Chapter-7-3D DIMENSIONAL-01-THEORY-I

System of Co-ordinates 7.1 Co-ordinates of a Point in Space. (1) Cartesian Co-ordinates : Let O be a fixed point, known as origin and let OX, OY and OZ be three mutually perpendicular lines, taken as x-axis, y-axis and z-axis respectively, in such a way that they form a right-handed system. The planes XOY, YOZ and ZOX are known as xy-plane, yz-plane and zx-plane respectively. Let P be a point in space and distances of P from yz, zx and xy-planes be x, y, z respectively (with proper signs), then we say that co-ordinates of P are (x, y, z). Also OA = x, OB = y, OC = z. The three co-ordinate planes (XOY, YOZ and ZOX) divide space into eight parts and these parts are called octants. Signs of co-ordinates of a point : The signs of the co-ordinates of a point in three dimension follow the convention that all distances measured along or parallel to OX, OY, OZ will be positive and distances moved along or parallel to OX, OY, OZ will be negative. The following table shows the signs of co-ordinates of points in various octants : Octant co-ordinate OXYZ OXYZ OXYZ OXYZ OXYZ OXYZ OXYZ OXYZ x + – + – + – + – y + + – – + + – – z + + + + – – – – (2) Other methods of defining the position of any point P in space : (i) Cylindrical co-ordinates : If the rectangular cartesian co-ordinates of P are (x, y, z), then those of N are (x, y, 0) and we can easily have the following relations : x = u cos, y = u sin and z = z. Hence, and . Cylindrical co-ordinates of P  (u, , z) (ii) Spherical polar co-ordinates : The measures of quantities r, ,  are known as spherical or three dimensional polar co-ordinates of the point P. If the rectangular cartesian co-ordinates of P are (x, y, z) then z = r cos, u = r sin  x = u cos = r sin cos, y = u sin = r sin sin and z = r cos Also and ; Note :  The co-ordinates of a point on xy-plane is (x, y, 0), on yz-plane is (0, y, z) and on zx-plane is (x, 0, z)  The co-ordinates of a point on x-axis is (x, 0, 0), on y-axis is (0, y, 0) and on z-axis is (0, 0, z)  Position vector of a point : Let be unit vectors along OX, OY and OZ respectively. Then position vector of a point P(x, y, z) is . 7.2 Distance Formula. (1) Distance formula : The distance between two points and is given by (2) Distance from origin : Let O be the origin and P(x, y, z) be any point, then . (3) Distance of a point from co-ordinate axes : Let P(x, y, z) be any point in the space. Let PA, PB and PC be the perpendiculars drawn from P to the axes OX, OY and OZ respectively. Then, Example: 1 The distance of the point (4, 3, 5) from the y-axis is (a) (b) 5 (c) (d) Solution: (c) Distance Example: 2 The points (5, –4, 2), (4, –3, 1), (7, –6, 4) and (8, –7, 5) are the vertices of (a) A rectangle (b) A square (c) A parallelogram (d) None of these Solution: (c) Let the points be A(5, –4, 2), B(4, –3, 1), C(7, –6, 4) and D(8, –7, 5). , , , Length of diagonals , i.e., AC  BD Hence, A, B, C, D are vertices of a parallelogram 7.3 Section Formulas. (1) Section formula for internal division : Let and be two points. Let R be a point on the line segment joining P and Q such that it divides the join of P and Q internally in the ratio . Then the co-ordinates of R are . (2) Section formula for external division : Let and be two points, and let R be a point on PQ produced, dividing it externally in the ratio . Then the co-ordinates of R are . Note :  Co-ordinates of the midpoint : When division point is the mid-point of PQ then ratio will be 1 : 1, hence co-ordinates of the mid point of PQ are .  