since x are || to plane.So, vector - , & are coplanar ( - ).( x )=0 .( x ) = .( x ) [ ]=[ ]Cartesian form: Eq. of plane passing through a point (x1, y1, z1) & || to two lineshaving direction ratios ( 1, 1, 1) &( 2, 2, 2) is:Illustration : Find eq. of plane passing through pointsP(1, 1, 1), Q(3, -1, 2), R(-3, 5, -4).Ans: Let eq. of plane is ax + by + cz + d = 0
  • 24. It passes through P(1, 1, 1). So,eq. of plane a(x - 1) + b(y - 1) + c(z - 1) = 0 .............................. (i)But it also passes through Q & R 2a - 2b + c = 0] x (- 2) - 4a + 4b - 5c = 0Similarly a = 6, b = 6, Putting these in eq. (i)We get, 6(x - 1) + 6(y - 1) = 0 x+y=2Angle b/w two planes : . = d1 . = d2 are two planers, then cos =Note: Angle b/w planes is defined as angle b/w their normals.Cartesion Form : Let planes are a1x + b1y + c1z + d1= 0 & a2x + b2y +c2z + d2= 0 cos =Condition for : . =0 or a1a2+ b1b2+ c1c2 = 0Condition for parallelism: orAngle b/w line & a plane:
  • 25. If , , be direction ratios of line & ax + by + cz + d = 0 be eq. ofplane in which normal has d.rs a, b, c. Ѳ is angle b/w line & plane. cos(900- )=Vector form: If is angle b/w line & plane . =dsin =Illustration : Find angle b/w line& 3x + 2y - 2z + 3 = 0.Ans: D.rs of line are 2, 3, 2 & d.rs of normal to plane are 3, 2, - 2 sin = = sin -1Eq. of plane passing through the Line of Intersection of planesa1x + b1y + c1z + d1= 0 & a2x + b2y + c2z + d2= 0 is(a1x + b1y + c1z + d1 ) + k(a2x + b2y + c2z + d2) = 0Dumb Question : How this eq. is that of required plane ?Ans: Let P( , , ) be point on line of intersection of two planes.
  • 26. So,it lies on both the planes.i.e. a1 + b1 + c1 + d1= 0& a2 + b2 + c2 + d2= 0So, point P( , , ) should lie on required plane.i.e. (a1 + b1+ c1 + d1) + k(a2 + b2 + c2 + d2) = 0 P( , , ) lieson. Plane (a1x + b1y + c1z + d1) + k(a2x + b2y + c2z + d2) = 0Vector Form: . = d1 & . = d2So, eq. of required plane is ( . - d1) + k( . - d2) = 0Illustration: Find eq. of plane constaining line of intersection ofplane x - y + z + 7 = 0 and x + 3y + 2z + 5 = 0 &passing through (1, 2, 2).Ans: Eq. of plane through line of intersection of given planes is, (x - y + z + 7) + (x + 3y + 2z + 5) = 0 .............. (i)It passes through (1, 2, 3) (1 - 2 + 2 + 7) + (1 + 3 x 2 + 2 x 2 + 5) = 0 8+ (7 + 4 + 5) =Putting =- in eq. (i)We get, (x - y + z + 7) + (- )(x + 3y + 2z +5) = 02x - 2y + 2z + 14 - x - 3y - 2z - 5 = 0 ⇨ x - 5y + 9 = 0Two sides of plane :If ax + by + cz + d = 0 be a plane then points (x1, y1, z1) & (x2, y2, z2)are points lies on Same side if >0&opposite side if <0**Distance of point from a plane:
  • 27. Length of from a point having P.V. to plane . = d isgiven by P=Proof: PM is length of from P to plane. Since line PM passesthrough P( ) & || to vector which is normal to plane. eq.of line = x .................................... (i)Since point M is intersection of line & plane. So, it lies on line aswell as planePutting in eq. (i) = + = P.V. of M - P.V. of P = + -Dumb Question: How P.V. of M is +
  • 28. Ans: M lies on line as well as plane. On solving value offor line eq. We get P.V. of M.So, this is P.V. of M PM = | |=Cartesian Form : Length of from point P(x1, y1, z1) to plane ax + by+ cz + d = 0, Then eq. of PM is =rDumb Question: How this eq. of PM comes ? Ans: It is passes through point (x1, y1, z1) & || to normal of planeso, we get this eq.]Coordinates of any point on PM are (x1+ ar, y1+ br, z1+ cr)But this also coordinate of M & M also lies on plane a(x1+ ar) + b(y1+ br) + c(z1+ cr) + d = 0i.e. r = PMDistance b/w parallel planes:Distance b/w || planes is difference of length of from origin totwo planes.Let ax + by + cz + d1= 0 & ax + by + cz + d2= 0 D=
  • 29. Vector Form: . = d1 & . =d2Illustration : Find distance b/w parallel planesx + 2y + 2z + 2 = 0 & 2x + 4y + 4z + 3 = 0Ans: Distance b/w ax + by + cz + d1= 0 & ax + by + cz + d2 = 0 isx + 2y + 2z + 2 = 0 & x + 2y + 2z + =0Equation of planes Bisecting Angle b/w Two planes:Eq. of planes bisceting angle b/w planes, a1x + b1y + c1z + d1 = 0 anda2x + b2y + c2z + d2 = 0 is,Proof: Let P(x, y, z) be point on plane bisecting angle b/w twoplanes PL & PM is length of from P to planes.By property of angle bisector.PL = PM
