IIT-JEE (2012) (Vector+3D+Probability) Solutions
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1 L.K. Gupta (Mathematic Classes) MOBILE: , 4677 PAPER -A IIT-JEE (0) (Vector+D+Probability) Solutions TOWARDS IIT- JEE IS NOT A JOURNEY, IT S A BATTLE, ONLY THE TOUGHEST WILL SURVIVE TIME: 60 MINS MAX. MARKS: 80 MARKING SCHEME In Section I (Total Marks: ), for each question you will be awarded marks if you darken ONLY the bubble corresponding to the correct answer and zero marks if no bubble is darkened. In all other cases, minus one ( ) mark will be awarded. In Section II (Total Marks: 6), for each question you will be awarded 4 marks if you darken ALL the bubble(s) corresponding to the correct answer(s) ONLY and zero marks otherwise. There are no negative marks in the section. In Section III (Total Marks : 5), for each question you will be awarded marks if you darken ONLY the bubble corresponding to the correct answer and zero marks if no bubble is darkened. In all other cases, minus one () mark will be awarded. In Section IV (Total Marks: 8), for each question you will be awarded 4 marks if you darken ONLY the bubble corresponding to the correct answer and zero marks otherwise, There are no negative marks in this section NAME OF THE CANDIDATE CONTACT NUMBER L.K. Gupta (Mathematics Classes) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH
2 L.K. Gupta (Mathematic Classes) MOBILE: , 4677 Section-I (Total Marks: ) (Single Correct Answer Type) This section contains 7 multiple choice questions. Each question has 4 choices.(a), (b), (c) and (d), out of which ONLY ONE is correct.. The line x y z intersects the curve (a) (b) xy c,z 0 if c is equal to. (c) 5 (d) none of these Sol. (c) we have z =0 for the point where the line intersects the curve x y 0 x y and x 5 and y Put these values in xy c we get 5 c c 5. The line joining the points (,, ) and (,, ) meets the plane x y z 6 at the point. (a) (,, ) (b) (,, ) (c) (,, ) (d) (,, ) Sol. (b),, and,, is The straight line joining the points x y z r (say) Point is r, r, r which lies on x y z 6 r r r 6 r Required point is,,.. If a and b (a) 48b (b) 48b and a.b = 0 then (a (a (a (a b))) is equal to (c) 48a (d) 48a PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH
3 L.K. Gupta (Mathematic Classes) MOBILE: , 4677 Sol. (a) a. b 0 are mutually perpendicular a and b Also a and b = Let a iˆ then b iˆ such that a iˆ, b ˆj a a a ab iˆ iˆ iˆ iˆ ˆj 48 ˆj 48bˆ 4. The value of the x for which the angle between the vectorsx i 4xj k and 7i j xk are obtuse and the angle between the z axis and 7i j xk is acute and Less than π 6 is given by (a) 0 x (b) x or x 0 (c) x 5 (d) there is no such value for x Sol. (d) Let a x iˆ 4xj ˆ kˆ and b 7iˆ ˆj xkˆ the angle between a and b is obtuse. a. b 0 4x 8x x 0 x x 0 x 0,... i Also it is given b. k x and b. k cos b 6 x 5 x x ii there is one common value for Eqs. (i) and (ii) 5. Let v i j k and w i k If u is a unit vector, then maximum value of the Scalar triple product [uvw] is. (a) (b) 0 6 (c) 59 (d) 60 Sol. (c) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH
4 L.K. Gupta (Mathematic Classes) MOBILE: , u. i ˆ 7 ˆj kˆ u 59 cos uvw u,cos uvw u v w Maximum If a, b 4 and a b 5, then a b is equal to (a) (b) 4 (c) 5 (d) 6 Sol. (c) a b a b a b a. b 9 6 a. b 5 a. b.(i) But a. b 5 (given) a b 5 a b a. b a. b 5 a. b 0 from eq.(i) a b A piece of wire of length 4L is bent at random, to from a rectangle. The probability L That its area is at most 4 is (a) (b) (c) 4 (d) 4 Sol. (c) x y 4L x y L The area of the rectangle = x y x L x But given x L x L 4 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 4
5 L.K. Gupta (Mathematic Classes) MOBILE: , 4677 L x Lx 0 4 L x L L 0 4 L x L 0 L L x L x L 0 x Lx L 0 x 0, L L,L Now, x takes value in the interval [0, L] Required Probability L L dx dx 0 L L L L L L 0 dx 0 L L L PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 5
