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12/03/2022, 18:03, , Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry – HSSLive …, , Menu, , HSSLive Guru, , Plus Two Maths Chapter Wise Questions and, Answers Chapter 11 Three Dimensional, Geometry, February 11, 2020 by Prasanna, , Students can Download Chapter 11 Three Dimensional Geometry Questions and, Answers, Plus Two Maths Chapter Wise Questions and Answers helps you to revise, the complete Kerala State Syllabus and score more marks in your examinations., , Kerala Plus Two Maths Chapter Wise Questions and, Answers Chapter 11 Three Dimensional Geometry, Plus Two Maths Three Dimensional Geometry Three Mark, Questions and Answers, Question 1., , 1. Write the Cartesian equation. (1), 2. Find the angle between the line. (2), Answer:, 1., , https://hssliveguru.com/plus-two-maths-chapter-wise-questions-and-answers-chapter-11/, , 1/25

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12/03/2022, 18:03, , Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry – HSSLive …, , 2. cosθ, , Question 2., Find the vector equation of the plane passing through the intersection of the planes r̄, .(i + j + k) = 6 and r̄ .(2i + 3 j + 4k) = -5 at the point (1,1,1)., Answer:, The Cartesian equation of the planes are x + y + z = 6 and 2x + 3y + 4z = – 5. Therefore, the equation of the plane passing through the intersection of these planes is, x + y + z – 6 + k(2x + 3y + 4z + 5) = 0, Since it pass through (1, 1, 1) we get,, 1 + 1 + 1 – 6 + k(2 + 3 + 4 + 5) = 0 ⇒ -3 + k14 ⇒ k =, , 3, 14, , ∴ the equation is, 3, x + y + z + -6 + 14 (2x + 3 y + 4z + 5) = 0, 14x + 14y + 14z – 84 + 6x + 9y + 12z + 15 = 0, 20x + 23y + 26z = 69, Vector equation is r̄ . (20i + 23j + 26k) = 69., , Question 3., Find the equation of the plane passing through the intersection of the planes x + y +, 4z + 5 = 0 and 2x – y + 3z + 6 = 0 and contains the point (1, 0, 0)., Answer:, The equation of the planes passing through the intersection of the planes, x + y + 4z + 5 = 0 and 2x – y + 3z + 6 = 0 is, x + y + 4z + 5 + k(2x – y + 3z + 6) = 0 ____(1), Since (1) pass through (1, 0, 0), ⇒ 1 + 0 + 0 + 5 + k(2 – 0 + 0 + 6) = 0, 3, , ⇒ 6 + 8k = 0 ⇒ k = – 4 ; Then (1), ⇒x+y+z+5+, , 3, 4, , (2x – y + 3z + 6) = 0, , ⇒ 4x + 4y + 16z + 20 – 6x + 3y – 9z – 18 = 0, ⇒ 2x – 7y – 7z = 2., https://hssliveguru.com/plus-two-maths-chapter-wise-questions-and-answers-chapter-11/, , 2/25

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12/03/2022, 18:03, , Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry – HSSLive …, , Plus Two Maths Three Dimensional Geometry Four Mark, Questions and Answers, Question 1., Consider the point (-1, -2, -3)., 1. In which octant, the above point lies.(1), 2. Find the direction cosines of the line joining (-1, -2, -3) and (3, 4, 5). (1), −, −, , 3. If P is any point such that OP = √50 and direction cosines of OP are, , 3, √50, , and, , 5, , ,, , 4, √50, , , then find the co-ordinate of P. (2), , √50, , Answer:, 1. The point lies in the octant X’OY’Z’., 2. Direction ratios of the line joining (-1, -2, -3) and (3, 4, 5) are (3 + 1), (4 + 2), (5 + 3) ⇒, 4, 6, 8 ⇒ 2, 3, 4., Therefore direction cosines are, , −, −, , 3. Given, OP = √50, , Therefore the point is (3, 4, 5)., , Question 2., Consider a cube of side ‘a’ unit has one vertex at the origin O., , https://hssliveguru.com/plus-two-maths-chapter-wise-questions-and-answers-chapter-11/, , 3/25