  • 30. Dumb Question: How PL = PM ?Ans: In fig 1. OPL & OPM are congruent by AAA symmetry. So, allsides are equal. So, PL = PM.]Note: (i) Eq. of bisector of angle b/w two planes containing origin is(ii) Bisector of acute & obtuse angles b/w:Let a1x + b1y + c1z + d1 = 0& a2x + b2y + c2z + d2 = 0 where d1, d2 > 0(a) If a1a2 + b1b2 + c1c2 > 0, origin lies in obtuse angle bisector & eq.of bisector of acute angle is(b) If a1a2 + b1b2 + c1c2 < 0, origin lies in atute angle bisector & eq. ofacute angle bisector isIllustration: Find eq. of bisector planes of angle b/w planes 2x + y -2z + 3 = 0 & 3x + 2y - 6z + 8 = 0 specify obtuse & acute anglebisectors.Ans: 2x + y - 2z + 3 = 0 & 3x + 2y - 6z + 8 = 0 where d1, d2 > 0
  • 31. Now a1a2 + b1b2 + c1c2 = 2 x 3 + 1 x 2 + 2 x 6 > 0 ............................ (i)(i) is obtuse angle bisector plane& ............................ (ii)(ii) is acute angle bisector plane. 14x + 7y - 14z + 21 = ± 9x + 6y - 18z + 24For obtuse angle bisector plane, Take -ve sign5x + y + 4z - 3 = 0Intersection of line & plane: = r & plane ax + by + cz + d = 0Let P be point of intersection coordinate of P(x1 + lr, y1 + mr, z1 + nr)But it satisfy eq. of plane.a(x1 + lr) + b(y1 + mr) + c(z1 + nr)
  • 32. Condition for line to be || to a plane: be || to plane ax + by + cz + d = 0if al + bm + cm = 0 or sin = 0Condition for a line to lie in the plane:line lie in plane ax + by + cz + d = 0if al + bm + cn = 0 & ax1 + by1 + cz1 + d = 0Dumb Question: How line be || to plane if al + bm + cn = 0 ?Ans: If is angle b/w line & plane, thensin = [Previously desired]line is || to plane if sin = 0 or al + bm + cn = 0Easy Type Q.1. Find ratio in which 2x + 3y + 5z = 1 divides linejoining the points (1, 0, 2) & (1, 2, 5)Ans: Let the ratio be k = 1 at point PThen, P = must satisfy 2x + 3y + 5z = 1 2(K + 1) + 6K + 25K + 10 = K + 1 K + 1 + 6K + 25K + 10 = 0 32K + 11 = 0 K=Thus line divides externally in rstio of 11:32Q.2. What are direction cosines of a line which is equally inclined toaxes ?
  • 33. Ans: If , , are angles, if line is equally inclined = = l2 + m2 + n2 cos2 + cos2 + cos2 =1 3 cos2 =1 cos =± = cos = cos d.cs arw ( , , ) or (- ,- ,- )Q.3. Find direction cosines of line which is to lines with d.r (1, -2, -2) & (0, 2, 1)Ans: Let l, m, n be d.cs line 1 to given line, d.cs are proportional tod.rs d.cr of lines are ( ,-2 ,-2 ) & (0, 2µ, µ)Since line is to given lines l + (- 2m ) + (- 2n )=0 l - 2m - 2n = 0& 0 + 2mµ + nµ = 0 2m + n = 0By cross multiplication, l = 2R, m = - R & n = 2R , l2 + m2 + n2 = 1 4R2 + R2 + 4R2 = 1 9R2 = 1
  • 34. Q.4. Prove that line x = ay + b, z = cy + d & x = ay + b, z = cy + d are if aa + cc + 1 = 0Ans: Ist line is x = ay + b, z = cy + d = y, =y = = ...........(i)and IInd line ................................. (ii)These lines are if aa + |x| + cc = 0 aa + cc + 1 = 0Dumb Question: For what value of k, lines & intersect?Ans: =and =µSince these lines have point of intersection in common. then(2 + 1, 3 - 1, 4 + 1) = (µ + 3, 2µ + k, µ)or 2 + 1 = µ + 3 ....... (i), 3 - 1 = 2µ + k ....... (ii) & 4 +1=µ....... (iii) on solving (i) & (iii), we get = - 3/2 & µ = - 5Substituting in (ii) - - 1 = - 10 + k k=Q.6. Show that points and areequidistant from plane .( )+9=0Ans: .( )=-9
  • 35. length of from ( )So, length of is equal.Q.7. Find the point in which the plane; =( - ) + m( + + ) + n( + - ) is cut by line through point 2 +3 &parallel to .Ans: eq. of line through point 2 +3 & || to = (2 +3 )+Since line cuts plane For one point, (2 +3 )+ = - + m( + + ) + n( + - )Equating coeff. of , & on both side, we get m + n = 1, m - n = 4, - m + n = m= ,n= & =-4
  • 36. P.V. of required point is: 2 +3 -4Q.8. Show that line of intersection of planes .( )=0 & .( ) = 0 is equally inclined to & .Ans: Note: Line of intersection of two planes will be to normals tothe planes. Hence it is || to vectorNow, &So, line is equally inclined to &Q.9. Projection of line segment on 3 axes 4, 5 & 13 respectively.Find length & direction cosines of line segment.Ans:
  • 37. = (x2- x1) + (y2- y1) + (z2- z1) Projection of on x-axis = . = (x2- x1) = 4Projection of on y-axis = . = (y2- y1) = 5Projection of on z-axis = . = (z2- z1) = 13 Length of AB = d.rs = , &Q.10. Find locus of mid point of chords of sphere r2- 2 . + k = 0 ifchords being drawn || to vector .Ans: r2- 2 . + k = 0 is sphere of centre chord AB || .
