Question 27 (Or 1st) - CBSE Class 12 Sample Paper for 2019 Boards

Last updated at Oct. 1, 2019 by Teachoo

Question 27 (OR 1
st
question)

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

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.