THREE DIMENSIONAL GEOMETRY - NCERT

1 THREE DIMENSIONAL GEOMETRY463 The moving power of mathematical invention is notreasoning but imagination. IntroductionIn Class XI, while studying Analytical GEOMETRY in twodimensions, and the introduction to THREE dimensionalgeometry, we confined to the Cartesian methods only. Inthe previous chapter of this book, we have studied somebasic concepts of vectors. We will now use vector algebrato THREE DIMENSIONAL GEOMETRY . The purpose of thisapproach to 3- DIMENSIONAL GEOMETRY is that it makes thestudy simple and elegant*.In this chapter, we shall study the direction cosinesand direction ratios of a line joining two points and alsodiscuss about the equations of lines and planes in spaceunder different conditions, angle between two lines, twoplanes, a line and a plane, shortest distance between twoskew lines and distance of a point from a plane. Most ofthe above results are obtained in vector form. Nevertheless, we shall also translatethese results in the Cartesian form which, at times, presents a more clear geometricand analytic picture of the Direction Cosines and Direction Ratios of a LineFrom Chapter 10, recall that if a directed line L passing through the origin makesangles , and with x, y and z-axes, respectively, called direction angles, then cosineof these angles, namely, cos , cos and cos are called direction cosines of thedirected line we reverse the direction of L, then the direction angles are replaced by their supplements, , , and.

2 Thus, the signs of the direction cosines are DIMENSIONAL GEOMETRY *For various activities in THREE DIMENSIONAL GEOMETRY , one may refer to the Book A Hand Book for designing Mathematics Laboratory in Schools , NCERT , 2005 Leonhard Euler(1707-1783)2019-20 MATHEMATICS464 Note that a given line in space can be extended in two opposite directions and so ithas two sets of direction cosines. In order to have a unique set of direction cosines fora given line in space, we must take the given line as a directed line. These uniquedirection cosines are denoted by l, m and If the given line in space does not pass through the origin, then, in order to findits direction cosines, we draw a line through the origin and parallel to the given take one of the directed lines from the origin and find its direction cosines as twoparallel line have same set of direction THREE numbers which are proportional to the direction cosines of a line arecalled the direction ratios of the line. If l, m, n are direction cosines and a, b, c aredirection ratios of a line, then a = l, b= m and c = n, for any nonzero R.

3 Note Some authors also call direction ratios as direction a, b, c be direction ratios of a line and let l, m and n be the direction cosines( s) of the line. Thenla = mb =nkc= (say), k being a = ak, m =bk, n = (1)Butl2 + m2 + n2 =1 Thereforek2 (a2 + b2 + c2) =1ork =2221abc ++Fig DIMENSIONAL GEOMETRY465 Hence, from (1), the s of the line are222222222,,abclmnabcabcabc= = = ++++++where, depending on the desired sign of k, either a positive or a negative sign is to betaken for l, m and any line, if a, b, c are direction ratios of a line, then ka, kb, kc; k 0 is also aset of direction ratios. So, any two sets of direction ratios of a line are also , for any line there are infinitely many sets of direction Relation between the direction cosines of a lineConsider a line RS with direction cosines l, m, n. Throughthe origin draw a line parallel to the given line and take apoint P(x, y, z) on this line. From P draw a perpendicularPA on the x-axis (Fig. ).Let OP = r.