  • 38. locus of M is ( - ). =0 . = . represents aplane.Dumb Question: Why is same as ?Ans: Since || . So, if is . So, it will also toMedium TypeQ.1. If a variable plane forms a tetrahedron of constant volume27k3 with coordinate planes, find locus of centroid of tetrahedron.Ans: sLet variable plane cuts coordinate axes at A(a, 0, 0), B(0, b, 0),C(0, 0, c)Then, eq. of plane will be =1Let P( , , ) be centroid of terahedron OABC, then, = , = , =Dumb Question : How this centroid tetrahedron OABC comes = , = , = ? Ans: Centroid of tetrahedrold
  • 39. where (x1, y1, z1), (x2, y2, z2), (x3, y3, z3) & (x4, y4, z4) are coordinateof tetrahedron.]Volume of tetrahedron = (Area of AOB).OC 27k3= (ab)c = 27k3=Required locus of P( , , ) isQ.2. Find vector eq. of straight line passing through intersection ofplane , , are non coplanar vectors.Ans: At points of intersection of two planes.Since , , are non coplanar, then - 1+ µ1 - µ2= 0, 1 + 1- 2- µ2= 0, µ1- 1 + 2= 0On solving, we get 1 = 0 & µ1 = µ2 = + 1( - ) + µ1( + )Since 1 =0 = + µ1( + ) is required eq. of straight line.Q.3. Prove that three lines from O with direction cosines l1, m1, n1;
  • 40. l2, m2, n2; l3, m3, n3 are coplaner if l1(m2n3 - n2m3) + m1(n2l3 - l2n3) + n1(l2m3 - l3m2) = 0Ans: Note: Three given lines are coplanar if they have commonperpendicular. Let d.cs of common be l, m, n ll1 + mm1 + nn1 = 0 ............................. (i) ll2 + mm2 + nn2 = 0 ............................. (ii) ll3 + mm3 + nn3 = 0 ............................. (iii)Soplving (ii) & (iii) by cross multiplication ......l = k(m2n3 - n2m3), m = k(n2l3 - n3l2), n = k(l2m3 - l3m2)Substituting in (i), we get k(m2n3 - n2m3)l1 + k(n2l3 - n3l2)m1 + k(l2m3 - l3m2)n1 = 0 l1(m2n3 - n2m3) + m1(n2l3 - n3l2) + n1(l2m3 - l3m2) = = 0Q.4. Solve the equation x +y +z =Ans: x +y +z = .................................. (i)Taking dot product by x , we get x .( x )+y .( x )+z .( x )= .( x ) x[ ]=[ ] x=
  • 41. & z= x +y +z = [ ] +[ ] +[ ] =[ ] isrequired solution.Q.5. If planes x - cy - bz = 0, cx - y + az = 0 & bx + ay - z = 0 passthrough a straight line, the find value of a2 + b2 + c2 + 2abc.Ans: Givem planes are: x - cy - bz = 0 ............................. (i) cx - y + az = 0 ............................. (ii)& bx + ay - z = 0 ............................. (iii)Eq. of plane passing through line of intersection of plane (i) & (ii) is (x - cy - bz) + (cx - y + az) = 0 x(1 + c) - (c + ) + z(- b + a ) = 0 ............................. (iv)If plane (iv) & (iii) are same, then,
  • 42. and a - a3 + bc - a2bc = a2bc + ac2 + ab2 + bc a(2abc + c2 + b2 + a2 - 1) = 0 a2 + b2 + c2 + 2abc = 1ANOTHER METHODSThe parametric vector equationLines can be specified in a variety of ways. One way is described asfollows. Select a point P0 on the line l, and a non-zero vector vparallel to the line. The line l is then the unique line passingthrough P0 and parallel to v.Now P lies on l if and only if is parallel to v. As
  • 43. this condition is equivalent to r - r0 = tv for some scalar t. Theequationis called a parametric vector equation of the line l. (It is not unique,as a different point P0 on the line could have been chosen, changingr0, and v can be replaced by any other non-zero vector parallel to l.)Each value of the parameter t determines a unique point P, withposition vector r = r0 + tv, on the line l. As t takes all possible values,P takes all possible positions on the line l. The parametric scalar equationsSuppose that the point P0 has coordinates x0,y0,z0, and the non-zerovector v has components a,b,c.Let x,y,z be the coordinates of an arbitrary point P on the line l.Thenand the position vectors of P0 and P are given byandrespectively. The parametric vector equation may be rewritten asthat is,
  • 44. with t R. These equations are called parametric scalar equationsof the line. The Cartesian equationsFor each value of the parameter t, there is a point P withcoordinateson the line.If a,b,c are all non-zero, then we can eliminate the parameter t inthe parametric (scalar) equations to obtainWe call The vector equation of a planeA plane can be described in many ways. The plane, for example,can be specified by three non-collinear points of the plane: there isa unique plane containing a given set of three non-collinear pointsin space.An alternative way to specify a plane is given as follows.Select a point P0 in the plane. There is a unique line through P0perpendicular to the plane. This line is called the normal to to the
  • 45. plane at P0. A vector n 0 parallel to this normal is called a normalvector for the plane.There is a unique plane which passes through P0 and has n as anormal vector.Now P lies in the plane through P0 perpendicular to n if and only if and n are perpendicular.As = r - r0, this condition is equivalent to The Cartesian equation of a planeSuppose that P0 has coordinates x0,y0,z0 and n has componentsa,b,c.
  • 46. Let P, with coordinates x,y,z, be an arbitrary point on the plane.The position vectors of P0 and P areandSubstitutinginto the vector equation, we obtainwhich, when multiplied out, givesThis is called a Cartesian equation of the plane.It simplifies to
  • 47. where d is the constant ax0 + by0 + cz0.An equation of the formwhere a,b,c and d are constants and not all a,b,c are zero, can betaken to be an equation of a plane in space.The coefficients a, b and c are the components of a normal vectorfor the plane described by the equation. The Cartesian equation of a planeSuppose that P0 has coordinates x0,y0,z0 and n has componentsa,b,c.Let P, with coordinates x,y,z, be an arbitrary point on the plane.The position vectors of P0 and P are
  • 48. andSubstitutinginto the vector equation, we obtainwhich, when multiplied out, givesThis is called a Cartesian equation of the plane.It simplifies towhere d is the constant ax0 + by0 + cz0.An equation of the formwhere a,b,c and d are constants and not all a,b,c are zero, can betaken to be an equation of a plane in space.The coefficients a, b and c are the components of a normal vectorfor the plane described by the equation.
