3D DIMENSIONAL -02-THEORY-II

Example: 31 The length of the perpendicular from the origin to line is (a) (b) 2 (c) (d) 6 Solution: (d) The length of PL is magnitude of i.e., Length of perpendicular . Example: 32 The image of point (1, 2, 3) in the line is (a) (5, –8, 15) (b) (5, 8, –15) (c) (–5, –8, –15) (d) (5, 8, 15) Solution: (d) Given that, , and Then, On solving, . Thus is the position vector of Q, which is the image of P in given line. Hence image of point (1, 2, 3) in the given line is (5, 8, 15). 7.16 Shortest distance between two straight lines. (1) Skew lines : Two straight lines in space which are neither parallel nor intersecting are called skew lines. Thus, the skew lines are those lines which do not lie in the same plane. (2) Line of shortest distance : If and are two skew lines, then the straight line which is perpendicular to each of these two non-intersecting lines is called the “line of shortest distance.” Note :  There is one and only one line perpendicular to each of lines and . (3) Shortest distance between two skew lines (i) Cartesian form : Let two skew lines be and Therefore, the shortest distance between the lines is given by (ii) Vector form : Let and be two lines whose equations are and respectively. Then, Shortest distance (4) Shortest distance between two parallel lines : The shortest distance between the parallel lines and is given by . (5) Condition for two lines to be intersecting i.e. coplanar (i) Cartesian form : If the lines and intersect, then . (ii) Vector form : If the lines and intersect, then the shortest distance between them is zero. Therefore,   Important Tips  Skew lines are non-coplanar lines.  Parallel lines are not skew lines.  If two lines intersect, the shortest distance (SD) between them is zero.  Length of shortest distance between two lines is always taken to be positive.  Shortest distance between two skew lines is perpendicular to both the lines. (6) To determine the equation of line of shortest distance : To find the equation of line of shortest distance, we use the following procedure : (i) From the given equations of the straight lines, i.e. (say) ……(i) and (say) ……(ii) Find the co-ordinates of general points on straight lines (i) and (ii) as and . (ii) Let these be the co-ordinates of P and Q, the two extremities of the length of shortest distance. Hence, find the direction ratios of PQ as . (iii) Apply the condition of PQ being perpendicular to straight lines (i) and (ii) in succession and get two equations connecting  and . Solve these equations to get the values of  and . (iv) Put these values of  and  in the co-ordinates of P and Q to determine points P and Q. (v) Find out the equation of the line passing through P and Q, which will be the line of shortest distance. Note :  The same algorithm may be observed to find out the position vector of P and Q, the two extremities of the shortest distance, in case of vector equations of straight lines. Hence, the line of shortest distance, which passes through P and Q, can be obtained. Example: 33 The shortest distance between the lines and is (a) (b) (c) (d) Solution: (b) S.D. . Example: 34 The shortest distance between the lines and is (a) 6 (b) 0 (c) 2 (d) 4 Solution: (b) S.D. . Hence, S.D. = 0 Example: 35 The line and are coplanar, if (a) k = 0 or –1 (b) k = 0 or 1 (c) k = 0 or –3 (d) k = 3 or –3 Solution: (c) Lines are coplanar, if     , Example: 36 The lines and will intersect if (a) (b) (c) (d) None of these Solution: (b) If lines are intersecting, then        Example: 37 If the straight lines and , with parameters s and t respectively, are co-planar, then  equals (a) 0 (b) –1 (c) (d) –2 Solution: (d) We have and i.e. Since, lines are co-planar, Then,  On solving, . The Plane 7.17 Definition of plane and its equations. If point P(x, y, z) moves according to certain rule, then it may lie in a 3-D region on a surface or on a line or it may simply be a point. Whatever we get, as the region of P after applying the rule, is called locus of P. Let us discuss about the plane or curved surface. If Q be any other point on it’s locus and all points of the straight line PQ lie on it, it is a plane. In other words if the straight line PQ, however small and in whatever direction it may be, lies completely on the locus, it is a plane, otherwise any curved surface. (1) General equation of plane : Every equation of first degree of the form represents the equation of a plane. The coefficients of x, y and z i.e. A, B, C are the direction ratios of the normal to the plane. (2) Equation of co-ordinate planes XOY-plane : z = 0 