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12/03/2022, 18:03, , Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry – HSSLive …, , 1. Write down the co-ordinate of 0, 0′, A and A’ (1), 2. Find the direction ratios of OO’ and AA’. (2), 3. Show that the angle between the main diagonals of the above cube is cos-1, (, , 1, 3, , ), , (1), , Answer:, 1. O(0, 0, 0), O'(a, a, a), A(a, 0, 0) and A'(0, a, a)., 2. Direction ratios along OO’ is a – 0, a – 0, a – 0, ⇒ a, a ,a ⇒ 1, 1, 1, 3. cosθ, Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three, Dimensional Geometry 7, Question 3., Consider two points A and B and a line L as shown in the figure., , ¯, ¯¯¯¯, ¯¯, ¯, , 1. Find AB (1), 2. Find the Cartesian equation of the line L. (1), 3. Find the foot of the perpendicular drawn from ( 2, 3, 4 ) to the line L. (2), Answer:, ¯, ¯¯¯¯, ¯¯, ¯, , 1. AB = (3 – 1)i + (- 3 – 3)j + (3 – 0)k = 2i – 6j + 3k., , https://hssliveguru.com/plus-two-maths-chapter-wise-questions-and-answers-chapter-11/, , 4/25

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12/03/2022, 18:03, , Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry – HSSLive …, , 2. The Cartesian equation of a line passing through the point (4, 0, -1 ) and parallel to, ¯, ¯¯¯¯, ¯¯, ¯, , the vector AB is, , 3. We take,, , x−4, 2, , =, , y, −6, , =, , z+1, 3, , = r then any point of the line can be taken as (2r + 4,, , -6r, 3r – 1). Assume that this point be the foot of the perpendicular drawn from (2, 3,, 4 ). The dr’s of the line is 2 : – 6 : 3 and dr’s of the perpendicular line L is, 2r + 4 – 2 : -6r – 3 : 3r – 1 – 4 ⇒ 2r + 2: -6r – 3 : 3r – 5, Since perpendicular,, 2(2 r + 2) – 6(-6 r -3) + 3(3r – 5) = 0, 49r = -7 ⇒ r =, , −7, 49, , = – 17 . Therefore the foot of the, , perpendicular is, , Question 4., Cartesian equation of two lines are, , (i) Write the vector equation of the lines. (2), (ii) Shortest distance between the lines. (2), Answer:, , https://hssliveguru.com/plus-two-maths-chapter-wise-questions-and-answers-chapter-11/, , 5/25

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12/03/2022, 18:03, , Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry – HSSLive …, , Question 5., Consider the points (1, 3, 4) & (-3, 5, 2), 1. Find the equation of the line through P and Q. (1), 2. At which point that the above line cuts the plane 2x + y + z + 3 = 0. (3), Answer:, 1. Equation of a line passing through( 1, 3, 4) and (-3, 5, 2) is given by,, , 2. Let, , x−1, −2, , =, , y−3, 1, , =, , z−4, −1, , =λ, , Then any point on the line is (-2λ + 1, λ + 3, -λ + 4), Since the plane 2x + y + z + 3 = 0 cuts the aboveline. We have,, ⇒ 2(-λ + 1) + λ + 3 – λ + 4 + 3 = 0, ⇒ -2λ + 2 + λ + 3 – λ + 4 + 3 = 0, ⇒ -2λ = -12 ⇒ λ = 6, ∴ point of intersection is (-2 × 6 + 1, 6 + 3, -6 + 4), ⇒ (-11, 9, -2)., , https://hssliveguru.com/plus-two-maths-chapter-wise-questions-and-answers-chapter-11/, , 6/25