  • 49. EXTRA QUESTIONS:Class – XII Subject – Mathematics (Three Dimensional Geometry)1. Find the d.c’s of X, Y and Z-axis.[example 1,0,0(x-axis)0,1,0(y-axis]2. The equation of a line is given by (4 – x)/2 = (y + 3)/3 = (z + 2)/6.Write the direction cosines of a line parallel to the aboveline. [Ans: – 2/7, 3/7, 6/7]3. The equation of a line is (2x – 5)/4 = (y + 4)/3 = (6 – z)/6. Find thedirection cosines of a line parallel to this line. [ Ans : 2/7, 3/7, – 6/7]4. Find coordinates of the foot of the per. drawn from the origin tothe plane 2x – 3y + 4z -6 = 0. [example ans. 12/29, -18/29, 24/29]5. Find the angle between the line (x + 1)/2 = y/3 = (z-3)/6 and theplane 10x + 2y – 11z = 3.[example ans. sinφ = | | = 8/21]*6. Find the image of the point (1, 6, 3) in the line x = (y – 1)/2 = (z –2)/3. [ Hint: it will be found that foot of per. from P to the line is N(1,3,5). If Q( ) is the image of P then N is the mid point of PQ⇨ ⇨ Q(1, 0, 7)]7. Find the foot of the perpendicular drawn from the point P(1, 6, 3)on the line x/1 = (y – 1)/2 = (z – 2)/3. Also find the distance fromP. [ Ans : √13 units., eqn. Of line is (x-1)/0=(y-6)/-3=(z+1)/2]8. Find the length and the foot of the perpendicular drawn from thepoint (2, – 1 , 5) to the line (x – 11)/10 = (y + 2)/– 4 = (z + 8)/ –11. [ Ans. = Point (1, 2, 3); distance = √14 units.]*9. Show that the angles between the diagonals of a cube is cos-1 (1/3).
  • 50. Ans. (same as example. 26 of misc.) Let the length of cube be a , infig. Of example : A,B,C are on x-axis, y-axis, z-axis resp.& F,G,D arein yz, xz ,xy –axis , O is origin , d.r’s of OE (E (a,a,a)) & AF are a,a,a &-a,a,a resp. ∴ cosѲ =(-a2+a2+a2)/√3a.√3a = 1/3(angle b/w two lines)10. Find the length of the perpendicular drawn from the point (2, 3,7) to the plane 3x – y – z = 7 . Also find the coordinates of the footof the perpendicular. [ same as Q. 8 ]*11. Find the point on the line (x + 2)/3 = (y + 1)/2 = (z – 3)/2 at adistance 3√2 from the point (1, 2, 3).Ans. (x + 2)/3 = (y + 1)/2 = (z – 3)/2 =k , any point on the line A(3k-2,2k-1, 2k+3) , if its distance from B(1,2,3) is 3√2, then AB² = (3√2)²⇨ K= 0 OR 30/17 , ∴ point A(-2,-1,3) OR A(56/17,43/17,111/17) [byputting value of k in A]12. Find the equation of the perpendicular drawn from the point (2,4, – 1) to the line (x + 5)/1 = (y + 3)/4 = (z – 6)/–9 .[ Ans : (x – 2)/6 = (y – 4)/3 = (z + 1)/2]*13. Find the equation of the line passing through the point (-1, 3, -2) and perpendicular to the lines x = y/2 = z/3 and (x + 2)/(-3) = (y –1)/2 = (z + 1)/5.Ans. eqn. Of line (x+1)/a=(y-3)/b=(z+2)/c ........(i) ,where a,b,c ared.r’s , since (i) is per. to lines ∴ a+2b+3c=0 & -3a+2b+5c=0, aftersolving , we get a/4=b/-14=c/8 =k(say) , so eqn. Of line (by puttingthe values of a=4k, b=-14k,c=8k) is (x+1)/4 = (y-3)/-14 = (z+2)/814. Find the foot of the per. drawn from the point (0, 2, 3) on theline (x + 3)/5 = (y – 1)/2 = (z + 4)/3 Also, find the length of theperpendicular. [ans..(2,3,-1), use distance formula, length= √21]
  • 51. *15. Find the equation of the plane passing through the lineintersection of the planes x – 2y + z = 1 and 2x + y + z = 8 andparallel to the line with direction ratios (1, 2, 1) Also, find theperpendicular distance of the point P(3, 1, 2) from this plane.Ans. equation of the plane passing through the line intersection ofthe planes x – 2y + z -1 +k(2x + y + z – 8)=0 or (1+2k)x+(-2+k)y+(1+k)z-1-8k=0 ........(i) is parallel to line ∴ (1+2k).1+(-2+k).2+(1+k).1=0 ⇨ k= 2/5(normal to the plane must be per. to theline), put k in (i) ⇨ 9x-8y+7z-21=0, per. dis. From P Is 22/√(194)*19. Find the shortest distance between the following lines :(x – 3)/1 = (y – 5)/–2 = (z – 7)/1 and(x + 1)/7 = (y + 1)/–6 = (z + 1)/1.[ncert Ans : 2√29 , use formula ]20. Find the equation of the plane passing through the points (1, 2,3) and (0, – 1, 0) and parallel to the line (x – 1)/2 = (y + 2)/3 = z/–3.[Ans :same as Q. 26 , 6x – 3y + z = 3]21. Find the shortest distance between the following pairs of lineswhose cartesian equations are : (x -1)/2 = (y + 1)/3 = z and (x + 1)/3= (y – 2) , z = 2. [ SAME as Q. 19 ]*22. Find the distance of the point P(2, 3, 4) from the plane 3x + 2y+2 z + 5=0 measured parallel to the line (x+3)/3 = (y-2)/6 = z/2Ans. Any point on line (x+3)/3 = (y-2)/6 = z/2=k isQ(3k+2,6k+3,2k+4) It lies on plane ⇨ k=-1 ∴ point Q( -1,-3,2),PQ=723. Find the equation of the plane passing through the points (0, –1, – 1), (4, 5, 1) and (3, 9, 4). [ Ans : 5x – 7y + 11z + 4 = 0]
