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Question 27 (OR 1
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Find the equation of the plane through the line (x - 1)/3 = (y - 4)/2 = (z - 4)/(-2) and parallel to the line (x + 1)/2 = (1 - y)/4 = (z + 2)/1 . Hence, find the shortest distance between the lines

CBSE Class 12 Sample Paper for 2019 Boards

Paper Summary

Question 1 Important

Question 2

Question 3

Question 4 (Or 1st) Important

Question 4 (Or 2nd)

Question 5

Question 6

Question 7 Important

Question 8 (Or 1st) Important

Question 8 (Or 2nd)

Question 9

Question 10 (Or 1st) Important

Question 10 (Or 2nd)

Question 11 Important

Question 12 (Or 1st)

Question 12 (Or 2nd)

Question 13 (Or 1st) Important

Question 13 (Or 2nd)

Question 14 Important

Question 15

Question 16 (Or 1st)

Question 16 (Or 2nd) Important

Question 17

Question 18

Question 19 Important

Question 20 Important

Question 21 (Or 1st)

Question 21 (Or 2nd) Important

Question 22

Question 23 Important

Question 24 (Or 1st)

Question 24 (Or 2nd) Important

Question 25

Question 26 (Or 1st) Important

Question 26 (Or 2nd)

Question 27 (Or 1st) Important You are here

Question 27 (Or 2nd) Important

Question 28

Question 29 Important

Class 12

Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Sept. 24, 2021 by Teachoo

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Question 27 (OR 1
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st
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question)
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Find the equation of the plane through the line (x - 1)/3 = (y - 4)/2 = (z - 4)/(-2) and parallel to the line (x + 1)/2 = (1 - y)/4 = (z + 2)/1 . Hence, find the shortest distance between the lines

Question 27 (OR 1st question) Find the equation of the plane through the line (π₯ β 1)/3 = (π¦ β 4)/2 = (π§ β 4)/(β2) and parallel to the line (π₯ + 1)/2 = (1 β π¦)/4 = (π§ + 2)/1 . Hence, find the shortest distance between the lines Given lines (π₯ β 1)/3 = (π¦ β 4)/2 = (π§ β 4)/(β2) (π₯ + 1)/2 = (1 β π¦)/4 = (π§ + 2)/1 Since the plane passes through line (1) All points of line (1) will be in the plane Now, for line (1) (1, 4, 4) lies in line (1) So, plane passes through (1, 4, 4). Thus, the equation of plane is a(x β 1) + b(y β 4) + c(z β 4) = 0 Where a, b, c are the direction ratios of normal to the plane Now, given that plane is parallel to line (2) β΄ Plane is parallel to line (1) and (2) Thus, Normal of plane is perpendicular to both line (1) and (2) β΄ Normal will be cross product of parallel vectors of line (1) and (2) Now, our lines are (π₯ β 1)/3 = (π¦ β 4)/2 = (π§ β 4)/(β2) (π₯ + 1)/2 = (1 β π¦)/4 = (π§ + 2)/1 For line (2), we write y β 1 (π₯ + 1)/2 = (π¦ β 1)/(β4) = (π§ + 2)/1 So, Normal of plane = |β 8(π Μ&π Μ&π Μ@3&2&β2@2&β4&1)| = π Μ (2 Γ 1 β (-4) Γ (-2)) β π Μ (3 Γ 1 β 2 Γ (-2)) + π Μ (3 Γ (-4) β 2 Γ 2) = π Μ (2 β 8) β π Μ (3 + 4) + π Μ (β12 β 4) = β6π Μ β 7π Μ β 16π Μ So, direction ratios of normal are β6, β7, β16 β΄ a = β6, b = β7, c = β16 And, equation of plane is a(x β 1) + b(y β 4) + c(z β 4) = 0 (β6)(x β 1) + (β7)(y β 4) + (β16)(z β 4) = 0 6(x β 1) + 7(y β 4) + 16(z β 4) = 0 6x β 6 + 7y β 28 + 16z β 64 = 0 6x + 7y + 16z β 98 = 0 Also, we need to find shortest distance between lines Now, equation of line (2) is (π₯ + 1)/2 = (π¦ β 1)/(β4) = (π§ + 2)/1 Shortest distance between lines = Perpendicular distance of point (β1, 1, β2) from plane 6x + 7y + 16z β 98 = 0 = |(π΄π₯_1 + γπ΅π¦γ_1 +γ πΆπ§γ_1 β π·)/β(π΄^2 + π΅^2 + πΆ^2 )| Putting π₯_1 = β1, π¦_1 = 1, π§_1 = β2 and A = 6, B = 7, C = 16, D = 98 = |((6 Γ β1) + (7 Γ 1) + (16 Γ β2) β 98 )/β(6^2 + 7^2 +16^2 )| = |(β6 + 7 β 32 β 98)/β(36 + 49 + 256)| = |(β129)/β341| = 129/β341 Thus, shortest distance πππ/βπππ units