YOZ -plane : x = 0 ZOX-plane : y = 0 (3) Vector equation of plane (i) Vector equation of a plane through the point and perpendicular to the vector n is or Note :  The above equation can also be written as , where . This is known as the scalar product form of a plane. (4) Normal form : Vector equation of a plane normal to unit vector and at a distance d from the origin is . Note :  If n is not a unit vector, then to reduce the equation to normal form we divide both sides by |n| to obtain or . (5) Equation of a plane passing through a given point and parallel to two given vectors : The equation of the plane passing through a point having position vector a and parallel to b and c is , where  and  are scalars. (6) Equation of plane in various forms (i) Intercept form : If the plane cuts the intercepts of length a, b, c on co-ordinate axes, then its equation is . (ii) Normal form : Normal form of the equation of plane is , where l, m, n are the d.c.’s of the normal to the plane and p is the length of perpendicular from the origin. (7) Equation of plane in particular cases (i) Equation of plane through the origin is given by . i.e. if D = 0, then the plane passes through the origin. (8) Equation of plane parallel to co-ordinate planes or perpendicular to co-ordinate axes (i) Equation of plane parallel to YOZ-plane (or perpendicular to x-axis) and at a distance ‘a’ from it is x = a. (ii) Equation of plane parallel to ZOX-plane (or perpendicular to y-axis) and at a distance ‘b’ from it is y = b. (iii) Equation of plane parallel to XOY-plane (or perpendicular to z-axis) and at a distance ‘c’ from it is z = c. Important Tips  Any plane perpendicular to co-ordinate axis is evidently parallel to co-ordinate plane and vice versa.  A unit vector perpendicular to the plane containing three points A, B, C is . (9) Equation of plane perpendicular to co-ordinate planes or parallel to co-ordinate axes (i) Equation of plane perpendicular to YOZ-plane or parallel to x-axis is . (ii) Equation of plane perpendicular to ZOX-plane or parallel to y axis is . (iii) Equation of plane perpendicular to XOY-plane or parallel to z-axis is . (10) Equation of plane passing through the intersection of two planes (i) Cartesian form : Equation of plane through the intersection of two planes and is , where  is the parameter. (ii) Vector form : The equation of any plane through the intersection of planes and is , where  is an arbitrary constant. (11) Equation of plane parallel to a given plane (i) Cartesian form : Plane parallel to a given plane is , i.e. only constant term is changed. (ii) Vector form : Since parallel planes have the common normal, therefore equation of plane parallel to plane is , where is a constant determined by the given condition. 7.18 Equation of plane passing through the given point. (1) Equation of plane passing through a given point : Equation of plane passing through the point is , where A, B and C are d.r.’s of normal to the plane. (2) Equation of plane through three points : The equation of plane passing through three non-collinear points , and is or . 7.19 Foot of perpendicular from a point A(, , ) to a given plane ax + by + cz + d = 0. If AP be the perpendicular from A to the given plane, then it is parallel to the normal, so that its equation is (say) Any point P on it is . It lies on the given plane and we find the value of r and hence the point P. (1) Perpendicular distance (i) Cartesian form : The length of the perpendicular from the point to the plane is . Note :  The distance between two parallel planes is the algebraic difference of perpendicular distances on the planes from origin.  Distance between two parallel planes and is . (ii) Vector form : The perpendicular distance of a point having position vector a from the plane is given by (2) Position of two points w.r.t. a plane : Two points and lie on the same or opposite sides of a plane according to and are of same or opposite signs. The plane divides the line joining the points P and Q externally or internally according to P and Q are lying on same or opposite sides of the plane. 7.20 Angle between two planes. (1) Cartesian form : Angle between the planes is defined as angle between normals to the planes drawn from any point. Angle between the planes and is Note :  If , then the planes are perpendicular to each other.  If , then the planes are parallel to each other. (2) Vector form : An angle  between the planes and is given by . 