  • 52. *24. Find the equation of the plane passing through the point (–1, –1, 2) and perpendicular to each of the following planes :2x + 3y – 3z = 2 and 5x – 4y + z = 6.[ Ans : 9x + 17y + 23z – 20 = 0, same as example of misc. Eqn. Ofplane a(x+1)+b(y+1)+c(z-2)=0........(i) By condition ofperpendicularity to the plane (i) with the planes 2a+3b-3c=0 & 5a-4b+c =0 , after solving , we get a=9c/23 & b=17c/23, put in (i) ]25. Find the equation of the plane passing through the point (1, 1, -1) and perpendicular to the planes x + 2y + 3z – 7=0 and 2x – 3y +4z = 0. [ SAME as Q. 24]26. Find the equation of the plane passing through the points (3, 4,1) and (0, 1, 0) and parallel to the line (x + 3)/2 = (y – 3)/7 = (z –2)/5.Ans : 8x – 13y + 15z + 13 = 0, eqn. Of plane a(x-3)+b(y-4)+c(z-1)=0.......(i), It passes through (0,1,0) ,then -3a-3b-c=0 // to line ∴2a+7b+5c=0, after solving a=8c/15, b=-13c/15, put in (i)*27. Find the image of the point P(1,2,3) in the plane x+2y+4Z=38[ Ans :Let M is foot of per. (mid point of PQ) , if Q ( ) , d.r’s ofa normal to the plane are 1,2,4, eqn. Of line PM is (x-1)/1=(y-2)/2=(z-3)/4 = k ⇨ any point on a line Q(1+K,2+2K,3+4K) ∴M{(2+k)/2,(4+2k)/2,(6+4k)/2} put in plane ⇨ k=2 ∴ Q(3,6,11) ]28. Find the co-ordinates of the image of the point (1, 3, 4) in theplane 2x – y + z + 3 = 0. [Ans : (–3, 5, 2)]*29. Find the distance between the point P(6, 5, 9) and the planedetermined by the points A(3, – 1 , 2), B(5, 2, 4) and C(– 1, – 1 , 6).Ans. [example28 of misc. Ans: 6/√(34),we can find eqn. Of planethrough A,B,C and find distance of P from plane or can be found asPD = . ( proj. Of on , D is foot of per. fromP) or find eqn. of plane passes through A,B,C , then find dist. ]
  • 53. 30. Find the co-ordinates of the point where the line (x + 1)/2 = (y +2)/3 = (z + 3)/4 meets the plane x + y + 4z = 6. [Ans: P(1, 1, 1)]*31. Find the distance of the point A(– 2, 3 – 4) from the line(x + 2)/3 = (2y + 3)/4 = (3z + 4)/5measured parallel to the plane 4x + 12y – 3z + 1 = 0. [Ans: |AP|= 17/2, eqn. Can be written as (x + 2)/3 = (y + 3/2)/2 =(z + 4/3)/5/3 = t ⇨ P{ -2+3t, (-3/2)+2t, (-4/3)+5t/3} be any point online D.R’S of AP ,3t, 2t-(9/2), (5t/3)+(8/3)} , since AP is parallel toplane 4.3t+ (12).[ 2t-(9/2)]+ (-3). (5t/3)+(8/3) = 0 ⇨ t=2 ∴ P(4,5/2,2) ]Ques. 32 Find the distance of the point P(-2, 3, -4) from the line(x+2)/3 = (2y+3)/4 = (3z+4)/6 measured // to the plane 4x + 12y -3 z+ 1=0.Ans. same as Q. 22, point on lineQ(3k-2,2k-(3/2),(5k/3)-(4/3)), d.r’sof PQ are (3k,2k-(9/2),(5k/3)+(8/3)) , since PQ is // to plane ∴ PQ isper. to normal to the plane ⇨ (3k).4+{2k-(9/2)}.12+{(5k/3)+(8/3)}.(-3)=0 ∴ k= 2 ⇨ Q(4,5/2,2), PQ = 17/2Question 33 find whether the lines = (i-j-k) +λ (2i+j) and =(2i-j) +μ (i+j-k) intersect or not . if intersecting, find their point ofintersection.Ans. - = I +k , = = -i+2j+k , itsmagnitude is √6Shortest distance = = 0/√6 = 0Hence lines intersect. Point of intersection is given by(i-j-k) +λ (2i+j) = (2i-j) +μ (i+j-k) ⇨ (1+2λ) = 2+μ , -1+λ - -1+μand -1 = -μ ⇨ λ = 1 and μ =1 satisfy μ = 2λ -1 , put the values of
  • 54. λ and μ in given lines , we obtain the position vector of point ofintersection of the two given lines as 3i- k i.e., the point ofintersection is (3,o,-1).Ques. 34 Show that the four points (0,-1,-1), (4,5,1), (3,9,4) and (-4,4,4) are coplanar. Also find the eqn. of the plane containing them.Ans. Eqn. of plane passes through (0,-1,-1) is a(x-0)+b(y+1)+(z+1)=0It passes through (4,5,1), (3,9,4), we get 4a+6b+2c=0, 3a+10b+5c=0⇨ eqn. of plane is 5x-7y+11z+4=0 and (-4,4,4) will satisff the eqn. ofplane to be coplanar.Q. 35 Show that lines (x+3)/-3 = (y-1)/1 = (z-5)/5 & (x+1)/-1=(y-2)/2=(z-5)/5 are coplanar . find the eqn. of plane.Ans.(as example 21), The given lines are coplanar if == 0 , so lines are coplanar. Eqn of plane = == x-2y+z=0.Q.36 Find the shortest distance and eqn. Of shortest distance b/wthe lines (x-6)/3 = (y-7)/-1 = (z-4)/1 & x/-3 = (y+9)/2 = (z-2)/4 . S (3p+6,-p+7,p+4) ......(i) D (-3q,2q-9,4q+2) .......(ii)
