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Consider three planes P(1):x-y+z=1 P(2...

Consider three planes `P_(1):x-y+z=1`
`P_(2):x+y-z=-1`
and `" "P_(3):x-3y+3z=2`
Let `L_(1),L_(2),L_(3)` be the lines of intersection of the planes `P_(2)` and `P_(3),P_(3)` and `P_(1),P_(1)` and `P_(2)` respectively.
Statement I Atleast two of the lines `L_(1),L_(2)` and `L_(3)` are non-parallel.
Statement II The three planes do not have a common point.

A

Statement I is true, Statement II is also true,
Statement II is the correct explanation of
Statement I

B

Statement I is true, Statement II is also true,
Statement II is not the correct explanation of
Statement I

C

Statement I is true, Statement II is false

D

Statement I is false, Statement II is true

Text Solution

Verified by Experts

The correct Answer is:
D
Given three planes are
`P_(1):x-y+z=1" "...(i)`
`P_(2):x+y-z=-1" "...(ii)`
and`" "P_(3):x-3y+3z=2" "...(iii)`
On solving Eqs. (i) and (ii), we get
`x=0,z=1+y`
which does not satisfy Eq. (iii).
As`" "x-3y+3z=0-3y+3(1+y)=3(ne2)`
So, Statement II is true.
Next, since we know that direction reatios of line of intersection of planes `a_(1)x+b_(1)y+c_(1)z+d_(1)=0`
and`" "a_(2)x+b_(2)y+c_(2)z+d_(1)=0`is
`b_(1)c_(2)-b_(2)c_(1),c_(1)a_(2)-a_(1)c_(2),a_(1)b_(2)-a_(2)b_(1)`
Using above result,
Direction ratios of lines `L_(1),L_(2)` and `L_(3)` are
0,2,2,0,-4,-4,0,-2,-2
Since, all the three lines `L_(1),L_(2)` and `L_(3)` are parallel pairwise.
Hence, Statement I is false.
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