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12/03/2022, 18:03, , Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry – HSSLive …, , Question 6., Let the equation of a plane be r̄ . (2i – 3j + 5k) = 7, then, 1. Find the Cartesian equation of the plane. (1), 2. Find the equation of a plane passing through the point (3, 4, -1) and parallel to, the given plane. (2), 3. Find the distance between the parallel planes. (1), Answer:, 1. Given, r̄ .(2i – 3j + 5k) = 7 and if we substitute r̄ = xi + yj + zk Then we get the, Cartesian equation as 2x – 3y + 5z – 7 = 0., 2. The equation of a plane parallel to the above plane differ only by a constant,, therefore let the equation be 2x – 3y + 5z + k = 0., ⇒ 6 – 12 – 5 + k = 0 ⇒ k = 11, Therefore the equation is 2x – 3y + 5z + 11 = 0, 3. The distance between the parallel planes, Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 16, Question 7., 1. State the condition for the line r̄ = ā + λ ¯, b is parallel to the plane r̄ .n̄ = d. (2), 2. Show that the line r̄ = i + j + λ(2i + j + 4k) is parallel to the plane r̄ . (-2i + k) = 5., (1), 3. Find the distance between the line and The Plane in (ii). (1), Answer:, 1. The line r̄ = ā + λ ¯, b is parallel to the plane r̄ .n̄ = d, if the normal of the plane is, perpendicularto the line., ∴¯, b.n̄ = 0., 2. Given,, , The line r̄ = i + j + λ(2i + j + 4k) is parallel to the plane r̄ . (-2i + 4k) = 5 ⇒ -2x + 4y = 5., , https://hssliveguru.com/plus-two-maths-chapter-wise-questions-and-answers-chapter-11/, , 7/25

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12/03/2022, 18:03, , Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry – HSSLive …, , 3. Distance = Distance between – 2x + 4y = 5 and point (1, 1, 0) on the line, , Question 8., Choose the correct answer from the bracket,, (i) If a line in the space makes angle α, β and γ with the coordinates axes, then cos2α +, cos2β + cos2γ is equal to (1), (a) 1, (b) 2, (c) 0, (d) 3, (ii) The direction ratios of the line are, , x−6, 1, , =, , 2−y, 2, , =, , z−2, 2, , (1), , (a) 6, -2, -2, (b) 1, 2, 2, (c) 6, 1, -2, (d) 0, 0, 0, (iii) If the vector equation of a line is r̄ = i + j + k + µ(2i – 3j – 4k), then the Cartesian, equation of the line, , (iv) If the Cartesian equation of a plane is x + y + z =12, then the vector equation of, the line is (1), (a) r̄ .(2i + j + k) = 12, (b) r̄ .(i + j + k) = 12, (C) r̄ .(i + y + 2k) = 12, (d) r̄ .(i + 3j + k) = 12, Answer:, (i) (a) 1, (ii) (b) 1, 2, 2, (iii) (b), , x−I, 2, , =, , y−1, −3, , =, , z−1, −4, , (iv) (b) r̄ .(i + j + k) = 12., https://hssliveguru.com/plus-two-maths-chapter-wise-questions-and-answers-chapter-11/, , 8/25

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12/03/2022, 18:03, , Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry – HSSLive …, , Question 9., Consider the lines r̄ = (i + 2j – 2k) + λ(i + 2 j) and r̄ = (i + 2j – 2k) + µ(2j – k), 1. Find the angle between the lines., 2. Find a vector perpendicular to both the lines., 3. Find the equation of the line passing through the point of intersection of lines, and perpendicular to both the lines., Answer:, ¯, 1. ¯, b1 = i + 2j; b2 = 2j – k, , ¯, 2. Perpendicular vector = ¯, b1 × b2, , = i(-2 – 0) -j(-1 – 0) + k(2 – 0), = -2i + j + k., 3. Equation of line is r̄ = (i + 2j – k) + µ(-2i + j + 2k)., , Question 10., Consider the line r̄ = (2i – j + k) + λ(i + 2j + 3k), 1. Find the Cartesian equation of the line., 2. Find the vector equation of the line passing through A (1, 0, 2) and parallel to, the above line., 3. Write two points on the line obtained in (ii) which are equidistant from A., , https://hssliveguru.com/plus-two-maths-chapter-wise-questions-and-answers-chapter-11/, , 9/25