Co-ordinates of the general point : The co-ordinates of any point lying on the line joining points and may be taken as , which divides PQ in the ratio k : 1. This is called general point on the line PQ. Example: 3 If the x-co-ordinate of a point P on the join of Q (2, 2, 1) and R (5, 1, –2) is 4, then its z-co-ordinate is (a) 2 (b) 1 (c) –1 (d) –2 Solution: (c) Let the point P be . ∵ Given that   z-co-ordinate of 7.4 Triangle. (1) Co-ordinates of the centroid (i) If and are the vertices of a triangle, then co-ordinates of its centroid are . (ii) If ; r = 1, 2, 3, 4, are vertices of a tetrahedron, then co-ordinates of its centroid are . (iii) If G (, , ) is the centroid of ABC, where A is , B is , then C is . (2) Area of triangle : Let , and be the vertices of a triangle, then , , Now, area of ABC is given by the relation . Also, (3) Condition of collinearity : Points and are collinear If 7.5 Volume of Tetrahedron. Volume of tetrahedron with vertices ; r = 1, 2, 3, 4, is Example: 4 If centroid of tetrahedron OABC, where A, B, C are given by (a, 2, 3), (1, b, 2) and (2, 1, c) respectively be (1, 2, –1), then distance of P(a, b, c) from origin is equal to (a) (b) (c) (d) None of these Solution: (a) (1, 2, –1) is the centroid of the tetrahedron   a = 1,  b = 5,  c = – 9.  (a, b, c) = (1, 5, –9). Its distance from origin Example: 5 If vertices of triangle are , and , then the area of triangle is (a) (b) (c) (d) Solution: (b) Example: 6 The points (5, 2, 4), (6, –1, 2) and (8, –7, k) are collinear, if k is equal to (a) –2 (b) 2 (c) 3 (d) –1 Solution: (a) If given points are collinear, then     7.6 Direction cosines and Direction ratio. (1) Direction cosines (i) The cosines of the angle made by a line in anticlockwise direction with positive direction of co-ordinate axes are called the direction cosines of that line. If , ,  be the angles which a given directed line makes with the positive direction of the x, y, z co-ordinate axes respectively, then cos, cos, cos are called the direction cosines of the given line and are generally denoted by l, m, n respectively. Thus, , and . By definition, it follows that the direction cosine of the axis of x are respectively , , i.e. (1, 0, 0). Similarly direction cosines of the axes of y and z are respectively (0, 1, 0) and (0, 0, 1). Relation between the direction cosines : Let OP be any line through the origin O which has direction cosines l, m, n. Let P = (x, y, z) and OP = r. Then .....(i) From P draw PA, PB, PC perpendicular on the co-ordinate axes, so that OA = x, OB = y, OC = z. Also, and . From triangle AOP,  Similarly and . Hence from (i),  or, , or, Note :  If OP = r and the co-ordinates of point P be (x, y, z), then d.c.’s of line OP are x/r, y/r, z/r.  Direction cosines of are .  Since –1 ≤ cosx ≤ 1, , hence values of l, m, n are such real numbers which are not less than – 1 and not greater than 1. Hence .  The direction cosines of a line parallel to any co-ordinate axis are equal to the direction cosines of the co-ordinate axis.  The number of lines which are equally inclined to the co-ordinate axes is 4.  If l, m, n are the d.c.’s of a line, then the maximum value of . Important Tips  The angles , ,  are called the direction angles of line AB.  The d.c.’s of line BA are cos ( – ), cos ( – ) and cos ( – ) i.e., –cos, –cos, –cos.  Angles , ,  are not coplanar.   +  +  is not equal to 360° as these angles do not lie in same plane.  If P(x, y,z) be a point in space such that has d.c.’s l, m, n then .  Projection of a vector r on the co-ordinate axes are .  