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A plane passes through the point (0,-1,0...

A plane passes through the point `(0,-1,0)` and `(0,0,1)` and makes an angle of `(pi)/(4)` with the plane `y-z=0` then the point which satisfies the desired plane is

A

`(-sqrt(2), 1, -4)`

B

`(sqrt(2),1,4)`

C

`(sqrt(2),-1,4)`

D

`(-sqrt(2),-1,-4)`

Text Solution

Verified by Experts

The correct Answer is:
B
Let equation of the plane be `a(x-0)+b(y+1)+c(z-0)=0`
`rArr ax+by+cz=-b` ......(1)
Now `a(0-0)+b(0+1)+c(1-0)=0`
`rArr b+c=0` ......(2)
and `(a xx 0 + b xx 1 + c xx (-1))/(sqrt(a^(2)+b^(2)+c^(2))sqrt(2))=cos .(pi)/(4)=(1)/(sqrt(2))`
`rArr b-c=sqrt(a^(2)+2b^(2))`
`2b=sqrt(a^(2)+2b^(2))`
`4b^(2)=a^(2)+2b^(2) rArr a^(2)=2b^(2) rArr a= pm sqrt(2)b`
So, equation of plane `pm sqrt(2)x+y-z=-1`
`(sqrt(2),+4)` satisfy.
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