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12/03/2022, 18:03, , Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry – HSSLive …, , Question 12., Consider the Cartesian equation of a line, , x−3, 2, , =, , y+1, 3, , =, , z−5, −2, , 1. Find the vector equation of the line. (1), 2. Find its intersecting point with the plane 5x + 2y – 6z – 7 = 0 (2), 3. Find the angle made by the line with the plane 5x + 2y – 6z – 7 = 0 (1), Answer:, 1. The vector equation is r̄ = (3i – j + 5k) + λ(2i + 3 j – 2k)., 2. Any point on the line is, x−3, 2, , =, , y+1, 3, , =, , z−5, −2, , =λ, , x = 2λ + 3, y = 3, λ – 1, z = -2λ + 5, Since this lies on the plane ,it satisfies the plane, 5(2λ + 3) + 2(3λ – 1) -6(-2λ + 5) – 7 = 0, 10λ + 6λ + 12λ + 15 – 2 – 30 – 7 = 0, 28λ = 24, λ = 6/7, 33, The point of intersection is [ 7 ,, , 11, 7, , ,, , 23, 7, , ], , ., , 3. Let θ be the angle between the line and the plane. The direction of the line and the, plane, ¯, b, , = 2i + 3j + k; m̄ = 5i + 2j – 6k, , Question 13., From the following figure, , https://hssliveguru.com/plus-two-maths-chapter-wise-questions-and-answers-chapter-11/, , 11/25

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12/03/2022, 18:03, , Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry – HSSLive …, ¯, ¯¯¯¯, ¯¯, ¯, , 1. Find AB. (1), 2. Find the vector equation of line L. (1), 3. Find a point on line L other than C. (2), Answer:, 1. P.v of A = i – j + 4k,, P.v. of B = 2i + j + 2k, ¯, ¯¯¯¯, ¯¯, ¯, , = p. v. of B – p. v. of A, = 2i + j + 2k -(i – j + 4k) = i + 2j – 2k., AB, , ¯, ¯¯¯¯, ¯¯, ¯, , 2. The line L passes through (1, -2, -3) and parallel to AB, , ∴ Vector equation of line L is r̄, , = ā + λm̄, , 3. From (1) of part (ii), we have, xi + yj + zk = (l + λ)i + (-2 + 2λ)j + (-3 – 2λ)k, Put λ = 1, ⇒ xi +yj + zk = (1 +1)i + (-2 + 2)j + (-3 – 2 )k, ⇒ xi + yj + zk = 2i + 0j – 5k, Therefore a point on line L is (2, 0, -5)., , Question 14., Find the vector equation of the plane which is at a distance of, , 6, √29, , from the origin, , with perpendicular vector 2i – 3j + 4k. Convert into Cartesian form. Also, find the foot, of the perpendicular drawn from the origin to the Plane., Answer:, , Perpendicular distance from origin = d =, , 6, √29, , ^ =d, The equation of the Plane is r̄ .n, Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 31, , Cartesian equation is 2x – 3y + 4z = 6, The direction cosines perpendicular to the Plane is, Perpendicular distance to the Plane is as, , 2, √29, , ,−, , 3, √29, , ,, , 4, √29, , ., , 6, √29, , https://hssliveguru.com/plus-two-maths-chapter-wise-questions-and-answers-chapter-11/, , 12/25