7.21 Equation of planes bisecting angle between two given planes . (1) Cartesian form : Equations of planes bisecting angles between the planes and are . Note :  If angle between bisector plane and one of the plane is less than 45o, then it is acute angle bisector, otherwise it is obtuse angle bisector.  If is negative, then origin lies in the acute angle between the given planes provided d1 and d2 are of same sign and if is positive, then origin lies in the obtuse angle between the given planes. (2) Vector form : The equation of the planes bisecting the angles between the planes and are or or . 7.22 Image of a point in a plane. Let P and Q be two points and let  be a plane such that (i) Line PQ is perpendicular to the plane , and (ii) Mid-point of PQ lies on the plane . Then either of the point is the image of the other in the plane . To find the image of a point in a given plane, we proceed as follows (i) Write the equations of the line passing through P and normal to the given plane as . (ii) Write the co-ordinates of image Q as . (iii) Find the co-ordinates of the mid-point R of PQ. (iv) Obtain the value of r by putting the co-ordinates of R in the equation of the plane. (v) Put the value of r in the co-ordinates of Q. 7.23 Coplanar lines. Lines are said to be coplanar if they lie in the same plane or a plane can be made to pass through them. (1) Condition for the lines to be coplanar (i) Cartesian form : If the lines and are coplanar Then . The equation of the plane containing them is or . (ii) Vector form : If the lines and are coplanar, then and the equation of the plane containing them is or . Note :  Every pair of parallel lines is coplanar.  Two coplanar lines are either parallel or intersecting.  The three sides of a triangle are coplanar. Important Tips  Division by plane : The ratio in which the line segment PQ, joining P(x1, y1, z1) and Q(x2, y2, z2), is divided by plane is .  Division by co-ordinate planes : The ratio in which the line segment PQ, joining P(x1, y1, z1) and Q(x2, y2, z2) is divided by co-ordinate planes are as follows : (i) By yz-plane : –x1/x2 (ii) By zx-plane : –y1/y2 (ii) By xy-plane : –z1/z2 Example: 38 The xy-plane divides the line joining the points (–1, 3, 4) and (2, –5, 6) (a) Internally in the ratio 2 : 3 (b) Internally in the ratio 3 : 2 (c) Externally in the ratio 2 : 3 (d) Externally in the ratio 3 : 2 Solution: (c) Required ratio  xy-plane divide externally in the ratio 2 : 3. Example: 39 The ratio in which the plane divides the line joining the point (–2, 4, 7) and (3, –5, 8) is (a) 10 : 3 (b) 3 : 1 (c) 3 : 10 (d) 10 : 1 Solution: (c) Required ratio . Example: 40 The equation of the plane, which makes with co-ordinate axes a triangle with its centroid (, , ), is (a) (b) (c) (d) Solution: (d) We know that ……(i) Centroid i.e.  From equation (i),  . Example: 41 The equation of plane passing through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane is (a) (b) (c) (d) None of these Solution: (c) We know that, equation of plane is It passes through (2, 2, 1)  ……(i) Plane (i) also passes through (9, 3, 6) and is perpendicular to the plane ……(ii) and ……(iii) or or (say) From equation (i), Hence, . Example: 42 The equation of the plane containing the line and perpendicular to the plane is (a) (b) (c) (d) Solution: (c) Since the required plane contains the line and is perpendicular to the plane .  It passes through the point a and parallel to vectors b and n. Hence, it is perpendicular to the vector .  Equation of the required plane is  . Example: 43 The equation of the plane through the intersection of the planes , and passing through the origin will be (a) (b) (c) (d) Solution: (b) Any plane through the given planes is It passes through (0, 0, 0)  = k = 4  Required plane is  . Example: 44 The vector equation of the plane passing through the origin and the line of intersection of plane and is (a) (b) (c) (d) Solution: (b) The equation of a plane through the line of intersection of plane and can be written as ……(i) This passes through the origin, therefore putting the value of k in (i),  . Example: 45 Angle between two planes and is (a) (b) (c) (d) Solution: (a) We know that, i.e. . Example: 46 Distance between two parallel planes and is (a) (b) (c) (d) Solution: (c) We have ……(i) and or ……(ii) Distance between the planes . Example: 47 A tetrahedron has vertices at O(0, 0, 0), A(1, 2, 1), B(2, 1, 3) and C(–1, 1, 2). Then the angle between the faces OAB and ABC will be (a) (b) (c) 30° (d) 90° Solution: (a) Angle between two plane faces is equal to the angle between the normals and to the planes. , the normal to the face OAB is given by ……(i) , the normal to the face ABC, is given by . ……(ii) If  be the angle