4 ThenOAcosOP =xr=. This gives x = ,y =mr and z = nrThusx2 + y2 + z2 =r2 (l2 + m2 + n2)Butx2 + y2 + z2 =r2 Hencel2 + m2 + n2 = Direction cosines of a line passing through two pointsSince one and only one line passes through two given points, we can determine thedirection cosines of a line passing through the given points P(x1, y1, z1) and Q(x2, y2, z2)as follows (Fig (a)).Fig (,,)xyzAOaAOPxaaFig MATHEMATICS466 Let l, m, n be the direction cosines of the line PQ and let it makes angles , and with the x, y and z-axis, perpendiculars from P and Q to XY-plane to meet at R and S. Draw aperpendicular from P to QS to meet at N. Now, in right angle triangle PNQ, PQN= (Fig (b).Therefore,cos =21 NQPQPQzz =Similarlycos =2121andcosPQPQ xxyy =Hence, the direction cosines of the line segment joining the points P(x1, y1, z1) andQ(x2, y2, z2) are21 PQxx , 21 PQyy , 21 PQzz wherePQ =()222212121()()xxyyzz + + Note The direction ratios of the line segment joining P(x1, y1, z1) and Q(x2, y2, z2)may be taken asx2 x1, y2 y1, z2 z1 or x1 x2, y1 y2, z1 z2 Example 1 If a line makes angle 90 , 60 and 30 with the positive direction of x, y andz-axis respectively, find its direction Let the 's of the lines be l , m, n.)

5 Then l = cos 900 = 0, m = cos 600 = 12,n = cos 300 = 2 If a line has direction ratios 2, 1, 2, determine its direction Direction cosines are222)2()1(22 + +, 222)2()1(21 + + , ()222)2(122 + + or 212,,333 Example 3 Find the direction cosines of the line passing through the two points( 2, 4, 5) and (1, 2, 3).2019-20 THREE DIMENSIONAL GEOMETRY467 Solution We know the direction cosines of the line passing through two pointsP(x1, y1, z1) and Q(x2, y2, z2) are given by212121,,PQPQPQ xxyyzz wherePQ =()212212212)()(zzyyxx + + Here P is ( 2, 4, 5) and Q is (1, 2, 3).SoPQ =222(1(2))(24)(3(5)) + + = 77 Thus, the direction cosines of the line joining two points is328,,777777 Example 4 Find the direction cosines of x, y and The x-axis makes angles 0 , 90 and 90 respectively with x, y and , the direction cosines of x-axis are cos 0 , cos 90 , cos 90 , 1,0, , direction cosines of y-axis and z-axis are 0, 1, 0 and 0, 0, 1 5 Show that the points A (2, 3, 4), B (1, 2, 3) and C (3, 8, 11) Direction ratios of line joining A and B are1 2, 2 3, 3 + 4 , 1, 5, direction ratios of line joining B and C are3 1, 8 + 2, 11 3, , 2, 10, is clear that direction ratios of AB and BC are proportional, hence, AB is parallelto BC.

6 But point B is common to both AB and BC. Therefore, A, B, C arecollinear a line makes angles 90 , 135 , 45 with the x, y and z-axes respectively, find itsdirection the direction cosines of a line which makes equal angles with the a line has the direction ratios 18, 12, 4, then what are its direction cosines ? that the points (2, 3, 4), ( 1, 2, 1), (5, 8, 7) are the direction cosines of the sides of the triangle whose vertices are(3, 5, 4), ( 1, 1, 2) and ( 5, 5, 2).2019-20 Equation of a Line in SpaceWe have studied equation of lines in two dimensions in Class XI, we shall now studythe vector and cartesian equations of a line in line is uniquely determined if(i)it passes through a given point and has given direction, or(ii)it passes through two given of a line through a given point and parallel to a given vector bLet a be the position vector of the given pointA with respect to the origin O of therectangular coordinate system. Let l be theline which passes through the point A and isparallel to a given vector b.

7 Let r be theposition vector of an arbitrary point P on theline (Fig ).Then AP is parallel to the vector b , ,AP = b , where is some real =OP OA b =ra Conversely, for each value of the parameter , this equation gives the positionvector of a point P on the line. Hence, the vector equation of the line is given by r = a+ (1)Remark If baibjck=++ , then a, b, c are direction ratios of the line and conversely,if a, b, c are direction ratios of a line, then =++ baibjckwill be the parallel tothe line. Here, b should not be confused with | b|.Derivation of cartesian form from vector formLet the coordinates of the given point A be (x1, y1, z1) and the direction ratios ofthe line be a, b, c. Consider the coordinates of any point P be (x, y, z). Thenkzjyixr ++= ; kzjyixa 111++= and baibjck=++ Substituting these values in (1) and equating the coefficients of ,ij and k , we getx = x1 + a; y =y1 + b; z = z1+ (2)Fig DIMENSIONAL GEOMETRY469 These are parametric equations of the line.

