Question 37 (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards
Last updated at Oct. 27, 2020 by Teachoo
Find the shortest distance between the lines r
β
Β = 3iΒ + 2jΒ - 4kΒ + Ξ» (iΒ + 2jΒ -2k )
And r = 5iΒ + 2jΒ - u (3iΒ + 2jΒ +6k). If the lines intersect find their point of intersection
Β
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Β
Note
: This
is similar
to
Ex 11.2, 14 ; Ex 11.2, 16 and
Misc
9
of NCERT β
Chapter 11 Class 12 Three Dimensional Geometry
Check the answer here
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Transcript
Question 37 (Choice 1) Find the shortest distance between the lines π β = 3π Μ + 2π Μ β 4π Μ + π (π Μ + 2π Μ + 2π Μ) And π β = 5π Μ β 2π Μ + π(3π Μ + 2π Μ + 6π Μ) If the lines intersect find their point of intersection
Shortest distance between lines with vector equations
π β = (π1) β + π (π1) β and π β = (π2) β + π(π2) β is |(((ππ) β Γ (ππ) β ).((ππ) β β (ππ) β ))/|(ππ) β Γ (ππ) β | |
π β = (3π Μ + 2π Μ β 4π Μ) + π (π Μ + 2π Μ + 2π Μ)
Comparing with π β = (π1) β + π(π1) β ,
(π1) β = 3π Μ + 2π Μ β 4π Μ
& (π1) β = 1π Μ + 2π Μ + 2π Μ
π β = (5π Μ β 2π Μ) + π (3π Μ + 2π Μ + 6π Μ)
Comparing with π β = (π2) β + π(π2) β ,
(π2) β = 5π Μ β 2π Μ + 0π Μ
& (π2) β = 3π Μ + 2π Μ + 6π Μ
Now,
((ππ) β β (ππ) β) = (5π Μ β 2π Μ + 0π Μ) β (3π Μ + 2π Μ β 4π Μ)
= (5 β 3) π Μ + (β2 β 2)π Μ + (0 β (β4)) π Μ
= 2π Μ β 4π Μ + 4π Μ
((ππ) β Γ (ππ) β) = |β 8(π Μ&π Μ&π Μ@1& 2&2@3&2&6)|
= π Μ [(2 Γ 6)β(2 Γ 2)] β π Μ [(1 Γ 6)β(3 Γ 2)] + π Μ [(1 Γ 2)β(3 Γ 2)]
= π Μ [ 12β4] β π Μ [6β6] + π Μ [2β6]
= π Μ (8) β π Μ (0) + π Μ(β4)
= 8π Μ β 4π Μ
Magnitude of (π1) β Γ (π2) β = β(8^2+0^2+γ(β4)γ^2 )
|(ππ) β Γ (ππ) β | = β(64+16) = β80 = 4β5
Also,
((ππ) βΓ(ππ) β ) . ((ππ) β β (ππ) β ) = (8π Μ β 4π Μ). (2π Μ β 4π Μ + 4π Μ)
= (8 Γ 2) + (0 Γ β4) + (β4 Γ 4)
= β16 + 0 + (β16)
= 0
Shortest distance = |(((π1) β Γ (π2) β ) . ((π2) β β (π1) β ))/|(π1) β Γ (π2) β | |
= |( 0)/(4β5)|
= 0
Therefore, the shortest distance between the given two lines is 0
Finding Point of Intersection
Since our lines are
π β = 3π Μ + 2π Μ β 4π Μ + π (π Μ + 2π Μ + 2π Μ) And π β = 5π Μ β 2π Μ + π(3π Μ + 2π Μ + 6π Μ)
Thus,
3π Μ + 2π Μ β 4π Μ + π (π Μ + 2π Μ + 2π Μ) = 5π Μ β 2π Μ + π(3π Μ + 2π Μ + 6π Μ)
(3 + π)π Μ + (2 + 2π) π Μ + (β4 + 2π) π Μ = (5 + 3π)π Μ + (β2 + 2π)π Μ + 6ππ Μ
Thus, 3 + π = 5 + 3π
2 + 2π = β2 + 2π
β4 + 2π = 6π
Solving (1) and (2)
π β 3π = 5 β 3
π β 3π = 2
And,
2 + 2π = β2 + 2π
2π β 2π = β4
π β π = β2
Solving both equations
π=βπ, π=βπ
Putting π=βπ in π β
π β = 3π Μ + 2π Μ β 4π Μ + π (π Μ + 2π Μ + 2π Μ)
π β = 3π Μ + 2π Μ β 4π Μ β 4(π Μ + 2π Μ + 2π Μ)
π β = 3π Μ + 2π Μ β 4π Μ β 4π Μ β 8π Μ β 8π Μ
π β = βπ Μ β 6π Μ β 12π Μ
So, Point of intersection is (β1, β6, β12)
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