and (2) Direction ratio (i) Three numbers which are proportional to the direction cosines of a line are called the direction ratio of that line. If a, b, c are three numbers proportional to direction cosines l, m, n of a line, then a, b, c are called its direction ratios. They are also called direction numbers or direction components. Hence by definition, we have (say)  l = ak, m = bk, n = ck   , , where the sign should be taken all positive or all negative. Note :  Direction ratios are not uniques, whereas d.c.’s are unique. i.e., (ii) Let be a vector. Then its d.r.’s are a, b, c If a vector r has d.r.’s a, b, c then (iii) D.c.’s and d.r.’s of a line joining two points : The direction ratios of line PQ joining and are , and (say). Then direction cosines are, i.e., . Example: 7 A line makes the same angle  with each of the x and z-axis. If the angle , which it makes with y-axis, is such that , then equals (a) (b) (c) (d) Solution: (b) We know that, . Since line makes angle  with x and z-axis and angle  with y-axis.   ……(i) Given that ……(ii) From (i) and (ii),    Example: 8 Direction cosines of the line that makes equal angles with the three axes in a space are (a) (b) (c) (d) Solution: (c) ∵  Now,   i.e., . Hence required d.c.’s are . Example: 9 A line which makes angle 60° with y-axis and z-axis, then the angle which it makes with x-axis is (a) 45° (b) 60° (c) 75° (d) 30° Solution: (a) Given that i.e. , ∵     Example: 10 A line passes through the points (6, –7, –1) and (2, –3, 1). The direction cosines of line, so directed that the angle made by it with the positive direction of x-axis is acute, are (a) (b) (c) (d) Solution: (a) Let l, m, n be the d.c.’s of a given line. Then, as it makes an acute angle with x-axis, therefore l>0. Direction ratios = 4, –4, –2 or 2, –2, –1 and Direction cosines = . Example: 11 If the direction cosines of a line are , then (a) c > 0 (b) (c) 0 < c < 1 (d) c > 2 Solution: (b) We know that    . Example: 12 If r is a vector of magnitude 21 and has d.r.’s 2, –3, 6. Then r is equal to (a) (b) (c) (d) Solution: (a) D.r.’s of r are 2, –3, 6. Therefore, its d.c.’s are  . 7.7 Projection. (1) Projection of a point on a line : The projection of a point P on a line AB is the foot N of the perpendicular PN from P on the line AB. N is also the same point where the line AB meets the plane through P and perpendicular to AB. (2) Projection of a segment of a line on another line and its length : The projection of the segment AB of a given line on another line CD is the segment AB of CD where A and B are the projections of the points A and B on the line CD. The length of the projection A B. (3) Projection of a line joining the points P(x1, y1, z1) and Q(x2, y2, z2) on another line whose direction cosines are l, m and n : Let PQ be a line segment where and and AB be a given line with d.c.’s as l, m, n. If the line segment PQ makes angle  with the line AB, then Projection of PQ is PQ = PQ cos Important Tips  For x-axis, l = 1, m =0, n=0. Hence, projection of PQ on x-axis = x2 – x1, Projection of PQ on y-axis = y2 – y1 and Projection of PQ on z-axis = z2 – z1  If P is a point (x1, y1, z1), then projection of OP on a line whose direction cosines are l, m, n, is l1x1 + m1¬y1 + n1z1, where O is the origin.  If l1, m1, n1 and l2, m2, n2 are the d.c.’s of two concurrent lines, then the d.c.’s of the lines bisecting the angles between them are proportional to l1  l2, m1  m2, n1  n2. Example: 13 If A, B, C, D are the points (3, 4, 5) (4, 6, 3), (–1, 2, 4) and (1, 0, 5), then the projection of CD on AB is (a) (b) (c) (d) None of these Solution: (b) Let l, m, n be the direction cosines of AB Then , . Similarly  The projection of CD on AB Example: 14 The projection of a line on co-ordinate axes are 2, 3, 6. Then the length of the line is (a) 7 (b) 5 (c) 1 (d) 11 Solution: (b) Let AB be the line and its direction cosines be cos, cos, cos. Then the projection of line AB on the co-ordinate axes are ABcos, ABcos, ABcos.  