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In R^(3), consider the planes P(1):y=0 a...

In `R^(3)`, consider the planes `P_(1):y=0` and `P_(2),x+z=1.` Let `P_(3)` be a plane, different from `P_(1)` and `P_(2)` which passes through the intersection of `P_(1)` and `P_(2)`, If the distance of the point (0,1,0) from `P_(3)` is 1 and the distance of a point `(alpha,beta,gamma)` from `P_(3)` is 2, then which of the following relation(s) is/are true? (a) `2alpha + beta + 2gamma +2 = 0 ` (b) `2alpha -beta + 2gamma +4=0` (c) `2alpha + beta - 2gamma- 10 =0` (d) `2alpha- beta+ 2gamma-8=0`

A

`2alpha+beta+2gamma+2=0`

B

`2alpha+beta+2gamma+4=0`

C

`2alpha+beta+2gamma-10=0`

D

`2alpha+beta+2gamma-8=0`

Text Solution

Verified by Experts

The correct Answer is:
B, D
b.,d
Clearly , planr `P_(3) is P_(2) + lamdaP_(1) =0` .
`implies x+ lamday+z-1=0`
Distance of this from point `(0,1,0)`is 1,
`=(0+lamda+0-1)/(sqrt(1+lamda^(2)+1))=+_1`
`therefore lamda =-(1)/(2)`
thus , equation of `P_(3)is 2x-y+2z-2=0`
DIstance of this plane from point `(alpha, beta,gamma)`is 2.
`=|(2alpha-beta+2gamma-2)/(3)|=2`
`implies 2alpha-beta+2gamma=2+-6`
thus options (b) and (d) are correct.
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