between and , Then  . Example: 48 The distance of the point (2, 1, –1) from the plane is (a) (b) (c) (d) Solution: (c) Distance of the plane from (2, 1, –1) . Example: 49 A unit vector perpendicular to plane determined by the points P(1, –1, 2), Q(2, 0, –1) and R(0, 2, 1) is (a) (b) (c) (d) Solution: (b) We know that, , and Hence, the unit vector is i.e. . Example: 50 The perpendicular distance from origin to the plane through the point (2, 3, –1) and perpendicular to vector is (a) (b) (c) 13 (d) None of these Solution: (a) We know, the equation of the plane is or   Hence, perpendicular distance of the plane from origin . Example: 51 If P = (0, 1, 0), Q =(0, 0, 1), then projection of PQ on the plane is (a) (b) 3 (c) (d) 2 Solution: (c) Given plane is . From point P and Q draw PM and QN perpendicular on the given plane and QR  MP. (i.e. R and P are the same point)  Example: 52 The reflection of the point (2, –1, 3) in the plane is (a) (b) (c) (d) Solution: (b) Let P be the point (2, –1, 3) and Q be its reflection in the given plane. Then, PQ is perpendicular to the given plane Hence, d.r.’s of PQ are 3, –2, 1 and consequently, equations of PQ are Any point on this line is Let this point be Q. Then midpoint of PQ This point lies in given plane i.e.    Hence, the required point Q is . Example: 53 A non-zero vector a is parallel to the line of intersection of the plane determined by the vectors i, i + j and the plane determined by the vectors i – j, i + k. The angle between a and the vector i – 2j + 2k is (a) or (b) or (c) or (d) None of these Solution: (a) Equation of plane containing i and i + j is    z = 0 ……(i) Equation of plane containing i – j and i + k is    …… (ii) Let . Since a is parallel to (i) and (ii) ,  , Thus a vector in the direction of a is . If  is the angle between a and . Then   or Example: 54 The d.r.’s of normal to the plane through (1, 0, 0) and (0, 1, 0) which makes an angle with plane , are (a) (b) (c) 1, 1, 2 (d) Solution: (b) Let d.r.’s of normal to plane (a, b, c) ……(i) It is passes through (0, 1, 0).   . D.r.’s of normal is (a, a, c) and d.r.’s of given plane is (1, 1, 0)    Then, d.r.’s of normal or . Line and plane 7.24 Equation of plane through a given line. (1) If equation of the line is given in symmetrical form as , then equation of plane is …… (i) where a, b, c are given by ……(ii) (2) If equation of line is given in general form as , then the equation of plane passing through this line is . (3) Equation of plane through a given line parallel to another line : Let the d.c.’s of the other line be . Then, since the plane is parallel to the given line, normal is perpendicular.  ……(iii) Hence, the plane from (i), (ii) and (iii) is . 7.25 Transformation from unsymmetric form of the equation of line to the symmetric form. If and are equations of two non-parallel planes, then these two equations taken together represent a line. Thus the equation of straight line can be written as . This form is called unsymmetrical form of a line. To transform the equations to symmetrical form, we have to find the d.r.’s of line and co-ordinates of a point on the line. 7.26 Intersection point of a line and plane. To find the point of intersection of the line and the plane . The co-ordinates of any point on the line are given by (say) or .....(i) If it lies on the plane , then   . Substituting the value of r in (i), we obtain the co-ordinates of the required point of intersection. Algorithm for finding the point of intersection of a line and a plane Step I : Write the co-ordinates of any point on the line in terms of some parameters r (say). Step II : Substitute these co-ordinates in the equation of the plane to obtain the value of r. Step III : Put the value of r in the co-ordinates of the point in step I. 7.27 Angle between line and plane. (1) Cartesian form : The angle  between the line , and the plane , is given by . (i) The line is perpendicular to the plane if and only if . (ii) The line is parallel to the plane if and only if . (iii) The line lies in the plane if and only if and . (2) Vector form : If  is the angle between a line and the plane , then . (i) Condition of perpendicularity : If the line is perpendicular to the plane, then it is parallel to the normal to the plane. Therefore b and n are parallel. So, or b = n for some scalar . (ii) Condition of parallelism : If the line is parallel to the plane, then it is perpendicular to the normal to the plane. Therefore b and n are perpendicular. So, b.n = 0. (iii) If the line lies in the plane r.n = d, then (i) b.n = 0 and (ii) a.n = d. 7.28 Projection of