AB cos = 2, AB cos = 3, AB cos = 6    AB = 7 7.8 Angle between Two lines. (1) Cartesian form : Let  be the angle between two straight lines AB and AC whose direction cosines are and respectively, is given by . If direction ratios of two lines and are given, then angle between two lines is given by . Particular results : We have,  , which is known as Lagrange’s identity. The value of sin can easily be obtained by the following form. When d.r.’s of the lines are given if and are d.r.’s of given two lines, then angle  between them is given by Condition of perpendicularity : If the given lines are perpendicular, then i.e. cos = 0  or Condition of parallelism : If the given lines are parallel, then i.e. sin = 0  , which is true, only when , and  . Similarly, . Note :  The angle between any two diagonals of a cube is .  The angle between a diagonal of a cube and the diagonal of a faces of the cube is .  If a straight line makes angles , , ,  with the diagonals of a cube, then  If the edges of a rectangular parallelopiped be a, b, c, then the angles between the two diagonals are (2) Vector form : Let the vector equations of two lines be and As the lines are parallel to the vectors and respectively, therefore angle between the lines is same as the angle between the vectors and . Thus if  is the angle between the given lines, then . Note :  If the lines are perpendicular, then .  If the lines are parallel, then and are parallel, therefore for some scalar . Example: 15 If d.c.’s of two lines are proportional to (2, 3, –6) and (3, –4,5), then the acute angle between them is (a) (b) (c) 90° (d) Solution: (b) D.c.’s of two lines are proportional to (2, 3, –6) and (3, – 4, 5) i.e. d.r.’s are (2, 3, –6) and (3, –4, 5)   Taking acute angle, Example: 16 If the direction ratio of two lines are given by and , then the angle between the lines is (a) (b) (c) (d) Solution: (a) We have, ……(i) ……(ii) From equation (i), Putting the value of l in equation (ii)      or ……(i) ……(iii) ……(iv) From equation (i) and equation (iii), From equation (i) and equation (iv), Thus, the direction ratios of two lines are and , , .Hence, the angle between them /2. Example: 17 If a line makes angles , , ,  with four diagonals of a cube, then the value of is (a) (b) 1 (c) (d) Solution: (c) Let side of the cube = a Then OG, BE and AD, CF will be four diagonals. d.r.’s of OG = a, a, a = 1, 1, 1 d.r.’s of BE = –a, –a, a = 1, 1, –1 d.r.’s of AD = –a, a, a = –1, 1, 1 d.r.’s of CF = a, –a, a = 1, –1, 1 Let d.r.’s of line be l, m, n. Therefore angle between line and diagonal   Example: 18 If and are d.c.’s of two lines inclined to each other at an angle , then the d.c.’s of the internal bisectors of angle between these lines are (a) (b) (c) (d) Solution: (b) Let OA and OB be two lines. D.c.’s of OA is and OB is . Let OA = OB = 1. Then the co-ordinates of A and B are and . Let OC be the bisector of AOB. Then C is the mid-point of AB and so its co-ordinates are .  D.r.’s of line OC are We have, . D.r.’s of line OC are . Example: 19 The angle between the lines and is (a) (b) (c) (d) Solution: (b) We have, and We know that, , , Hence, acute angle i.e. The Straight Line 7.9 Straight line in Space. Every equation of the first degree represents a plane. Two equations of the first degree are satisfied by the co-ordinates of every point on the line of intersection of the planes represented by them. Therefore, the two equations together represent that line. Therefore and together represent a straight line. (1) Equation of a line passing through a given point (i) Cartesian form or symmetrical form : Cartesian equation of a straight line passing through a fixed point and having direction ratios a, b, c is . (ii) Vector form : Vector equation of a straight line passing through a fixed point with position vector a and parallel to a given vector b is . Important Tips  The parametric equations of the line are , where  is the parameter.  The co-ordinates of any point on the line are , where   R.  Since the direction cosines of a line are also direction ratios, therefore equation of a line passing through (x1, y¬1, z1) and having direction cosines l, m, n is .  