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Let P(1)=x+y+z+1=0, P(2)=x-y+2z+1=0,P(3)...

Let `P_(1)=x+y+z+1=0, P_(2)=x-y+2z+1=0,P_(3)=3x+y+4z+7=0` be three planes. Find the distance of line of intersection of planes `P_(1)=0` and `P_(2) =0` from the plane `P_(3) =0.`

A

`(2)/sqrt(26)`

B

`(4)/sqrt(26)`

C

`sqrt((1)/(26))`

D

`(7)/sqrt(26)`

Text Solution

Verified by Experts

The correct Answer is:
B
Let direction ratios of line are a,b,c
Now `a+b+c=0`
`a-b+2c=0`
`(a)/(3)=(b)/(-1)=(c )/(-2)`
So direction ratios of line are `(3,-1,-2)`
`x+y+z+1+0….(1)`
`x-y+2z+1=0….(2)`
Put z=0, we get `x=1, y=0`
Equation of line of intersection of `P_(1)=0 P_(2)=0` is parallel to the plane `P_(3)=0`. Sodistance of point `(-1,0,0)` from `P_(3)=0` is `|(-3+7)/sqrt(9+1+16)| = (4)/sqrt(26)`
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Knowledge Check

  • The equation of the plane passing through the line of intersection of the planes x+y+z+3 =0 and 2x-y + 3z +2 =0 and parallel to the line (x)/(1) = (y)/(2) = (z)/(3) is

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    `x+5y-3z=5`
    D
    `x+5y-3=5`

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