JEE Mains 06-Sep-2020 Shift 1 Solved Paper

Line of intersection of planes
x+y+z+1=0 ....(1)
2x−y+z+3=0 ....(2)
eliminate y
3x+2z+4=0
x=

−2z−4

3

....(3)
put in equation (1)
z=−3y+1 ....(4)
from (3) and (4)

3x+4

−2

=−3y+1=z

x−(− 4 3 )

− 2 3

=

y− 1 3

− 1 3

=

z−0

1

now shortest distance between skew lines

x−1

0

=

y+1

−1

=

z

1

x−(− 4 3 )

−23

=

y−( 1 3 )

− 1 3

=

z−0

1

S.D. =|

( → b − → a ).( → c × → d )

| → c × → d |

|
where

→

a

=(1,−1,0)

→

b

=(−

4

3

,

1

3

,0)

→

c

=(0,−1,1)

→

d

=(−

2

3

,−

1

3

,1)
⇒S.D=

1

√3