Question

For 3 planes P1 : x + y + z = 3, P2 : x – y + z = 3 & P3 : x + z = 2, if distance between the line of intersection of P1, P2 and the line of intersection of P2, P3 is d, then  is equal to
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Answer 1

Step-by-step explanation:

Given three planes are

P

1

:x−y+z=1 ...(1)

P

1

:x+y−z=−1 ....(2)

and P

1

:x−3y+3z=2 ...(3)

Solving Eqs. (1) and (2), we have

x=0,z=1+y

which does not satisfy Eq. (3)

As, x−3y+3z=0−3y+3(1+y)=3(



=2)

∴ Statement II is true.

Nest,since we know that direction ratio's 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

2

=0

b

1

c

2

−b

2

c

1

,c

1

a

2

−a

1

c

2

,a

1

b

1

Using above result, we get

Direction ratio's of lines L

1

,L

2

and L

3

are o,2,2;0,−4,−4;0,−2,−2 respectively.

⇒ All the three lines L

1

,L

2

and L

3

are parallel pairwise.

∴ Statement I is false.