HLO EVERYONE :)
SOLVE THE QUESTION IN THE ATTACHMENT.
CLASS 12 ( THREE DIMENSIONAL GEOMETRY)
Answers
r= 8i -9j +10 k +a( 3i -16j +7k)
i have taken a instead of lambda
r= 15i +29 j +5k +u( 3i +8j -5k)
As shortest that is perpendicular distance we have to find
Here both lines parallel vectors given
3i -16 j +7k and 3i +8j -5k
as vector perpendicular to their plane will be perpendicular to both
take cross
3i -16j +7k × (3i +8j -5k)
24 k +15j +48 k +80 i +21 j -56 i
24i + 36j + 72k
12( 2i +3j +6 k)
now perpendicular distance will be parallel to this vector
lets take one point of one line and lets find another point
eq first we should find
(8,-9,10) first line passing
r= 8 i -9 j +10 k+ s( 2i +3j +6k)
now its intersection point lets find with
another line
( 8 + 2s)i +j( -9 +3s) +k( 10+6s)
8+2s = 15+ 3u
2s - 3u = 7 × 3
6s - 9u = 21
-9+3s = 29 +8u
3s -8u = 38 × 2
6s -16 u = 76
subtract
--9u +16u = 21-76
7u = - 55
u= -55/7
3u +15 = -165/7. +15 = -60/7
29 +8u = 29 - 440/7 = 203-440)/7 = -237/7
(5-5u) = 5( 1+55/7) = 310/7
(-60/7, -237/7, 310/7)
and (8,-9,10)
distance
= √( 8 + 60/7)^2 +( -9+ 237/7)^2 + (10-310/7)^2
= 1/7) √( 106)^2 + (174)^2 + (240)^2
=1/7 ( √( 11236 + 30276 + 57600)
= 1/7 )√( 99112)
= 314.8/7
= 44.9 finally
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