  • 55. Ans. let SD be the shortest dist. , then co-ordinates of eqn.(i) are (3p+6, -p+7, p+4) & eqn.(ii) are (-3q, 2q-9, 4q+2)d.c’s of SD are proportional to -3q-3p-6, 2q+p-16, 4q-p-2 and if SD isat right angles to both lines , we get 3.( -3q-3p-6)+(-1).( 2q+p-16)+ 1.(4q-p-2)=0 and -3.(-3q-3p-6)+2.( 2q+p-16)+4.( 4q-p-2) =0 ⇨7q+11p+4=0 & 29q+7p-22=0 ⇨ p=-1 , q=1 and SD=3√(30) , eqn. Willbe (x-3)/-6 = (y-8)/-15 = (z-3)/3Q.37 Find the shortest distance and eqn. b/w the lines = (8+3λ)i-(9+16λ)j+(10+7λ)k & = (15i+29j+5k)+μ(3i+8j-5k).Ans. let SD be the shortest dist. b/w two lines , let position vectorof S be (8i-9j+10k)+λ(3i-16j+7k) and position vector of D is(15i+29j+5k+μ(3i+8j-5k), = = (3μ-3λ+7)i+(8μ+16λ+38)j+(-7λ-5μ-5)k since SD is perpendicular on line (i) & (ii) then*(3μ-3λ+7)i+(8μ+16λ+38)j+(-7λ-5μ-5)k]. (3i-16j+7k)=0 ⇨157λ+77μ+311=0...(iii) & 77λ+49μ+175=0...(iv), by solving λ = -1,μ=-2∴ = = 4i+6j+12k ⇨| |= 14 & eqn. Is =(5i+7j+3k)+t[(9i+13j+15k)-(5i+7j+3k)] = (5i+7j+3k)+t(4i+6j+12k)where =5i+7j+3k, =9i+13j+15k & use = +t( ).EX.11.2(ncert) QUESTION 5 Find the equation of the line in vector and in Cartesianform that passes through the point with position vectorand is in the direction .Ans.: Eqn of line passing through apoint = , parallel to = is = +λ(vector form)Cartesian form of line is xi+yj+zk = ( ) +λ ( )=(2+λ)i+(-1+2λ)j+(4-λ)k ⇨ λ = (x-2)/1 =(y+1)/2 = (z-4)/-1
  • 56. Ques. 7 The Cartesian equation of a line is . Write itsvector formAns. vector form is = +λ = (5i-4j+6k) +λ (3i+7j+2k)Ques.9 Find the vector and the Cartesian equations of the line thatpasses through the points (3, −2, −5), (3, −2, 6).Ans. vector form of a line is = +λ =(3i-2j-5k)+λ( 11k) =(3-3)i+(-2+2)j+(6+5)k)Cartesian form of a line is (x-x1)/a = (y-y1)/b = (z-z1)/c (x-3)/0 = (y+2)/0 = (z+5)/11Question12 Find the values of p so the line and are at right angles.Ans: write above lines in standard form , the d,r’s of given lines are-3, 2p/7, 2 & -3p/7, 1, -5. are perpendicular to each other, hencetheir dot product should be equal to 0 ⇨ p=70/11.Ques. 13 Show that the lines and areperpendicular to each other.Ans.: diff. Method , let , be two vwctors parallel to given lines = 7i-5j+k, = i+2j+3k⇨ cosφ =( . )/ .| | = п/2Ques.14 Find the shortest distance between the lines
  • 57. Ans.: = + & = + , shortest distant (d) =Nr. = (i-3j-2k).(-3i+3k) = -9 & Dr. = 3√2, so d= 3√2/2Ques.16 Find the shortest distance between the lines whose vectorequations areANS.: Same as above Nr. = (3i+3j+3k).(-9i+3j+9k) = 9, Dr. = 3√19.EX 11.3 Ques.4 In the following cases, find the coordinates of thefoot of the perpendicular drawn from the origin.(a) (b)(c) (d) [ co-ordinates of foot of per.(ld, md, nd)]Ans. (a) let foot of per.P (x1,y1,z1) & eqn. Of plane in normal form Lx+my+nz=d ⇨2x/√29 + 3y/√29 + 4z/√29 = 12/√29, so, 2/√29,3/√29 , 4/√29 are d.c’s of OP (O is orgin, P is foot of per.) & D.R’S ofOP is (x1,y1,z1) , they are proportional to each other (d.c’s & d.r’s ofa line) ⇨ = = = k , put the values of x1,y1,z1 in plane , weget k = 12/√29 ⇨foot of per. Is ( 24/29, 36/29, 48/29)(b) eqn. Of plane is 3y/5+ 4z/5 = 6/5, d.c’s of OP are 0, 3/5, 4/5 &d.r’s of OP are x1,y1,z1 , they are prop. ⇨ = = = k , we getk= 6/5 , foot of per. (0, 18/25, 24/25)
  • 58. (c) k= 1/√3 , foot of per. (1/3, 1/3, 1/3) (d) k= -8/5 , foot of per. (0,-8/5,0)Ques.5 Find the vector and Cartesian equation of the planes(a) that passes through the point (1, 0, −2) and the normal to theplane is .(b) that passes through the point (1, 4, 6) and the normal vector tothe plane is .Ans. (a) eqn. Of plane (vector ) ( - ). =0 ⇨[ - (i-2k)].(i+j-k) =0⇨ .( i+j-k) – (i-2k). (i+j-k)=0 ⇨ . (i+j-k)-3 = 0Cartesian form x+y-z=3 [ eqn. Of plane 1(x-1)+1(y-0)-1(z+2)=0(b) ) eqn. Of plane (vector ) ( - ). =0 ⇨[ - (i+4j+6k)].(i-2j+k) =0⇨ .( i-2j+k) – (i+4j+6k. (i-2j+k) =0 ⇨ . (i-2j+k)+1 = 0Cartesian form is x-2y+z+1-0 [ eqn. Of plane 1(x-1)-2(y-4)+1(z-6)=0]Ques.6 Find the equations of the planes that passes through threepoints.(a) (1, 1, −1), (6, 4, −5), (−4, −2, 3)(b) (1, 1, 0), (1, 2, 1), (−2, 2, −1)Ans. let the eqn. Of plane passes through (1,1,-1) bea(x-1)+b(y-1)+c(z+1)=0......(i) , it passes through (6, 4, −5), (−4, −2, 3)⇨ 5a+3b-4c=0....(ii) & 5a+3b-4c=0....(iii) , from (ii) & (iii) ⇨nounique plane can be drawn.(b) let the eqn. Of plane passes through (1,1,0) bea(x-1)+b(y-1)+cz=0......(i) , it passes through (1,2,1), (-2,2,-1)