a line on a plane. If P be the point of intersection of given line and plane and Q be the foot of the perpendicular from any point on the line to the plane then PQ is called the projection of given line on the given plane. Image of line about a plane : Let line is , plane is . Find point of intersection (say P) of line and plane. Find image (say Q) of point about the plane. Line PQ is the reflected line. Example: 55 The sine of angle between the straight line and the plane is (a) (b) (c) (d) Solution: (b) We know that Hence, Example: 56 Value of k such that the line is perpendicular to normal to the plane is (a) (b) (c) 4 (d) None of these Solution: (a) We have, or vector form of equation of line is i.e. and normal to the plane, . Given that,    . Example: 57 The equation of line of intersection of the planes , can be written as (a) (b) (c) (d) Solution: (c) Let equation of line ……(i) We have ……(ii) and ……(iii) Let z = 0. Now putting z = 0 in (ii) and (iii), we get, , , on solving these equations, we get . Equation of line passing through (1, 2, 0) is From equation (i) and (ii), and  i.e. . Hence, equation of line is . Example: 58 The equation of the plane containing the two lines and is (a) (b) (c) (d) None of these Solution: (a) Any plane through the first line may be written as ……(i) where, ……(ii) It will pass through the second line, if the point (0, 2, –1) on the second line also lies on (i) i.e. if , i.e., ……(iii) Solving (ii) and (iii), we get i.e.  Required plane is  . Example: 59 The plane which passes through the point (3, 2, 0) and the line is (a) (b) (c) (d) Solution: (a) Any plane through the line is ……(i) where, ……(ii) Plane (i) passes through (3, 2, 0), if i.e. ……(iii) From equation (ii) and (iii), .  .  Required plane is i.e. i.e. . Trick :  . Example: 60 The distance of point (–1, –5, –10) from the point of intersection of the line and plane is (a) 10 (b) 8 (c) 21 (d) 13 Solution: (d) Any point on the line is This lies on , then i.e. r = 0.  Point is (2, –1, 2). Its distance from (–1, –5, –10) is . Example: 61 The value of k such that lies in the plane is (a) 7 (b) –7 (c) No real value (d) 4 Solution: (a) Given, point (4, 2, k) is on the line and it also passes through the plane   . Example: 62 The distance between the line and the plane is (a) (b) (c) (d) Solution: (d) The given line is , Given plane,  Since  The line is parallel to plane. Thus the distance between line and plane is equal to length of perpendicular from a point on line to given plane. Hence, required distance . Sphere A sphere is the locus of a point which moves in space in such a way that its distance from a fixed point always remains constant. The fixed point is called the centre and the constant distance is called the radius of the sphere. 7.29 General equation of sphere. The general equation of a sphere is with centre (–u, –v, –w) i.e. (–(1/2) coeff. of x, –(1/2) coeff. of y, –(1/2) coeff. of z) and, radius From the above equation, we note the following characteristics of the equation of a sphere : (i) It is a second degree equation in x, y, z; (ii) The coefficients of are all equal; (iii) The terms containing the products xy, yz and zx are absent. Note :  The equation represents, (i) A real sphere, if . (ii) A point sphere, if . (iii) An imaginary sphere, if . Important Tips  If , then the radius of sphere is imaginary, whereas the centre is real. Such a sphere is called “pseudo-sphere” or a “virtual sphere.  The equation of the sphere contains four unknown constants u, v, w and d and therefore a sphere can be found to satisfy four conditions. 7.30 Equation in sphere in various forms. (1) Equation of sphere with given centre and radius (i) Cartesian form : The equation of a sphere with centre (a, b, c) and radius R is ……(i) If the centre is at the origin, then equation (i) takes the form , which is known as the standard form of the equation of the sphere. (ii) Vector form : The equation of sphere with centre at C(c) and radius ‘a’ is . (2) Diameter form of the equation of a sphere (i) Cartesian form : If and are the co-ordinates of the extremities of a diameter of a sphere, then its equation is . (ii) Vector form : If the position vectors of the extremities of a diameter of a sphere are a and b, then its equation is or . 