If (2,-1,3) is the foot of the perpendic...

If (2,-1,3) is the foot of the perpendicular down from the origin to the plane, then the equation of the plane is

A

`2x+y-3z+6=0`

B

`2x-y+3z-14=0`

C

`2x-y+3z-13=0`

D

`2x+y+3z-10=0`

Text Solution

Verified by Experts

The correct Answer is:

B

Let the equation of any plane passing through
P(2,-1,3) is
`a(x-2)+b(y+1)+c(z-3)=0` . . . (i)
`:.` DR's of OP=(2,-1,3)
Since, the line OP is perpendicular to the plane, therefore the DR's of the normal to the plane is proportional to the DR's of OP.
`:.` Required equation of plane is
`2(x-2)-1(y+1)+3(z-3)=0`
`rArr" "3x-y+3z-14=0`

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