  • 59. ⇨ b+c=0....(ii) & 3a-b+c=0....(iii) , from (ii) & (iii) a = -2c/3, put theValues of a & b in terms of c in (i) ⇨ 2x+3y-3z=5.Ques. 10 Find the vector eqn. Of the plane through the line ofintersection of the planes . (2i+2j-3k) = 7 & . (2i+5j+3k) = 9 andthrough the plane (2, 1, 3).Ans. Eqn. Of the plane through the line of intersection of the planes . (2i+2j-3k) - 7 +λ* . (2i+5j+3k) – 9]=0⇨ . *(2+2λ)i+(2+5λ)j+(-3+3λ)k+-7-9λ = 0 .........(i) It passes (2,1,3)⇨(2i+j+3k). *(2+2λ)i+(2+5λ)j+(-3+3λ)k+-7-9λ = 0 ⇨ λ =10/9, put in (i)We get, . (38i+68j+3k) = 153.Ques.11 Find the equation of the plane through the line ofintersection of the planes and which isperpendicular to the planeAns. Eqn. of the plane through the line of intersection of the planesx+y+z-1+λ(2x+3y+4z-5) = 0 or (1+2λ)x+(1+3λ)y+(1+4λ)z-1-5λ=0.....(i)D.R’s of normal of the plane (i) are (1+2λ), (1+3λ), (1+4λ)Eqn. (i) is per. to , its D.R’s of normal are 1, -1, 1⇨ 1. (1+2λ)+ (-1).(1+3λ)+ 1. (1+4λ) =0 ⇨ λ =-1/3 , put in (i) ⇨x-z+2=0 Ques.13 In the following cases, determine whether the givenplanes are parallel or perpendicular, and in case they are neither,find the angles between them.(a)(b)(c)
  • 60. (d)(e)Ans. (a) d.r’s of the normal of the planes are 7,5,6 & 3, -1,-10We shall check a1.a2+b1.b2+c1.c2 =0 & a1/a2 = b1/b2 = c1/c2 for per. &parallel , neither per. nor parallel, now find angle b/w them cosѲ = ⇨ = -2/5(b) per. planes [(c)&(d)] parallel planes (e) cosѲ = 1/√2 ⇨Ѳ=п/4Ques.14 In the following cases, find the distance of each of thegiven points from the corresponding given plane.Point Plane(a) (0, 0, 0)(b) (3, −2, 1)(c) (2, 3, −5)(d) (−6, 0, 0)Ans. Use distance of point from plane(a) per. distance = | = 3/13(b) 13/3, (c) 3, (d) 2.MISCELLANOUS **Ques.2 If l1, m1, n1 and l2, m2, n2 are the directioncosines of two mutually perpendicular lines, show that the
  • 61. direction cosines of the line perpendicular to both of these are m1n2− m2n1, n1l2 − n2l1, l1m2 − l2m1.Ans Let l, m, n be the d.c’s of the line per. to each one of the givenlines. Then ll1 +mm1+ nn1 = 0 & ll2 +mm2+ nn2 = 0 , after solvingthese two results, we get (by cross mult.) = = = =Where Ѳ is the angle b/w the given lines. But Ѳ= п/2(given)∴sinѲ=1 ⇨ l = m1n2 − m2n1, m = n1l2 − n2l1, n = l1m2 − l2m1.Ques.7 Find the vector equation of the plane passing through (1, 2,3) and perpendicular to the planeAns. d.r’s of the normal of plane are 1, 2, -5.Eqn. Of line passes through (1,2,3) and having d.r’s 1,2,-5 is =(i+2j+3k)+λ (i+2j-5k) [ ∵ eqn. Of line passing through with d.r’s is = +λ ]Ques.10 Find the coordinates of the point where the line through(5, 1, 6) and (3, 4, 1) crosses the YZ-planeAns. Eqn. Of line passes through (5, 1, 6) and (3, 4, 1)(x-5)/2 = (y-1)/-3 = (z-6)/5 = λ , point on the line is (5+2λ, 1-3λ, 6+5λ)Crosses the YZ-plane [ eqn. Of YZ-plane is X = 0] ⇨ λ = -5/2,Required point is (0,17/2,-13/2)Ques.12 Find the coordinates of the point where the line through(3, −4, −5) and (2, − 3, 1) crosses the plane 2x + y + z = 7.Ans. Eqn. Of line passes through (3, −4, −5) and (2, − 3, 1)(x-3)/-1 = (y+4)/1 = (z+5)/6 = λ, point on the line is (3-λ,-4+λ,-5+6λ)
  • 62. Lies on 2x + y + z = 7 ⇨ λ = 2 ,required point is (1,-2,7)Ques.15 Find the equation of the plane passing through the line ofintersection of the planes and andparallel to x-axis.Ans. Equation of the plane passing through the line of intersectionof the planes (2i+3j-k)+4+ λ* . (i+j+k)-1] = 0 .*(2+λ)i+(3+λ)j+(-1+λ)k++4-λ = 0.......(i), X-axis is parallel to plane (i)⇨ its normal is per. to x-axis , d.r’s of x-axis are 1,0,0 & d.r’s ofnormal are 2+λ, 3+λ, -1+λ and they are per. ⇨ λ = -2, put in (i) ,weget . (j-3k)+6 =0Ques. 16 If O be orgin and the coordinates of P be (1,2,-3), then findthe eqn. Of the plane passing through P and per. to P.Ans. Eqn. Of plane passes through (x1,y1,z1) isA(x-x1)+b(y-y1)+c(z-z1) = 0, where a,b,c are d.r’s of normalEqn. Of plane is 1.(x-1)+2(y-2)-3(z+3) = 0⇨ x+2y-3z-14=0Ques.17 Find the equation of the plane which contains the line ofintersection of the planes , andwhich is perpendicular to the plane .Ans. equation of the plane which contains the line of intersection ofthe planes +λ( )......(i) .*(1+2λ)i+(2+λ)j+(3-λ)k+-4+5λ=0 is per. to(1+2λ).5+(2+λ).3+(3-λ).(-6) =0 ⇨ λ = 7/19 , put in (i), we get . (33i+45j+50k) – 41 = 0