7.31 Section of a sphere by a plane. Consider a sphere intersected by a plane. The set of points common to both sphere and plane is called a plane section of a sphere. The plane section of a sphere is always a circle. The equations of the sphere and the plane taken together represent the plane section. Let C be the centre of the sphere and M be the foot of the perpendicular from C on the plane. Then M is the centre of the circle and radius of the circle is given by The centre M of the circle is the point of intersection of the plane and line CM which passes through C and is perpendicular to the given plane. Centre : The foot of the perpendicular from the centre of the sphere to the plane is the centre of the circle. (radius of circle)2 = (radius of sphere)2 – (perpendicular from centre of spheres on the plane)2 Great circle : The section of a sphere by a plane through the centre of the sphere is a great circle. Its centre and radius are the same as those of the given sphere. 7.32 Condition of tangency of a plane to a sphere. A plane touches a given sphere if the perpendicular distance from the centre of the sphere to the plane is equal to the radius of the sphere. (1) Cartesian form : The plane touches the sphere , if (2) Vector form : The plane touches the sphere if . Important Tips  Two spheres and with centres and and radii and respectively (i) Do not meet and lies farther apart iff (ii) Touch internally iff (iii) Touch externally iff (iv) Cut in a circle iff (v) One lies within the other if . When two spheres touch each other the common tangent plane is and when they cut in a circle, the plane of the circle is ; coefficients of being unity in both the cases.  Let p be the length of perpendicular drawn from the centre of the sphere to the plane , then (i) The plane cuts the sphere in a circle iff p < r and in this case, the radius of circle is . (ii) The plane touches the sphere iff . (iii) The plane does not meet the sphere iff p > r.  Equation of concentric sphere : Any sphere concentric with the sphere is , where  is some real which makes it a sphere. 7.33 Intersection of straight line and a sphere. Let the equations of the sphere and the straight line be .....(i) And (say) .....(ii) Any point on the line (ii) is . If this point lies on the sphere (i) then we have, or, ....(iii) This is a quadratic equation in r and so gives two values of r and therefore the line (ii) meets the sphere (i) in two points which may be real, coincident and imaginary, according as root of (iii) are so. Note :  If l, m, n are the actual d.c.’s of the line, then and then the equation (iii) can be simplified. 7.34 Angle of intersection of two spheres. The angle of intersection of two spheres is the angle between the tangent planes to them at their point of intersection. As the radii of the spheres at this common point are normal to the tangent planes so this angle is also equal to the angle between the radii of the spheres at their point of intersection. If the angle of intersection of two spheres is a right angle, the spheres are said to be orthogonal. Condition for orthogonality of two spheres Let the equation of the two spheres be .....(i) and .....(ii) If the sphere (i) and (ii) cut orthogonally, then which is the required condition. Note :  If the spheres and cut orthogonally, then .  Two spheres of radii and cut orthogonally, then the radius of the common circle is . Example: 63 The centre of sphere passing through four points (0, 0, 0), (0, 2, 0), (1, 0, 0) and (0, 0, 4) is (a) (b) (c) (d) Solution: (a) Let the equation of sphere be It passes through (0, 0, 0), Also, It passes through (0, 2, 0) i.e., Also, It passes through (1, 0, 0) i.e., Also, it passes through (0, 0, 4) i.e.,  Centre (–u, –v, –w) = (1/2, 1, 1/2) Example: 64 The equation represents a (a) Plane (b) Sphere of radius 4 (c) Sphere of radius 3 (d) None of these Solution: (b) The given equation is  , which is the equation of sphere, whose centre is (1, 2 –1) and radius . Example: 65 The intersection of the spheres and is the same as the intersection of one of the sphere and the plane (a) (b) (c) (d) Solution: (a) We have the spheres and Required plane is  i.e.  . Example: 66 The radius of the circle in which the sphere is cut by the plane is (a) 1 (b) 2 (c) 3 (d) 4 Solution: (c) For sphere , Centre O is (–1, 1, 2) and radius , Now, OL = length of perpendicular from O to plane is , i.e. . In . Example: 67 The radius of circular section of the sphere by the plane is (a) 2 (b) 3 (c) 4 (d) 6 Solution: (b) Radius of the sphere =5 Given plane is Length of the perpendicular from the centre (0, 0, 0) of the sphere to the plane = Hence, radius of circular section . Example: 68 The shortest distance from the plane to the sphere (a) 26 (b) (c) 13 (d) 39 Solution: (c) Centre of sphere is (–2, 1, 3) Radius of sphere is Distance of centre from plane  Plane cuts the sphere and hence . .