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12/03/2022, 18:03, , Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry – HSSLive …, , Hence the foot of the perpendicular is, , Question 15., Consider the Plane r̄ .(-6i -3j – 2k) + 1 = 0, find the direction cosines perpendicular to, the Plane and perpendicular distance from the origin., Answer:, Convert the equation of the plane into normal form, , Direction cosines perpendicular to the Plane is, Perpendicular distance from the origin is, , 1, 7, , 6, 7, , ,, , 3, 7, , ,, , 2, 7, , ., , Question 16., Consider three points (6, -1, 1), (5, 1, 2) and (1, – 5, -4) on space., 1. Find the Cartesian equation of the plane passing through these points. (2), 2. Find direction ratios normal to the Plane.(1), 3. Find a unit vector normal to the Plane. (1), Answer:, 1. Equation of a plane passing through the points (6, -1, 1),(5, 1, 2) and (1, -5, 4), , https://hssliveguru.com/plus-two-maths-chapter-wise-questions-and-answers-chapter-11/, , 13/25

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12/03/2022, 18:03, , Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry – HSSLive …, , ⇒ (x – 6)(-10 + 4) – (y + 1)(5 + 5) + (z – 1)(4 + 10) = 0, ⇒ (x – 6)(-6) – (y + 1)(10) + (z – 1)(14) = 0, ⇒ -6x + 36 – 10y – 10 + 14z – 14 = 0, ⇒ 6x +10y – 14z -12 = 0., 2. Dr’s normal to the plane are 6 : 10 : -14 ⇒ 3 : 5 : -7., 3. Since the dr’s normal to the plane are 3 : 5 : -7, a unit vector in this direction is, , Question 17., Consider a straight line through a fixed point with position vector 2i – 2j + 3k and, parallel to i – j + 4k., 1. Write down the vector equation of the straight line. (1), 2. Show that the straight line is parallel to the plane r̄ .(i + 5y + k) = 5 (1), 3. Find the distance between the line and plane. (2), Answer:, 1. Vector equation of a straight line is r̄, , ¯, = ā + λ b, , where a is ā fixed point and ¯, b is a, , vector parallel to the line. Here ā = 2i – 2y + 3 it and ¯, b = i – j + 4k. Therefore vector, equation of the line r̄ = 2i – 2j + 3k + λ(i – j + 4k)., 2. The vector parallel to the line is i – j + 4k and vector normal to the plane is i + 5j + k., Then, (i – j + 4k). (i + 5j + k) = 1 – 5 + 4 = 0, implies that straight line and plane are parallel., , https://hssliveguru.com/plus-two-maths-chapter-wise-questions-and-answers-chapter-11/, , 14/25

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12/03/2022, 18:03, , Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry – HSSLive …, , 3. A point on the line is 2i – 2j + 3k. Then the distance of 2i – 2j + 3k to the given plane, , Question 18., Consider the vector equation of two planes r̄ .(2i + j + k) = 3, r̄ .(i – j – k) = 4, 1. Find the vector equation of any plane through the intersection of the above, two planes. (2), 2. Find the vector equation of the plane through the intersection of the above, planes and the point (1, 2, -1 ) (2), Answer:, 1. The cartesian equation are 2x + y + z – 3 = 0 and x – y – z – 4 = 0 Required equation, of the plane is, (2x + y + z – 3) + λ(x – y – z – 4) = 0, (2+ λ)x + (1 – λ)y + (1 – λ)z + (-3 – 4λ) = 0., 2. The above plane passes through (1, 2, -1), (2+ λ)1 + (1 – λ)2 + (1 – λ)(-1) + (-3 – 4λ) = 0, 3 – 3 + 4λ = 0, λ=0, Equation of the plane is 2x + y + z – 3 = 0, r̄ .(2i + j + k) = 3., Question 19., (i) Distance of the point(0, 0, 1) from the plane x + y + z = 3, (a) 1 units, √3, , (b), , 2, √3, , units, , –, , (c) √3 units, (d), , √3, 2, , units, , (ii) Find the equation of the plane through the line of intersection of the planes x + y, + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to x – y + z = 0 (3), , https://hssliveguru.com/plus-two-maths-chapter-wise-questions-and-answers-chapter-11/, , 15/25

