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Find the equation of plane which is at a...

Find the equation of plane which is at a distance `(4)/(sqrt(14))` from the origin and is normal to vector `2hati+hatj-3hatk`.

Text Solution

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Here `vecn=2hati+hatj-3hatk.` Then `(vecn)/(|vecn|)=(2hati+hatj-3hatk)/(sqrt(2^(2)+1^(2)+(-3)^(2)))=(2hati+hatj-3hatk)/(sqrt(14)`
Hence, required equation of plane is `vecr*(1)/(sqrt(14))(2hati+hatj-3hatk)=pm(1)/(sqrt(14))`
or `" "vecr*(2hati+hatj-3hatk)=pm1` (vector form)
or `" "2x+y-3z=pm 1` (Cartesian form)
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