If the foot of the perpendicular from `O(0,0,0)` to a plane is `P(1,2,2)`. Then the equation of the plane isA. `-x+2y+8z-9=0`B. `x+2y+2z-9=0`C. `x+y+z-5=0`D. `x+2y-3z+1=0`

If the foot of the perpendicular from `O(0,0,0)` to a plane is `P(1,2,

1.	If the foot of the perpendicular from `O(0,0,0)` to a plane is `P(1,2,2)`. Then the equation of the plane isA. `-x+2y+8z-9=0`B. `x+2y+2z-9=0`C. `x+y+z-5=0`D. `x+2y-3z+1=0`
Answer» Correct Answer - B The plane passes through `P(1,2,2)` and is normal to `vec(OP)=hati+2hatj+2hatk`. So its vector equation is `vecr.vec(OP)=(hati+2hatj+2hatk).vec(OP)` [ Using `vecr.vecn=veca.vecn`] `impliesvecr.(hati+2hatj+2hatk)=(hati+2hatj+2hatk).(hati+2hatj+2hatk)` `impliesvecr.(hati+2hatj+hatk)=9impliesx+2y+2z=9`

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If the foot of the perpendicular from `O(0,0,0)` to a plane is `P(1,2,2)`. Then the equation of the plane isA. `-x+2y+8z-9=0`B. `x+2y+2z-9=0`C. `x+y+z-5=0`D. `x+2y-3z+1=0`

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