6 L.K. Gupta (Mathematic Classes) MOBILE: , 4677 Section II (Total Marks: 6) (Multiple Correct Answers Type) This section contains 4 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONE or MORE may be correct. 8. The lines x y z 4 x and y 4 z 5 are coplanar if k k (a) k = 0 (b) k = (c) k = (d) k = Sol. (a), (c) The given lines are coplanar, if k k k k Applying C C C and C C C k k k k or k k k 0 or k k 0 k 0, 9. The probabilities that a student in Mathematics, Physics and Chemistry are, and respectively. Of these subjects, a student has a 75% chance of passing in atleast one, a 50% chance of passing in atleast two and a 40% chance of passing n exactly two subjects. Which of the following relations are true? (a) 9/0 (b) 7 / 0 (c) / 0 (d) /4 Sol. (b,c) Here,P M, P P and P C The probability of passing in atleast one subject = 0.75 (given) P MPC 0.75 P M P P P C 0.75 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 6
7 L.K. Gupta (Mathematic Classes) MOBILE: , or (i) 4 The probability of passing in atleast two subjects = 0.50 (given) or PM P C P M P C PM P C PM P C 0.50 PM PP PC PM PPPC PMPPPCPM PPPC (ii) and the probability of passing in exactly two subjects = 0.40 (given) PM PC PM P C PM P C 5 PM PP PC PMPPPC PMPPPC 5 5 (iii) 5 Eqs. (ii) and (iii) 5 (c) 5 0 From Eq. (ii), (iv) Substituting the values of and in Eq. (i), then (b) The letters of the word PROBABILITY are written down at random in a row, Let E denotes the event that two I s are together and E denotes the event that two B s are together, then (a) PE PE (b) PE E 55 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 7
8 L.K. Gupta (Mathematic Classes) MOBILE: , (c) PE E (d) PE / E 55 5 Sol. (a, b, c, d) Option (a) : PE PE 0!! Option (b) : PE E 9!!!! !!! = probability of two I s are together and two B s are together Option (c) : PE E PE PE PE E Option (d) : PE / E PE E 55 P E 5. The vector c directed along the bisectors of the angle between the vectors a 7i 4j 4k and b i j k if c 6 is given by (a) i 7 j k (b) i 7 j k (c) i 7 j k (d) i 7 j k Sol. (a, c) 7i 4j 4k i j k c 9 c i 7j k (i) 9 c Putting 9 in Eq. (i) Then, c i 7 j k PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 8
9 L.K. Gupta (Mathematic Classes) MOBILE: , 4677 Section III (Total Marks: 5) (Paragraph Type) This section contains two paragraphs. Based upon one of the paragraph, multiple choice questions and based on the other paragraph multiple choice questions have to be answered. Each of these questions has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct. Passage for questions Nos. and Let a,b,c be three vector such that a b c 4 and angle between aandb is π /, angle Between b and c is π / and angle between c and a is π /. On the basis of above information, answer the following questions:. The height of the parallelepiped whose adjacent edges are represented by vectors a,bandc is (a) 4 (b) (c) 4 (d) Sol. (a) Volume of the parallelepiped = (base area ) (height ) 44sin h 8 h h 4. The volume of the tetrahedron whose adjacent edges are represented by the vectors a,bandc is 4 (a) (b) 8 (c) 6 (d) 6 Sol.(d) Volume of the tetrahedron 6 abc PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 9
10 L.K. Gupta (Mathematic Classes) MOBILE: , Paragraph for question 4 to 6 Let two planes P :x y z and P : x y z are given. On the basis of above information, answer the following questions 4. The equation of the lane through the intersection of P andp and the point (,,) is (a) x y z 9 0 (b) x y z 0 (c) x y z 0 (d) 4x y z 8 0 Sol. (b) The equation of any plane through the intersection of P and P is P P 0 x y z x y z 0... i Since it passes through (,,) then From Eq. () x y z 0 which is the required plane. 5. Equation of the plane which passes through the point (,, ) and is perpendicular to each of the planes P andp is (a) x y 5z 0 (b) x y 5z 8 0 (c) x y 5z 0 0 (d) x y + 5z = 0 Sol. (c) The equation of any plane through,, is a x b y c z 0 (ii) if this plane (ii) is perpendicular to P then a b c 0... ii and if the plane (ii) is perpendicular to P then a b c 0... iii From Eqs. (ii) and (iii) we get a b c 5 Substituting these Proportionate values of a, b, c in Eq. (ii) we get the required x y 5 z 0 equation as or x y 5z 0 0 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 0