  • 63. Ques.18 Find the distance of the point (−1, −5, −10) from the pointof intersection of the line and the plane .Ans. put the value of from first given eqn. in second eqn. , we get[2i-j+2k+λ(3i+4j+2k)+.(i-j+k) = 5 ⇨ λ =0, then point of intersection ofplane & line is 2i-j+2k (2,-1,2) & distance b/w two points is 13Ques.20 Find the vector equation of the line passing through thepoint (1, 2, − 4) and perpendicular to the two lines:Ans. The vector equation of the line passing through the point (1, 2,− 4) is = i+2j-4k+λ(b1i+b2j+b3k) or (x-1)/b1=(y-2)/b2=(z+4)/b3........(i) Where b1, b2, b3 are d.r’s of above line which is per. togiven two lines whose d.r’s are 3, -16, 7 & 3, 8, -5 ⇨ 3b1+(-16) b2+7b3 = 0..........(ii)& 3b1+8 b2+(-5) b3 =0.......(iii) (by condition of perpendicularity)By solving (ii) & (iii) ,we get b1/2 = b2/3 = b3/6 = p, put in (i)Ques.21 Prove that if a plane has the intercepts a, b, c and is at adistance of P units from the origin, thenAns. use formula , we getP= ⇨ 1/ P² =(For class xi)QUES. 1:.Show that the points (-2, 3 , 5 ) , ( 1, 2 , 3 )and ( 7, 0 , -1 ) are collinear.
  • 64. QUESTION 2: Find the equation of the set of points which areequidistant from the points (1, 2 , 3) and ( 3 , 2 , -1)3- DIM. GEOMETRY FOR XI class (ncert)WITH HINTSEx12.2 Question 4: Find the equation of the set of points which areequidistant from the points (1, 2, 3) and (3, 2, –1).Ans.Let P(x,y,z) be a point which is equidistant ∴ PA =PB ⇨ PA² = PB² ⇨ x – 2z = 0 (by distance formula P1P2 = (x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2.)Ex 12.3 Question 4: Using section formula, show that the points A(2, –3, 4), B (–1, 2, 1) and are collinear.Ans. Integral division: If C(x, y, z) is point dividing join of A(x1, y1, z1) & B(x2, y2,z3) in ratio of k:1.Then, x = ,y= ,z=, we get ( , , ) = (0, 1/3, 2) ⇨ k=2Question 5: Find the coordinates of the points which trisect the linesegment joining the points P (4, 2, –6) and Q (10, –16, 6).Ans. Let A, B be two points which trisect PQ ∴ PA = AB = BQ A divides PQ in the ratio 1:2 , ∴ A ( 6, -4, -2) ( by using sectionformula) , B divides PQ in the raio 2:1, ∴ B(8, -10, 2)Misc.Question 1: Three vertices of a parallelogram ABCD are A (3, –1, 2), B (1, 2, –4) andC (–1, 1, 2). Find the coordinates of the fourthvertex.Ans. let D(x,y,z) be the coordinates of the fourth vertex.by propertyof //gm. We know that diagonals of //gm. Are bisect each other ,so mid point of AC & BD are same ⇨ mid point of AC is (1, 0, 2)=mid point of BD is ((x+1)/2, (y+1)/2, (z-4)/2) ⇨ x=1, y=-2, z= 4Question 2: Find the lengths of the medians of the triangle withvertices A (0, 0, 6), B (0, 4, 0) and C(6, 0, 0).Ans. Take AD, BE& CF are medians (D,E,F are mid points on sidesopposite to A,B,C resp.)by distance formula ⇨ AD=7,BE=√34& CF=7
  • 65. Question 4: If the origin is the centroid of the triangle PQR withvertices P (2a, 2, 6), Q (–4, 3b, –10) and R (8, 14, 2c), then find thevalues of a, b and c.Ans. co-ordinates of centriod= {(x1+x2+x3)/3,(y1+y2+y3)/3,(z1+z2+z3)/3 } ⇨ a=-2, b= -16/3, c= 2Question 5: Find the coordinates of a point on y-axis which are at adistance of from the point P (3, –2, 5).Ans. the coordinates of a point on y-axis is A(0,y,0) by distanceformula ⇨ PA = ⇨ Y = -6, 2Question 6: A point R with x-coordinate 4 lies on the line segmentjoining the pointsP (2, –3, 4) and Q (8, 0, 10). Find the coordinatesof the point R.[Hint suppose R divides PQ in the ratio k: 1. Thecoordinates of the point R are given by⇨ (8K+2)/(K+1) = 4 ⇨ K = ½ , SO R divides PQ in the ratio 1:2 , putthe value of k , we get coordinates of R ( 4, -2, 6)Question 7: If A and B be the points (3, 4, 5) and (–1, 3, –7),respectively, find the equation of the set of points P such that PA2 +PB2 = k2, where k is a constant.Ans. Let P(x,y,z), use distance formula (x-3)2+(y-4)2+(z-5)2+(x+1)2+(y-3)2+(z+7)2= k2 ⇨ 2x2+2y2+2z2-4x-14y+4z = k2 -109

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