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12/03/2022, 18:03, , Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry – HSSLive …, , Answer:, (i) (b), , 2, √3, , units., , (ii) Equation of the plane passing through the intersection is of the form, x + y + z – 1 + λ(2x + 3y + 4z – 5) = 0 _____(1), (1 + 2λ)x + (1 + 3λ)j + (1 + 4λ)z – 1 – 5λ = 0, Thr Dr’s of the required plane is, (1 + 2λ), (1 + 3λ), (1 + 4λ), Thr Dr’s of the Perpendicular plane is 1, -1, 1, ⇒ (1 + 2λ)(1) + (1 + 3λ)(-1) + (1 + 4λ)(1) = 0, ⇒ 1 + 2λ – 1 – 3λ + 1 + 4λ = 0, ⇒ 3λ + 1 = 0 ⇒ λ =, (1) ⇒ x + y + z –, , 1, 3, , −1, 3, , (2x + 3y + 4z – 5) = 0, , ⇒ 3x + 3y + 3z – 2x – 3y – 4z + 5 = 0, ⇒ x – z + 2 = 0., , Question 20., Consider a plane r̄ .(6i – 3j – 2k) + 1 = 0, 1. Find dc’s perpendicular to the plane. (2), 2. Find a vector of magnitude 14 units perpendicular to given plane. (1), 3. Find the equation of a line parallel to the above vector and passing through, the point (1, 2, 1 ). (1), Answer:, 1. Given, r̄ .(6i – 3j – 2k) + 1 = 0 ____(1), −, −, −, −, −−, −, −, −, , Now, |6i – 3j – 2k| = √36 + 9 + 4 = 7, 6, 3, ∴ 7 i − 7 j − 27 k is a unit perpendicular to the plane (1), ⇒ the dc’s perpendicular to the plane (1) are, 2. We have,, , 6, 7, , i −, , 3, 7, , j −, , 2, 7, , k, , 6, 7, , ,−, , 3, 7, , ,−, , 2, 7, , ., , is a unit perpendicular to the Plane (1). Therefore, a, 6, , vector of magnitude 14 units perpendicular to the Plane (1) is 14( 7 i −, , 3, 7, , j −, , 2, 7, , ), , k, , ⇒ 12i – 6j – 4k., 3. Equation of a line parallel to the vector 12i – 6j – 4k and passing through the point, (1, 2, 1 )is given by, , https://hssliveguru.com/plus-two-maths-chapter-wise-questions-and-answers-chapter-11/, , 16/25

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12/03/2022, 18:03, , Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry – HSSLive …, , Plus Two Maths Three Dimensional Geometry Six Mark, Questions and Answers, Question 1., Consider the pair of lines whose equations are, , 1. Write the direction ratios of the lines. (1), 2. Find the shortest distance between the above skew lines. (4), 3. Find the angle between these two lines. (1), Answer:, 1. The direction ratios are 2, 5, – 3 and – 1, 8, 4., 2. The given lines are r̄ = (2i + j – 3k) + λ(2i + 5j – 3k), ¯¯¯¯, , ¯¯¯, ¯, i.e. r̄ = ¯a, 1 + λb 1 ,, ¯¯¯, ¯, where ¯a, = 2i + j – 3k) + λ(2i + 5j – 3k), 1, , and r̄ =(-i + 4j + 5k) + µ(-i + 8j + 4k), , https://hssliveguru.com/plus-two-maths-chapter-wise-questions-and-answers-chapter-11/, , 17/25