8 Eliminating the parameter from (2),we get1x xa =11y yz z= (3)This is the Cartesian equation of the line. Note If l, m, n are the direction cosines of the line, the equation of the line is1x xl =11y yz z=mnExample 6 Find the vector and the Cartesian equations of the line through the point(5, 2, 4) and which is parallel to the vector 328ijk+ .Solution We havea = 524and328ijkbijk+ =+ Therefore, the vector equation of the line isr = 524(328)ijkijk+ + + Now, r is the position vector of any point P(x, y, z) on the , xiyjzk++ = 524(328)+ + + ijkijk= (53)(22)(48)+ ++ + ijk Eliminating , we get53x =2428yz += which is the equation of the line in Cartesian Equation of a line passing through two given pointsLet a and b be the position vectors of twopoints A(x1, y1, z1) and B(x2, y2, z2),respectively that are lying on a line (Fig ).Let r be the position vector of anarbitrary point P(x, y, z), then P is a point onthe line if and only if APra= andABba= are collinear vectors.

9 Therefore,P is on the line if and only if()raba = Fig MATHEMATICS470or()raba=+ =+ =+ =+ , (1)This is the vector equation of the of cartesian form from vector formWe have111 ,rxiyjzkaxiyjzk=++=++ and 222 ,bxiyjzk=++ Substituting these values in (1), we get 111212121[()()()]xiyjzkxiyjzkxxiyyjzzk++ =+++ + + Equating the like coefficients of kji , , , we getx = x1 + (x2 x1); y = y1 + (y2 y1); z = z1 + (z2 z1)On eliminating , we obtain111212121xxyyzzxxyyzz ======== which is the equation of the line in Cartesian 7 Find the vector equation for the line passing through the points ( 1, 0, 2)and (3, 4, 6).Solution Let a and b be the position vectors of the point A( 1, 0, 2) and B(3, 4, 6).Then 2aik= + and 346bijk=++ Therefore 444baijk =++ Let r be the position vector of any point on the line. Then the vector equation ofthe line is 2(444)rikijk= ++ ++ Example 8 The Cartesian equation of a line is356242xyz+ +==Find the vector equation for the Comparing the given equation with the standard form111xxyyzzabc ==We observe thatx1 = 3, y1 = 5, z1 = 6; a = 2, b = 4, c = DIMENSIONAL GEOMETRY471 Thus, the required line passes through the point ( 3, 5, 6) and is parallel to thevector 242ijk++.

10 Let r be the position vector of any point on the line, then thevector equation of the line is given by (356)rijk= + + (242)ijk++ Angle between Two LinesLet L1 and L2 be two lines passing through the originand with direction ratios a1, b1, c1 and a2, b2, c2,respectively. Let P be a point on L1 and Q be a pointon L2. Consider the directed lines OP and OQ asgiven in Fig Let be the acute angle betweenOP and OQ. Now recall that the directed linesegments OP and OQ are vectors with componentsa1, b1, c1 and a2, b2, c2, respectively. Therefore, theangle between them is given bycos =121212222222111222aabbccabcabc+++++++++ +++++++++++++++.. (1)The angle between the lines in terms of sin is given bysin =21cos =()()2121212222222111222()1aabbccabcabc+ + ++++=()()()()()2222222111222121212222222 111222abcabcaabbccabcabc++++ ++++++=222122112211221222222111222()()() + + ++++ (2) Note In case the lines L1 and L2 do not pass through the origin, we may takelines 12 LandL which are parallel to L1 and L2 respectively and pass throughthe MATHEMATICS472If instead of direction ratios for the lines L1 and L2, direction cosines, namely,l1, m1, n1 for L1 and l2, m2, n2 for L2 are given, then (1) and (2) takes the following form:cos =|l1 l2 + m1m2 + n1n2| (as 2221111lmn++=222222lmn=++).