Find the equation of the plane which is at a distance of 6√29 from the origin and its normal vector from the origin is 2^i−3^j+4^k
A
2x+3y+4z=6
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B
3x−4y+2z=6
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C
2x−3y+4z=6
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D
2x−3y−4z=6
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Solution
The correct option is C2x−3y+4z=6 Let →n=2ˆi−3ˆj+4ˆkˆn=2ˆi−3ˆj+4ˆk√4+9+16=2ˆi−3ˆj+4ˆk√29 And let r be a position vector→r=xˆi+yˆj+zˆk Hence, the required equation of the plane is →r.ˆn=6√29(xˆi+yˆj+zˆk).(2ˆi−3ˆj+4ˆk√29)=6√292x−3y+4z√29=6√292x−3y+4z=6