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12/03/2022, 18:03, , Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry – HSSLive …, , 3. Let L1 ⇒ r̄ = 3i + 4j – 2k + λ(-i + 2j + k) is of the form r̄, , ¯¯¯¯, , ¯¯¯, ¯, = ¯, a, 1 + λb 1, , where, , Question 3., Consider the points A (2, 2, -1), B (3, 4, 2) and C (7, 0, 6)., 1. Are A, B, and C collinear? Explain., 2. Find the vector and Cartesian equation of the plane passing these three, points. (2), 3. Find the angle between the above plane and the line r̄ = (i + 2j – k) + λ(i – j + k), (2), Answer:, 1. Direction ratios along A and B is 3 -2, 4 -2, 2 + 1 ⇒ 1, 2, 3, Direction ratios along B and C is, 7 -3, 0 -4, 6 -2 ⇒ 4, -4, 4, Since the direction ratios are not proportional they are not collinear., 2. Cartesian equation of the Plane is, Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 43, ⇒ (x – 2)(14 + 6) -(y – 2)(7 – 15) + (z + 1)(-2 -10) = 0, ⇒ 20(x – 2) + 8(y – 2) – 12(z + 1) = 0, https://hssliveguru.com/plus-two-maths-chapter-wise-questions-and-answers-chapter-11/, , 19/25

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12/03/2022, 18:03, , Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry – HSSLive …, , ⇒ 20x – 40 + 8y – 16 – 12z – 12 = 0, ⇒ 20x + 8y – 12z = 68, ⇒ 5x + 2y – 3z = 17, Vector Equation is r̄ .(5i + 2j – 3k) = 17., 3. Angle between the Plane and the Line, r̄ = (i + 2j – k) + λ(i – j + k), , Question 4., Consider three points on space ( 2, 1, 0 ), (3, -2, -2)and(3, 1, 7), 1. Find the Cartesian equation of the plane passing through the above points. (2), 2. Convert the above equation into vector form., 3. Hence, find a unit vector perpendicular to the above plane and also find the, perpendicular distance of the plane from the origin. (2), Answer:, 1. Equation of the plane is, , ⇒ (x – 2)(-21) – (y – 1)(7 + 2) + z(0 + 3) = 0, ⇒ 21x + 42 – 9y + 9 + 3z = 0 ⇒ -21x – 9y + 3z + 51 = 0, ⇒ 7x + 3y – z = 17., 2. Vector form is r̄ .(7i + 3j – k) = 17 _____(1), −, −, −, −, −−, −, −, −, , 3. Now, |7i + 3j – k| = √49 + 9 + 1 =, −, −, Dividing equation (1) by √59 , we get, , −, −, √59, , https://hssliveguru.com/plus-two-maths-chapter-wise-questions-and-answers-chapter-11/, , 20/25

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12/03/2022, 18:03, , Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry – HSSLive …, , Therefore the above equation is the normal form of the plane. Then, unit vector perpendicular to the plane and, , 17, √59, , 7i+3j−k, √59, , is the, , is the perpendicular distance from, , the origin., , Question 5., Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 48, ¯, ¯¯¯¯¯, ¯, ¯, , = i + 2j + 3k, , ¯, ¯¯¯¯¯¯, ¯, , = i – 2j + 4k, , OA, , OB, , ¯, ¯¯¯¯¯¯, ¯, , = 2i + 3j + k, are adjacent sides of the parallelopiped., OC, , 1. Find the base area of the parallelopiped. (2), ¯, ¯¯¯¯¯, ¯, ¯, , ¯, ¯¯¯¯¯¯, ¯, , (Base determined by OA and OB), 2. Find the volume of the parallelopiped. (2), 3. Find the height of the parallelopiped. (2), Answer:, ¯, ¯¯¯¯¯, ¯, ¯, , ¯, ¯¯¯¯¯¯, ¯, , 1. OA × OB =, , = 14i – j – 4k, Base area = |l4i – j – 4k|, −−−−−−−−−, −, −, −, −, = √196 + 1 + 16 = √213, , 2. Volume of the parallelopiped is, , = (14i – j – 4k).(2i + 3 j + k), = 28 – 3 – 4 = 21., 3. Height =, , volume, base area, , =, , 21, , ., , √213, , https://hssliveguru.com/plus-two-maths-chapter-wise-questions-and-answers-chapter-11/, , 21/25