11 L.K. Gupta (Mathematic Classes) MOBILE: , The equation of the acute angle bisector of planes P andp is (a) x y z 0 (b) x y 5 0 (c) x y z 0 (d) x z 7 0 Sol. (a) The given plane can be written as x y z 0 and x y z 0 Here 0 Equation of bisectors x y z x y z 4 4 Acute angle bisector is x y z x y z x y z 0 Section IV (Total Marks: 8) (Integer Answer Type) This section contains 7 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9. The bubble corresponding to the correct answer is to be darkened in the Answer sheet. 7. ABCD is a parallelogram. A and B are the mid points of sides BC and CD respectively. If AA AB AC, then the value of must be Ans. 4 Sol. Let position vectors of A, B, D be O, b and d respectively. AC AB BC AB AD b d AB AC AA PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH
12 L.K. Gupta (Mathematic Classes) MOBILE: , 4677 d b AD AC b and AB d AA AB b d AC Then, A, B, C, D are any four points in the space. If AB CD BC AD CA BD (area of ABC) then the value of must be Ans. 4 Sol. Let PV of A, B, C and D be a, b, c and O AB CD b a c b c a c BC AD c b a c a b a and CA BD a c b a b c b Adding all we get AB CD BC AD CA BD a b b c c a AB CD BC AD CA BD a b b c c a c a b a AC AB 4. AC AB 4 (Area of ABC) 4 Then, PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH
13 L.K. Gupta (Mathematic Classes) MOBILE: , Let a and b be unit vectors such that a b, then the value of 00a 5b. a b a b 894 must be Ans. 6 Sol. a b a b a b a. b a. b 00 a 5b. a b a b 00 6a a.b 0 5 a. b 5b If angle between i and line of intersection of the planes r. i j k 0 and r. i j k 0 is cos 090 then the value of must be Ans. 8 Sol. Line of intersection of r. i j k 0 and r. i j k 0 will be parallel to i j k i j k, ie, 7 i 8j k, then required angle is. 7 cos cos or cos PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH
14 L.K. Gupta (Mathematic Classes) MOBILE: , If the distance of the point B from the line which is passing through i j k A 4i j k and which is parallel to the vector c i j 6k is, then the value of must be Ans. Sol. AB OB OA i j k 4i j k i 0j k AB Now, AM = Projection of AB along c AB. c c i 0j k. i j 6k BA AM 7 Perpendicular distance of B from AM = AB 0 (given) =. If the area of the triangle whose vertices are A (,, ), B (,, ) and C(,, 4) is sq unit then Ans. 0 Sol. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 4
15 L.K. Gupta (Mathematic Classes) MOBILE: , 4677 The coordinates of the projections of A, B, C on the yz-plane are (0,, ), (0,, ) and (0,, 4) respectively = area of projection of ABC on yz-plane x 4 Applying R R R and R R R 0 5 then x sq unit 0 Similarly, the projection of A, B and C on zx and xy-planes are (, 0, ), (, 0, ), (, 0, 4) and (,, 0), (,, 0), (,, 0) respectively. Also let y and z be the areas of the projection of the ABC on zx and xy-planes respectively. Then, y 4 Applying R R R and R R R Then, y sq unit and z Applying C C C 0 : Then, z 0 : 0 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 5
16 L.K. Gupta (Mathematic Classes) MOBILE: , 4677 The required area sq unit 7 0 sq unit 0 x y z If the distance of the point P (4,, 5) from the axis of y is unit, then the value of 5 04 must be Ans. Sol. The equations of y-axis are x y z, 0 0 Any point N on y-axis is (0, r, 0).(i) The direction cosines of the line PN are 0 4, r, 0 5 ie, 4, r, 5..(ii) Let N be the foot of the perpendicular from P to y-axis, then PN is to the y-axis whose direction cosines are 0,, 0 and so from Eq. (ii), we have 0. ( 4) +. (r ) + 0. ( 5) = 0 r = From Eq. (i) the coordinates of N are (0,, 0) Required distance = PN unit PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 6
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