Since x, y and z-axes pass through the origin and have direction cosines 1, 0, 0; 0, 1, 0 and 0, 0, 1 respectively. Therefore, the equations are x-axis : or y = 0 and z = 0. y-axis : or x = 0 and z = 0; z-axis : or x = 0 and y = 0.  In the symmetrical form of equation of a line, the coefficients of x, y, z are unity. 7.10 Equation of Line passing through Two given points. (i) Cartesian form : If be two given points, the equations to the line AB are The co-ordinates of a variable point on AB can be expressed in terms of a parameter  in the form  being any real number different from –1. In fact, (x, y, z) are the co-ordinates of the point which divides the join of A and B in the ratio  : 1. (ii) Vector form : The vector equation of a line passing through two points with position vectors a and b is 7.11 Changing Unsymmetrical form to Symmetrical form. The unsymmetrical form of a line Can be changed to symmetrical form as follows : Example: 20 The equation to the straight line passing through the points (4, –5, –2) and (–1, 5, 3) is (a) (b) (c) (d) Solution: (a) We know that equation of a straight line is of the form D.r.’s of the line = i.e., or . Hence the equation is i.e., Example: 21 The d.c.’s of the line are (a) (b) (c) 1, 2, 3 (d) None of these Solution: (b) We have    d.r.’s of line are (1, 2, 3). Hence d.c.’s of line are Example: 22 The vector equation of line through the point A(3, 4, –7) and B(1, –1, 6) is (a) (b) (c) (d) Solution: (c) Position vector of A is and that of B is We know that equation of line in vector form, , . 7.12 Angle between Two lines. Let the cartesian equations of the two lines be .....(i) and .....(ii) Condition of perpendicularity : If the lines are perpendicular, then Condition of parallelism : If the lines are parallel, then . Example: 23 If the lines and are at right angles, then k = (a) –10 (b) 10/7 (c) –10/7 (d) –7/10 Solution: (a) We have and Since lines are  to each other. So,    . Example: 24 The lines , and , are perpendicular to each other, if (a) (b) (c) (d) Solution: (b) We have, , ,  ……(i) and , ,  ……(ii) ∵ Given, lines (i) and (ii) are perpendicular  , Example: 25 The direction ratio of the line which is perpendicular to the lines and are (a) < 4, 5, 7 > (b) < 4, –5, 7 > (c) < 4, –5, –7 > (d) < –4, 5, 7 > Solution: (a) Let d.r.’s of line be l, m, n. ∵ line is perpendicular to given line  ……(i) ……(ii) From equation (i) and (ii) or . Hence, d.r.’s of line (< 4, 5, 7 >) 7.13 Reduction of Cartesian form of the Equation of a line to Vector form and Vice versa. Cartesian to vector : Let the Cartesian equation of a line be ……(i) This is the equation of a line passing through the point and having direction ratios a, b, c. In vector form this means that the line passes through point having position vector and is parallel to the vector . Thus, the vector form of (i) is or , where  is a parameter. Vector to cartesian : Let the vector equation of a line be ……(ii) Where and  is a parameter. To reduce (ii) to Cartesian form we put and equate the coefficients of i, j and k as discussed below. Putting and in (ii), we obtain Equating coefficients of i, j and k, we get or Example: 26 The cartesian equations of a line are . The vector equation of the line is (a) (b) (c) (d) None of these Solution: (a) The given line is  This show that the given line passes through (1/3, –1/3) and has direction ratio 1, 2, 3. Position vector and is parallel to vector . Hence, . 