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12/03/2022, 18:03, , Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry – HSSLive …, , Question 6., 1. Find the equation of the line passing through the point (2, 1, 0) and (3, 2, -1) (3), 2. Find the shortest distance of the above line from the line r̄ = (i – j + 2k) + λ(2i +, j – 3k) (3), Answer:, 1., , 2., , = i(-3 + 1) – j(-3 + 2) + k(1 – 2), = -2i + j – k, , https://hssliveguru.com/plus-two-maths-chapter-wise-questions-and-answers-chapter-11/, , 22/25

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12/03/2022, 18:03, , Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry – HSSLive …, , Question 7., The equation of two lines are, Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 55, 1. Find the dr’s of the given lines. (2), 2. Find the angle between the given lines. (2), 3. Find the equation of the line passing through (2, 1, 3) and perpendicular to the, given lines. (2), Answer:, 1. The given lines are, , The dr’s of (1) are 2, 2, 3 and dr’s of (2) are -3, 2, 5., 2. The angle between (1) and (2) is given by, , 3. Let a, b, c be the dr’s of the line perpendicular to lines (1) and (2)., ∴ 2a + 2b + 3c = 0, -3a + 2b + 5c = 0, Solving by the rule of cross-multiplication, we get, , ∴ dr’s of the required line are 4, -19, 10 and the line passes through (2, 1, 3)., ∴ Equation of the required line is, x−2, 4, , =, , y−1, −19, , =, , z−3, 10, , ., , Question 8., https://hssliveguru.com/plus-two-maths-chapter-wise-questions-and-answers-chapter-11/, , 23/25

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12/03/2022, 18:03, , Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry – HSSLive …, , 1. Find the direction cosines of the vector 2i + 2j – k. (1), 2. Find the distance of the point (2, 3, 4) from the plane r̄ .(3i – 6j + 2 k) = -11. (2), 3. Find the shortest distance between the lines r̄ = (2i – j – k)+ λ(3i – 5 j + 2k) an r̄, = (i+ 2 j + k)+ µ(i – j + k) (3), Answer:, 1. Direction ratios of the vector 2i + 2j – k is 2, 2, -1, Direction cosines of the vector 2i + 2j – k is, , 2. The equation of the plane in the Cartesian form is 3x – 6y + 2z + 11 = 0 . Then, distance from the point (2, 3, 4) is, , 3. The given lines are r̄ = (2i – j – k) + λ(3i – 5j + 2k), Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 61, Plus Two, Plus One Bussiness Studies Chapter Wise Questions and Answers Chapter 4, Business Services, Plus Two Maths Chapter Wise Questions and Answers Chapter 12 Linear, Programming, , Search …, , Recent Posts, HSSLive Plus Two Study Material / Question Bank Kerala, Plus One Malayalam Textbook Answers Unit 4 Chapter 6 Shasthrakriya, Plus One Accountancy Chapter Wise Questions and Answers Kerala, https://hssliveguru.com/plus-two-maths-chapter-wise-questions-and-answers-chapter-11/, , 24/25

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12/03/2022, 18:03, , Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry – HSSLive …, , Plus One Economics Chapter Wise Questions and Answers Chapter 4 Presentation of, Data, Kerala Syllabus 9th Standard Maths Solutions Chapter 13 Statistics in Malayalam, Plus One Malayalam Textbook Answers Unit 4 Chapter 5 Samkramanam, Plus One Malayalam Textbook Answers Unit 4 Chapter 4 Vasanavikrithi, Plus One Physics Notes Chapter 5 Law of Motion, Kerala Syllabus 10th Standard Social Science Solutions Chapter 5 Culture and, Nationalism in Malayalam, Plus One Malayalam Textbook Answers Unit 4 Chapter 3 Muhyadheen Mala, Plus One Malayalam Textbook Answers Unit 4 Chapter 2 Anukampa, , Copyright © 2022 HSSLive Guru, , https://hssliveguru.com/plus-two-maths-chapter-wise-questions-and-answers-chapter-11/, , 25/25