7.14 Intersection of Two lines. Determine whether two lines intersect or not. In case they intersect, the following algorithm is used to find their point of intersection. Algorithm for cartesian form : Let the two lines be ……(i) And ……(ii) Step I : Write the co-ordinates of general points on (i) and (ii). The co-ordinates of general points on (i) and (ii) are given by and respectively. i.e., and Step II : If the lines (i) and (ii) intersect, then they have a common point. and . Step III : Solve any two of the equations in  and  obtained in step II. If the values of  and  satisfy the third equation, then the lines (i) and (ii) intersect, otherwise they do not intersect. Step IV : To obtain the co-ordinates of the point of intersection, substitute the value of  (or ) in the co-ordinates of general point (s) obtained in step I. Example: 27 If the line and intersect, then k = (a) 2/9 (b) 9/2 (c) 0 (d) –1 Solution: (b) We have, (Let) i.e. point is and (Let) i.e. point is . If the lines are intersecting, then they have a common point.  On solving, Hence, k = 9/2. Example: 28 A line with direction cosines proportional to 2, 1, 2 meets each of the lines and . The co-ordinates of each of the points of intersection are given by (a) (2a, 3a, 3a) (2a, a, a) (b) (3a, 2a, 3a) (a, a, a) (c) (3a, 2a, 3a) (a, a, 2a) (d) (3a, 3a, 3a) (a, a, a) Solution: (b) Given lines are (say)  Point is P(, – a, ) and i.e. (say)  Point Q(2 – a, , ) Since d.r.’s of given lines are 2, 1, 2 and d.r.’s of PQ = (2 – a – ,  –  + a,  – ) According to question, Then ,  = a. Therefore, points of intersection are P(3a, 2a, 3a) and Q(a, a, a). Alternative method : Check by option i.e.  a = a = a and i.e.  a = a = a. Hence (b) is correct. 7.15 Foot of perpendicular from a point A(, , ) to the line . (1) Cartesian form Foot of perpendicular from a point A(, , ) to the line : If P be the foot of perpendicular, then P is . Find the direction ratios of AP and apply the condition of perpendicularity of AP and the given line. This will give the value of r and hence the point P which is foot of perpendicular. Length and equation of perpendicular : The length of the perpendicular is the distance AP and its equation is the line joining two known points A and P. Note :  The length of the perpendicular is the perpendicular distance of given point from that line. Reflection or image of a point in a straight line : If the perpendicular PL from point P on the given line be produced to Q such that PL = QL, then Q is known as the image or reflection of P in the given line. Also, L is the foot of the perpendicular or the projection of P on the line. (2) Vector form Perpendicular distance of a point from a line : Let L is the foot of perpendicular drawn from on the line . Since r denotes the position vector of any point on the line . So, let the position vector of L be . Then The length PL, is the magnitude of , and required length of perpendicular. Image of a point in a straight line : Let is the image of P in Then, Example: 29 The co-ordinates of the foot of the perpendicular drawn from the point A(1, 0, 3) to the join of the points B(4, 7, 1) and C(3, 5, 3) are (a) (5/3, 7/3, 17/3) (b) (5, 7, 17) (c) (5/3, –7/3, 17/3) (d) (–5/3, 7/3, –17/3) Solution: (a) Equation of BC, i.e. (say) Any point on the given line is Then, d.r.’s of AD = i.e. d.r.’s of and d.r.’s of BC = (–1, –2, 2) Since AD is  to given line,      D is {4 – (7/3), 7– (14/3), (14/3)+1} i.e. D is (5/3, 7/3, 17/3). Example: 30 The image of the point (1, 6, 3) in the line is (a) (1, 0, 7) (b) (–1, 0, 7) (c) (1, 0, –7) (d) None of these Solution: (a) Let P(1, 6, 3) be the given point, and let L be the foot of the perpendicular from P to the given line. The co-ordinates of a general point on the given line are given by i.e. . Let the co-ordinates of L be (, 2 + 1, 3 + 2) ……(i) So, direction ratios of PL are i.e. . Direction ratios of the given line are 1, 2, 3 which is perpendicular to PL.    So, co-ordinates of L are (1, 3, 5). Let be the image of in the given line. Then L is the mid-point of PQ.  and  and . Hence the image of P(1, 6, 3) in the given line is (1, 0, 7).